### Section SLT  Surjective Linear Transformations

From A First Course in Linear Algebra
Version 2.23
http://linear.ups.edu/

The companion to an injection is a surjection. Surjective linear transformations are closely related to spanning sets and ranges. So as you read this section reflect back on Section ILT and note the parallels and the contrasts. In the next section, Section IVLT, we will combine the two properties.

As usual, we lead with a definition.

Definition SLT
Surjective Linear Transformation
Suppose T : U → V is a linear transformation. Then T is surjective if for every v ∈ V there exists a u ∈ U so that T\left (u\right ) = v.

Given an arbitrary function, it is possible for there to be an element of the codomain that is not an output of the function (think about the function y = f(x) = {x}^{2} and the codomain element y = −3). For a surjective function, this never happens. If we choose any element of the codomain (v ∈ V ) then there must be an input from the domain (u ∈ U) which will create the output when used to evaluate the linear transformation (T\left (u\right ) = v). Some authors prefer the term onto where we use surjective, and we will sometimes refer to a surjective linear transformation as a surjection.

#### Subsection ESLT: Examples of Surjective Linear Transformations

It is perhaps most instructive to examine a linear transformation that is not surjective first.

Example NSAQ
Not surjective, Archetype Q
Archetype Q is the linear transformation

 T : {ℂ}^{5} → {ℂ}^{5},\quad T\left (\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} \cr {x}_{5} } \right ]\right ) = \left [\array{ −2{x}_{1} + 3{x}_{2} + 3{x}_{3} − 6{x}_{4} + 3{x}_{5} \cr −16{x}_{1} + 9{x}_{2} + 12{x}_{3} − 28{x}_{4} + 28{x}_{5} \cr −19{x}_{1} + 7{x}_{2} + 14{x}_{3} − 32{x}_{4} + 37{x}_{5} \cr −21{x}_{1} + 9{x}_{2} + 15{x}_{3} − 35{x}_{4} + 39{x}_{5} \cr −9{x}_{1} + 5{x}_{2} + 7{x}_{3} − 16{x}_{4} + 16{x}_{5} } \right ]

We will demonstrate that

 v = \left [\array{ −1\cr 2 \cr 3\cr −1 \cr 4 } \right ]

is an unobtainable element of the codomain. Suppose to the contrary that u is an element of the domain such that T\left (u\right ) = v. Then

\eqalignno{ \left [\array{ −1\cr 2 \cr 3\cr −1 \cr 4 } \right ] & = v = T\left (u\right ) = T\left (\left [\array{ {u}_{1} \cr {u}_{2} \cr {u}_{3} \cr {u}_{4} \cr {u}_{5} } \right ]\right ) & & \cr & = \left [\array{ −2{u}_{1} + 3{u}_{2} + 3{u}_{3} − 6{u}_{4} + 3{u}_{5} \cr −16{u}_{1} + 9{u}_{2} + 12{u}_{3} − 28{u}_{4} + 28{u}_{5} \cr −19{u}_{1} + 7{u}_{2} + 14{u}_{3} − 32{u}_{4} + 37{u}_{5} \cr −21{u}_{1} + 9{u}_{2} + 15{u}_{3} − 35{u}_{4} + 39{u}_{5} \cr −9{u}_{1} + 5{u}_{2} + 7{u}_{3} − 16{u}_{4} + 16{u}_{5} } \right ] & & \cr & = \left [\array{ −2 &3& 3 & −6 & 3\cr −16 &9 &12 &−28 &28 \cr −19&7&14&−32&37\cr −21 &9 &15 &−35 &39 \cr −9 &5& 7 &−16&16 } \right ]\left [\array{ {u}_{1} \cr {u}_{2} \cr {u}_{3} \cr {u}_{4} \cr {u}_{5} } \right ] & & }

Now we recognize the appropriate input vector u as a solution to a linear system of equations. Form the augmented matrix of the system, and row-reduce to

 \left [\array{ \text{1}&0&0&0&−1&0 \cr 0&\text{1}&0&0&−{4\over 3}&0 \cr 0&0&\text{1}&0&−{1\over 3}&0\cr 0&0 &0 &\text{1 } &−1 &0 \cr 0&0&0&0& 0 &\text{1}\cr } \right ]

With a leading 1 in the last column, Theorem RCLS tells us the system is inconsistent. From the absence of any solutions we conclude that no such vector u exists, and by Definition SLT, T is not surjective.

Again, do not concern yourself with how v was selected, as this will be explained shortly. However, do understand why this vector provides enough evidence to conclude that T is not surjective.

To show that a linear transformation is not surjective, it is enough to find a single element of the codomain that is never created by any input, as in Example NSAQ. However, to show that a linear transformation is surjective we must establish that every element of the codomain occurs as an output of the linear transformation for some appropriate input.

Example SAR
Surjective, Archetype R
Archetype R is the linear transformation

 T : {ℂ}^{5} → {ℂ}^{5},\quad T\left (\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} \cr {x}_{5} } \right ]\right ) = \left [\array{ −65{x}_{1} + 128{x}_{2} + 10{x}_{3} − 262{x}_{4} + 40{x}_{5} \cr 36{x}_{1} − 73{x}_{2} − {x}_{3} + 151{x}_{4} − 16{x}_{5} \cr −44{x}_{1} + 88{x}_{2} + 5{x}_{3} − 180{x}_{4} + 24{x}_{5} \cr 34{x}_{1} − 68{x}_{2} − 3{x}_{3} + 140{x}_{4} − 18{x}_{5} \cr 12{x}_{1} − 24{x}_{2} − {x}_{3} + 49{x}_{4} − 5{x}_{5} } \right ]

To establish that R is surjective we must begin with a totally arbitrary element of the codomain, v and somehow find an input vector u such that T\left (u\right ) = v. We desire,

\eqalignno{ T\left (u\right ) & = v & & \cr \left [\array{ −65{u}_{1} + 128{u}_{2} + 10{u}_{3} − 262{u}_{4} + 40{u}_{5} \cr 36{u}_{1} − 73{u}_{2} − {u}_{3} + 151{u}_{4} − 16{u}_{5} \cr −44{u}_{1} + 88{u}_{2} + 5{u}_{3} − 180{u}_{4} + 24{u}_{5} \cr 34{u}_{1} − 68{u}_{2} − 3{u}_{3} + 140{u}_{4} − 18{u}_{5} \cr 12{u}_{1} − 24{u}_{2} − {u}_{3} + 49{u}_{4} − 5{u}_{5} } \right ] & = \left [\array{ {v}_{1} \cr {v}_{2} \cr {v}_{3} \cr {v}_{4} \cr {v}_{5} } \right ] & & \cr \left [\array{ −65&128&10&−262& 40\cr 36 &−73 &−1 & 151 &−16 \cr −44& 88 & 5 &−180& 24\cr 34 &−68 &−3 & 140 &−18 \cr 12 &−24&−1& 49 & −5 } \right ]\left [\array{ {u}_{1} \cr {u}_{2} \cr {u}_{3} \cr {u}_{4} \cr {u}_{5} } \right ] & = \left [\array{ {v}_{1} \cr {v}_{2} \cr {v}_{3} \cr {v}_{4} \cr {v}_{5} } \right ] & & }

We recognize this equation as a system of equations in the variables {u}_{i}, but our vector of constants contains symbols. In general, we would have to row-reduce the augmented matrix by hand, due to the symbolic final column. However, in this particular example, the 5 × 5 coefficient matrix is nonsingular and so has an inverse (Theorem NI, Definition MI).

 { \left [\array{ −65&128&10&−262& 40\cr 36 &−73 &−1 & 151 &−16 \cr −44& 88 & 5 &−180& 24\cr 34 &−68 &−3 & 140 &−18 \cr 12 &−24&−1& 49 & −5 } \right ]}^{−1} = \left [\array{ −47& 92 & 1 &−181&−14 \cr 27 &−55& {7\over 2} & {221\over 2} & 11\cr −32 & 64 &−1 &−126 &−12 \cr 25 &−50& {3\over 2} & {199\over 2} & 9 \cr 9 &−18& {1\over 2} & {71\over 2} & 4 } \right ]

so we find that

\eqalignno{ \left [\array{ {u}_{1} \cr {u}_{2} \cr {u}_{3} \cr {u}_{4} \cr {u}_{5} } \right ] & = \left [\array{ −47& 92 & 1 &−181&−14 \cr 27 &−55& {7\over 2} & {221\over 2} & 11\cr −32 & 64 &−1 &−126 &−12 \cr 25 &−50& {3\over 2} & {199\over 2} & 9 \cr 9 &−18& {1\over 2} & {71\over 2} & 4 } \right ]\left [\array{ {v}_{1} \cr {v}_{2} \cr {v}_{3} \cr {v}_{4} \cr {v}_{5} } \right ] & & \cr & = \left [\array{ −47{v}_{1} + 92{v}_{2} + {v}_{3} − 181{v}_{4} − 14{v}_{5} \cr 27{v}_{1} − 55{v}_{2} + {7\over 2}{v}_{3} + {221\over 2} {v}_{4} + 11{v}_{5} \cr −32{v}_{1} + 64{v}_{2} − {v}_{3} − 126{v}_{4} − 12{v}_{5} \cr 25{v}_{1} − 50{v}_{2} + {3\over 2}{v}_{3} + {199\over 2} {v}_{4} + 9{v}_{5} \cr 9{v}_{1} − 18{v}_{2} + {1\over 2}{v}_{3} + {71\over 2} {v}_{4} + 4{v}_{5} } \right ] & & }

This establishes that if we are given any output vector v, we can use its components in this final expression to formulate a vector u such that T\left (u\right ) = v. So by Definition SLT we now know that T is surjective. You might try to verify this condition in its full generality (i.e. evaluate T with this final expression and see if you get v as the result), or test it more specifically for some numerical vector v (see Exercise SLT.C20).

Let’s now examine a surjective linear transformation between abstract vector spaces.

Example SAV
Surjective, Archetype V
Archetype V is defined by

 T : {P}_{3} → {M}_{22},\quad T\left (a + bx + c{x}^{2} + d{x}^{3}\right ) = \left [\array{ a + b&a − 2c \cr d & b − d } \right ]

To establish that the linear transformation is surjective, begin by choosing an arbitrary output. In this example, we need to choose an arbitrary 2 × 2 matrix, say

 v = \left [\array{ x&y\cr z &w } \right ]

and we would like to find an input polynomial

 u = a + bx + c{x}^{2} + d{x}^{3}

so that T\left (u\right ) = v. So we have,

\eqalignno{ \left [\array{ x&y\cr z &w } \right ] & = v & & \cr & = T\left (u\right ) & & \cr & = T\left (a + bx + c{x}^{2} + d{x}^{3}\right ) & & \cr & = \left [\array{ a + b&a − 2c\cr d & b − d } \right ] & & }

Matrix equality leads us to the system of four equations in the four unknowns, x,y,z,w,

\eqalignno{ a + b & = x & & \cr a − 2c & = y & & \cr d & = z & & \cr b − d & = w & & }

which can be rewritten as a matrix equation,

 \left [\array{ 1&1& 0 & 0\cr 1&0 &−2 & 0 \cr 0&0& 0 & 1\cr 0&1 & 0 &−1 } \right ]\left [\array{ a\cr b \cr c\cr d } \right ] = \left [\array{ x\cr y \cr z\cr w } \right ]

The coefficient matrix is nonsingular, hence it has an inverse,

 { \left [\array{ 1&1& 0 &0\cr 1&0 & −2 &0 \cr 0&0& 0 &1\cr 0&1 &0 − 1 } \right ]}^{−1} = \left [\array{ 1& 0 &−1&−1\cr 0 & 0 & 1 & 1 \cr {1\over 2}&−{1\over 2}&−{1\over 2}&−{1\over 2} \cr 0& 0 & 1 & 0 } \right ]

so we have

\eqalignno{ \left [\array{ a\cr b \cr c\cr d } \right ] & = \left [\array{ 1& 0 &−1&−1\cr 0 & 0 & 1 & 1 \cr {1\over 2}&−{1\over 2}&−{1\over 2}&−{1\over 2} \cr 0& 0 & 1 & 0 } \right ]\left [\array{ x\cr y \cr z\cr w } \right ] & & \cr & = \left [\array{ x − z − w\cr z + w \cr {1\over 2}(x − y − z − w)\cr z } \right ] & & }

So the input polynomial u = (x − z − w) + (z + w)x + {1\over 2}(x − y − z − w){x}^{2} + z{x}^{3} will yield the output matrix v, no matter what form v takes. This means by Definition SLT that T is surjective. All the same, let’s do a concrete demonstration and evaluate T with u,

\eqalignno{ T\left (u\right ) & = T\left ((x − z − w) + (z + w)x + {1\over 2}(x − y − z − w){x}^{2} + z{x}^{3}\right ) & & \cr & = \left [\array{ (x − z − w) + (z + w)&(x − z − w) − 2({1\over 2}(x − y − z − w)) \cr z & (z + w) − z } \right ] & & \cr & = \left [\array{ x&y\cr z &w } \right ] & & \cr & = v & & }

#### Subsection RLT: Range of a Linear Transformation

For a linear transformation T : U → V , the range is a subset of the codomain V . Informally, it is the set of all outputs that the transformation creates when fed every possible input from the domain. It will have some natural connections with the column space of a matrix, so we will keep the same notation, and if you think about your objects, then there should be little confusion. Here’s the careful definition.

Definition RLT
Range of a Linear Transformation
Suppose T : U → V is a linear transformation. Then the range of T is the set

 ℛ\kern -1.95872pt \left (T\right ) = \left \{\left .T\left (u\right )\right \vert u ∈ U\right \}

(This definition contains Notation RLT.)

Example RAO
Range, Archetype O
Archetype O is the linear transformation

 T : {ℂ}^{3} → {ℂ}^{5},\quad T\left (\left [\array{ {x}_{1}\cr {x}_{2} \cr {x}_{3} } \right ]\right ) = \left [\array{ −{x}_{1} + {x}_{2} − 3{x}_{3} \cr −{x}_{1} + 2{x}_{2} − 4{x}_{3} \cr {x}_{1} + {x}_{2} + {x}_{3} \cr 2{x}_{1} + 3{x}_{2} + {x}_{3} \cr {x}_{1} + 2{x}_{3} } \right ]

To determine the elements of {ℂ}^{5} in ℛ\kern -1.95872pt \left (T\right ), find those vectors v such that T\left (u\right ) = v for some u ∈ {ℂ}^{3},

\eqalignno{ v & = T\left (u\right ) & & \cr & = \left [\array{ −{u}_{1} + {u}_{2} − 3{u}_{3} \cr −{u}_{1} + 2{u}_{2} − 4{u}_{3} \cr {u}_{1} + {u}_{2} + {u}_{3} \cr 2{u}_{1} + 3{u}_{2} + {u}_{3} \cr {u}_{1} + 2{u}_{3} } \right ] & & \cr & = \left [\array{ −{u}_{1} \cr −{u}_{1} \cr {u}_{1} \cr 2{u}_{1} \cr {u}_{1} } \right ] + \left [\array{ {u}_{2} \cr 2{u}_{2} \cr {u}_{2} \cr 3{u}_{2} \cr 0 } \right ] + \left [\array{ −3{u}_{3} \cr −4{u}_{3} \cr {u}_{3} \cr {u}_{3} \cr 2{u}_{3} } \right ] & & \cr & = {u}_{1}\left [\array{ −1\cr −1 \cr 1\cr 2 \cr 1 } \right ] + {u}_{2}\left [\array{ 1\cr 2 \cr 1\cr 3 \cr 0 } \right ] + {u}_{3}\left [\array{ −3\cr −4 \cr 1\cr 1 \cr 2 } \right ] & & }

This says that every output of T (v) can be written as a linear combination of the three vectors

\eqalignno{ \left [\array{ −1\cr −1 \cr 1\cr 2 \cr 1 } \right ] & &\left [\array{ 1\cr 2 \cr 1\cr 3 \cr 0 } \right ] & &\left [\array{ −3\cr −4 \cr 1\cr 1 \cr 2 } \right ] & & & & & & }

using the scalars {u}_{1},\kern 1.95872pt {u}_{2},\kern 1.95872pt {u}_{3}. Furthermore, since u can be any element of {ℂ}^{3}, every such linear combination is an output. This means that

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ −1\cr −1 \cr 1\cr 2 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 1\cr 2 \cr 1\cr 3 \cr 0 } \right ],\kern 1.95872pt \left [\array{ −3\cr −4 \cr 1\cr 1 \cr 2 } \right ]\right \}\right \rangle

The three vectors in this spanning set for ℛ\kern -1.95872pt \left (T\right ) form a linearly dependent set (check this!). So we can find a more economical presentation by any of the various methods from Section CRS and Section FS. We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to Theorem BRS, so we can describe the range of T with a basis,

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ 1\cr 0 \cr −3\cr −7 \cr −2 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 2\cr 5 \cr 1 } \right ]\right \}\right \rangle

We know that the span of a set of vectors is always a subspace (Theorem SSS), so the range computed in Example RAO is also a subspace. This is no accident, the range of a linear transformation is always a subspace.

Theorem RLTS
Range of a Linear Transformation is a Subspace
Suppose that T : U → V is a linear transformation. Then the range of T, ℛ\kern -1.95872pt \left (T\right ), is a subspace of V .

Proof   We can apply the three-part test of Theorem TSS. First, {0}_{U} ∈ U and T\left ({0}_{U}\right ) = {0}_{V } by Theorem LTTZZ, so {0}_{V } ∈ℛ\kern -1.95872pt \left (T\right ) and we know that the range is non-empty.

Suppose we assume that x,\kern 1.95872pt y ∈ℛ\kern -1.95872pt \left (T\right ). Is x + y ∈ℛ\kern -1.95872pt \left (T\right )? If x,\kern 1.95872pt y ∈ℛ\kern -1.95872pt \left (T\right ) then we know there are vectors w,\kern 1.95872pt z ∈ U such that T\left (w\right ) = x and T\left (z\right ) = y. Because U is a vector space, additive closure (Property AC) implies that w + z ∈ U. Then

\eqalignno{ T\left (w + z\right ) & = T\left (w\right ) + T\left (z\right ) & &\text{@(a href="fcla-jsmath-2.23li51.html#definition.LT")Definition LT@(/a)} & & & & \cr & = x + y & &\text{Definition of $w$ and $z$} & & & & }

So we have found an input, w + z, which when fed into T creates x + y as an output. This qualifies x + y for membership in ℛ\kern -1.95872pt \left (T\right ). So we have additive closure.

Suppose we assume that α ∈ {ℂ}^{} and x ∈ℛ\kern -1.95872pt \left (T\right ). Is αx ∈ℛ\kern -1.95872pt \left (T\right )? If x ∈ℛ\kern -1.95872pt \left (T\right ), then there is a vector w ∈ U such that T\left (w\right ) = x. Because U is a vector space, scalar closure implies that αw ∈ U. Then

\eqalignno{ T\left (αw\right ) & = αT\left (w\right ) & &\text{@(a href="fcla-jsmath-2.23li51.html#definition.LT")Definition LT@(/a)} & & & & \cr & = αx & &\text{Definition of $w$} & & & & }

So we have found an input (αw) which when fed into T creates αx as an output. This qualifies αx for membership in ℛ\kern -1.95872pt \left (T\right ). So we have scalar closure and Theorem TSS tells us that ℛ\kern -1.95872pt \left (T\right ) is a subspace of V .

Let’s compute another range, now that we know in advance that it will be a subspace.

Example FRAN
Full range, Archetype N
Archetype N is the linear transformation

 T : {ℂ}^{5} → {ℂ}^{3},\quad T\left (\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} \cr {x}_{5} } \right ]\right ) = \left [\array{ 2{x}_{1} + {x}_{2} + 3{x}_{3} − 4{x}_{4} + 5{x}_{5} \cr {x}_{1} − 2{x}_{2} + 3{x}_{3} − 9{x}_{4} + 3{x}_{5} \cr 3{x}_{1} + 4{x}_{3} − 6{x}_{4} + 5{x}_{5} } \right ]

To determine the elements of {ℂ}^{3} in ℛ\kern -1.95872pt \left (T\right ), find those vectors v such that T\left (u\right ) = v for some u ∈ {ℂ}^{5},

\eqalignno{ v & = T\left (u\right ) & & \cr & = \left [\array{ 2{u}_{1} + {u}_{2} + 3{u}_{3} − 4{u}_{4} + 5{u}_{5} \cr {u}_{1} − 2{u}_{2} + 3{u}_{3} − 9{u}_{4} + 3{u}_{5} \cr 3{u}_{1} + 4{u}_{3} − 6{u}_{4} + 5{u}_{5} } \right ] & & \cr & = \left [\array{ 2{u}_{1} \cr {u}_{1} \cr 3{u}_{1}} \right ] + \left [\array{ {u}_{2} \cr −2{u}_{2} \cr 0} \right ] + \left [\array{ 3{u}_{3} \cr 3{u}_{3} \cr 4{u}_{3}} \right ] + \left [\array{ −4{u}_{4} \cr −9{u}_{4} \cr −6{u}_{4} } \right ] + \left [\array{ 5{u}_{5} \cr 3{u}_{5} \cr 5{u}_{5}} \right ] & & \cr & = {u}_{1}\left [\array{ 2\cr 1 \cr 3 } \right ] + {u}_{2}\left [\array{ 1\cr −2 \cr 0 } \right ] + {u}_{3}\left [\array{ 3\cr 3 \cr 4 } \right ] + {u}_{4}\left [\array{ −4\cr −9 \cr −6 } \right ] + {u}_{5}\left [\array{ 5\cr 3 \cr 5 } \right ] & & \cr & & }

This says that every output of T (v) can be written as a linear combination of the five vectors

\eqalignno{ \left [\array{ 2\cr 1 \cr 3 } \right ] & &\left [\array{ 1\cr −2 \cr 0 } \right ] & &\left [\array{ 3\cr 3 \cr 4 } \right ] & &\left [\array{ −4\cr −9 \cr −6 } \right ] & &\left [\array{ 5\cr 3 \cr 5 } \right ] & & & & & & & & & & }

using the scalars {u}_{1},\kern 1.95872pt {u}_{2},\kern 1.95872pt {u}_{3},\kern 1.95872pt {u}_{4},\kern 1.95872pt {u}_{5}. Furthermore, since u can be any element of {ℂ}^{5}, every such linear combination is an output. This means that

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ 2\cr 1 \cr 3 } \right ],\kern 1.95872pt \left [\array{ 1\cr −2 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 3\cr 3 \cr 4 } \right ],\kern 1.95872pt \left [\array{ −4\cr −9 \cr −6 } \right ],\kern 1.95872pt \left [\array{ 5\cr 3 \cr 5 } \right ]\right \}\right \rangle

The five vectors in this spanning set for ℛ\kern -1.95872pt \left (T\right ) form a linearly dependent set (Theorem MVSLD). So we can find a more economical presentation by any of the various methods from Section CRS and Section FS. We will place the vectors into a matrix as rows, row-reduce, toss out zero rows and appeal to Theorem BRS, so we can describe the range of T with a (nice) basis,

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ 1\cr 0 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1 } \right ]\right \}\right \rangle = {ℂ}^{3}

In contrast to injective linear transformations having small (trivial) kernels (Theorem KILT), surjective linear transformations have large ranges, as indicated in the next theorem.

Theorem RSLT
Range of a Surjective Linear Transformation
Suppose that T : U → V is a linear transformation. Then T is surjective if and only if the range of T equals the codomain, ℛ\kern -1.95872pt \left (T\right ) = V .

Proof   () By Definition RLT, we know that ℛ\kern -1.95872pt \left (T\right ) ⊆ V . To establish the reverse inclusion, assume v ∈ V . Then since T is surjective (Definition SLT), there exists a vector u ∈ U so that T\left (u\right ) = v. However, the existence of u gains v membership in ℛ\kern -1.95872pt \left (T\right ), so V ⊆ℛ\kern -1.95872pt \left (T\right ). Thus, ℛ\kern -1.95872pt \left (T\right ) = V .

() To establish that T is surjective, choose v ∈ V . Since we are assuming that ℛ\kern -1.95872pt \left (T\right ) = V , v ∈ℛ\kern -1.95872pt \left (T\right ). This says there is a vector u ∈ U so that T\left (u\right ) = v, i.e. T is surjective.

Example NSAQR
Not surjective, Archetype Q, revisited
We are now in a position to revisit our first example in this section, Example NSAQ. In that example, we showed that Archetype Q is not surjective by constructing a vector in the codomain where no element of the domain could be used to evaluate the linear transformation to create the output, thus violating Definition SLT. Just where did this vector come from?

The short answer is that the vector

 v = \left [\array{ −1\cr 2 \cr 3\cr −1 \cr 4 } \right ]

was constructed to lie outside of the range of T. How was this accomplished? First, the range of T is given by

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ 1\cr 0 \cr 0\cr 0 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 0\cr 0 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1\cr 0 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 0\cr 1 \cr 2 } \right ]\right \}\right \rangle

Suppose an element of the range {v}^{∗} has its first 4 components equal to − 1, 2, 3,−1, in that order. Then to be an element of ℛ\kern -1.95872pt \left (T\right ), we would have

 { v}^{∗} = (−1)\left [\array{ 1\cr 0 \cr 0\cr 0 \cr 1 } \right ]+(2)\left [\array{ 0\cr 1 \cr 0\cr 0 \cr −1 } \right ]+(3)\left [\array{ 0\cr 0 \cr 1\cr 0 \cr −1 } \right ]+(−1)\left [\array{ 0\cr 0 \cr 0\cr 1 \cr 2 } \right ] = \left [\array{ −1\cr 2 \cr 3\cr −1 \cr −8 } \right ]

So the only vector in the range with these first four components specified, must have − 8 in the fifth component. To set the fifth component to any other value (say, 4) will result in a vector (v in Example NSAQ) outside of the range. Any attempt to find an input for T that will produce v as an output will be doomed to failure.

Whenever the range of a linear transformation is not the whole codomain, we can employ this device and conclude that the linear transformation is not surjective. This is another way of viewing Theorem RSLT. For a surjective linear transformation, the range is all of the codomain and there is no choice for a vector v that lies in V , yet not in the range. For every one of the archetypes that is not surjective, there is an example presented of exactly this form.

Example NSAO
Not surjective, Archetype O
In Example RAO the range of Archetype O was determined to be

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ 1\cr 0 \cr −3\cr −7 \cr −2 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 2\cr 5 \cr 1 } \right ]\right \}\right \rangle

a subspace of dimension 2 in {ℂ}^{5}. Since ℛ\kern -1.95872pt \left (T\right )\mathrel{≠}{ℂ}^{5}, Theorem RSLT says T is not surjective.

Example SAN
Surjective, Archetype N
The range of Archetype N was computed in Example FRAN to be

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{\left [\array{ 1\cr 0 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 0 } \right ],\kern 1.95872pt \left [\array{ 0\cr 0 \cr 1 } \right ]\right \}\right \rangle

Since the basis for this subspace is the set of standard unit vectors for {ℂ}^{3} (Theorem SUVB), we have ℛ\kern -1.95872pt \left (T\right ) = {ℂ}^{3} and by Theorem RSLT, T is surjective.

#### Subsection SSSLT: Spanning Sets and Surjective Linear Transformations

Just as injective linear transformations are allied with linear independence (Theorem ILTLI, Theorem ILTB), surjective linear transformations are allied with spanning sets.

Theorem SSRLT
Spanning Set for Range of a Linear Transformation
Suppose that T : U → V is a linear transformation and S = \left \{{u}_{1},\kern 1.95872pt {u}_{2},\kern 1.95872pt {u}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {u}_{t}\right \} spans U. Then

\eqalignno{ R = \left \{T\left ({u}_{1}\right ),\kern 1.95872pt T\left ({u}_{2}\right ),\kern 1.95872pt T\left ({u}_{3}\right ),\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt T\left ({u}_{t}\right )\right \} & & }

spans ℛ\kern -1.95872pt \left (T\right ).

Proof   We need to establish that ℛ\kern -1.95872pt \left (T\right ) = \left \langle R\right \rangle , a set equality. First we establish that ℛ\kern -1.95872pt \left (T\right ) ⊆\left \langle R\right \rangle . To this end, choose v ∈ℛ\kern -1.95872pt \left (T\right ). Then there exists a vector u ∈ U, such that T\left (u\right ) = v (Definition RLT). Because S spans U there are scalars, {a}_{1},\kern 1.95872pt {a}_{2},\kern 1.95872pt {a}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {a}_{t}, such that

 u = {a}_{1}{u}_{1} + {a}_{2}{u}_{2} + {a}_{3}{u}_{3} + \mathrel{⋯} + {a}_{t}{u}_{t}

Then

\eqalignno{ v & = T\left (u\right ) & &\text{@(a href="#definition.RLT")Definition RLT@(/a)} & & & & \cr & = T\left ({a}_{1}{u}_{1} + {a}_{2}{u}_{2} + {a}_{3}{u}_{3} + \mathrel{⋯} + {a}_{t}{u}_{t}\right ) & &\text{@(a href="fcla-jsmath-2.23li39.html#definition.TSVS")Definition TSVS@(/a)} & & & & \cr & = {a}_{1}T\left ({u}_{1}\right ) + {a}_{2}T\left ({u}_{2}\right ) + {a}_{3}T\left ({u}_{3}\right ) + \mathop{\mathop{…}} + {a}_{t}T\left ({u}_{t}\right ) & &\text{@(a href="fcla-jsmath-2.23li51.html#theorem.LTLC")Theorem LTLC@(/a)} & & & & \cr & & & & }

which establishes that v ∈\left \langle R\right \rangle (Definition SS). So ℛ\kern -1.95872pt \left (T\right ) ⊆\left \langle R\right \rangle .

To establish the opposite inclusion, choose an element of the span of R, say v ∈\left \langle R\right \rangle . Then there are scalars {b}_{1},\kern 1.95872pt {b}_{2},\kern 1.95872pt {b}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {b}_{t} so that

\eqalignno{ v & = {b}_{1}T\left ({u}_{1}\right ) + {b}_{2}T\left ({u}_{2}\right ) + {b}_{3}T\left ({u}_{3}\right ) + \mathrel{⋯} + {b}_{t}T\left ({u}_{t}\right ) & &\text{@(a href="fcla-jsmath-2.23li38.html#definition.SS")Definition SS@(/a)} & & & & \cr & = T\left ({b}_{1}{u}_{1} + {b}_{2}{u}_{2} + {b}_{3}{u}_{3} + \mathrel{⋯} + {b}_{t}{u}_{t}\right ) & &\text{@(a href="fcla-jsmath-2.23li51.html#theorem.LTLC")Theorem LTLC@(/a)} & & & & }

This demonstrates that v is an output of the linear transformation T, so v ∈ℛ\kern -1.95872pt \left (T\right ). Therefore \left \langle R\right \rangle ⊆ℛ\kern -1.95872pt \left (T\right ), so we have the set equality ℛ\kern -1.95872pt \left (T\right ) = \left \langle R\right \rangle (Definition SE). In other words, R spans ℛ\kern -1.95872pt \left (T\right ) (Definition TSVS).

Theorem SSRLT provides an easy way to begin the construction of a basis for the range of a linear transformation, since the construction of a spanning set requires simply evaluating the linear transformation on a spanning set of the domain. In practice the best choice for a spanning set of the domain would be as small as possible, in other words, a basis. The resulting spanning set for the codomain may not be linearly independent, so to find a basis for the range might require tossing out redundant vectors from the spanning set. Here’s an example.

Example BRLT
A basis for the range of a linear transformation
Define the linear transformation T : {M}_{22} → {P}_{2} by

 T\left (\left [\array{ a&b\cr c&d } \right ]\right ) = \left (a + 2b + 8c + d\right )+\left (−3a + 2b + 5d\right )x+\left (a + b + 5c\right ){x}^{2}

A convenient spanning set for {M}_{22} is the basis

 S = \left \{\left [\array{ 1&0\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 0&1\cr 0&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 1&0 } \right ],\kern 1.95872pt \left [\array{ 0&0\cr 0&1 } \right ]\right \}

So by Theorem SSRLT, a spanning set for ℛ\kern -1.95872pt \left (T\right ) is

\eqalignno{ R & = \left \{T\left (\left [\array{ 1&0\cr 0&0 } \right ]\right ),\kern 1.95872pt T\left (\left [\array{ 0&1\cr 0&0 } \right ]\right ),\kern 1.95872pt T\left (\left [\array{ 0&0\cr 1&0 } \right ]\right ),\kern 1.95872pt T\left (\left [\array{ 0&0\cr 0&1 } \right ]\right )\right \} & & \cr & = \left \{1 − 3x + {x}^{2},\kern 1.95872pt 2 + 2x + {x}^{2},\kern 1.95872pt 8 + 5{x}^{2},\kern 1.95872pt 1 + 5x\right \} & & }

The set R is not linearly independent, so if we desire a basis for ℛ\kern -1.95872pt \left (T\right ), we need to eliminate some redundant vectors. Two particular relations of linear dependence on R are

\eqalignno{ (−2)(1 − 3x + {x}^{2}) + (−3)(2 + 2x + {x}^{2}) + (8 + 5{x}^{2}) & = 0 + 0x + 0{x}^{2} = 0 & & \cr (1 − 3x + {x}^{2}) + (−1)(2 + 2x + {x}^{2}) + (1 + 5x) & = 0 + 0x + 0{x}^{2} = 0 & & }

These, individually, allow us to remove 8 + 5{x}^{2} and 1 + 5x from R with out destroying the property that R spans ℛ\kern -1.95872pt \left (T\right ). The two remaining vectors are linearly independent (check this!), so we can write

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{1 − 3x + {x}^{2},\kern 1.95872pt 2 + 2x + {x}^{2}\right \}\right \rangle

and see that \mathop{ dim}\nolimits \left (ℛ\kern -1.95872pt \left (T\right )\right ) = 2.

Elements of the range are precisely those elements of the codomain with non-empty preimages.

Theorem RPI
Range and Pre-Image
Suppose that T : U → V is a linear transformation. Then

 v ∈ℛ\kern -1.95872pt \left (T\right )\text{ if and only if }{T}^{−1}\left (v\right )\mathrel{≠}∅

Proof   () If v ∈ℛ\kern -1.95872pt \left (T\right ), then there is a vector u ∈ U such that T\left (u\right ) = v. This qualifies u for membership in {T}^{−1}\left (v\right ), and thus the preimage of v is not empty.

() Suppose the preimage of v is not empty, so we can choose a vector u ∈ U such that T\left (u\right ) = v. Then v ∈ℛ\kern -1.95872pt \left (T\right ).

Theorem SLTB
Surjective Linear Transformations and Bases
Suppose that T : U → V is a linear transformation and B = \left \{{u}_{1},\kern 1.95872pt {u}_{2},\kern 1.95872pt {u}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {u}_{m}\right \} is a basis of U. Then T is surjective if and only if C = \left \{T\left ({u}_{1}\right ),\kern 1.95872pt T\left ({u}_{2}\right ),\kern 1.95872pt T\left ({u}_{3}\right ),\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt T\left ({u}_{m}\right )\right \} is a spanning set for V .

Proof   () Assume T is surjective. Since B is a basis, we know B is a spanning set of U (Definition B). Then Theorem SSRLT says that C spans ℛ\kern -1.95872pt \left (T\right ). But the hypothesis that T is surjective means V = ℛ\kern -1.95872pt \left (T\right ) (Theorem RSLT), so C spans V .

() Assume that C spans V . To establish that T is surjective, we will show that every element of V is an output of T for some input (Definition SLT). Suppose that v ∈ V . As an element of V , we can write v as a linear combination of the spanning set C. So there are are scalars, {b}_{1},\kern 1.95872pt {b}_{2},\kern 1.95872pt {b}_{3},\kern 1.95872pt \mathop{\mathop{…}},\kern 1.95872pt {b}_{m}, such that

 v = {b}_{1}T\left ({u}_{1}\right ) + {b}_{2}T\left ({u}_{2}\right ) + {b}_{3}T\left ({u}_{3}\right ) + \mathrel{⋯} + {b}_{m}T\left ({u}_{m}\right )

Now define the vector u ∈ U by

 u = {b}_{1}{u}_{1} + {b}_{2}{u}_{2} + {b}_{3}{u}_{3} + \mathrel{⋯} + {b}_{m}{u}_{m}

Then

\eqalignno{ T\left (u\right ) & = T\left ({b}_{1}{u}_{1} + {b}_{2}{u}_{2} + {b}_{3}{u}_{3} + \mathrel{⋯} + {b}_{m}{u}_{m}\right ) & & & & \cr & = {b}_{1}T\left ({u}_{1}\right ) + {b}_{2}T\left ({u}_{2}\right ) + {b}_{3}T\left ({u}_{3}\right ) + \mathrel{⋯} + {b}_{m}T\left ({u}_{m}\right ) & &\text{@(a href="fcla-jsmath-2.23li51.html#theorem.LTLC")Theorem LTLC@(/a)} & & & & \cr & = v & & & & }

So, given any choice of a vector v ∈ V , we can design an input u ∈ U to produce v as an output of T. Thus, by Definition SLT, T is surjective.

#### Subsection SLTD: Surjective Linear Transformations and Dimension

Theorem SLTD
Surjective Linear Transformations and Dimension
Suppose that T : U → V is a surjective linear transformation. Then \mathop{ dim}\nolimits \left (U\right ) ≥\mathop{ dim}\nolimits \left (V \right ).

Proof   Suppose to the contrary that m =\mathop{ dim}\nolimits \left (U\right ) <\mathop{ dim}\nolimits \left (V \right ) = t. Let B be a basis of U, which will then contain m vectors. Apply T to each element of B to form a set C that is a subset of V . By Theorem SLTB, C is spanning set of V with m or fewer vectors. So we have a set of m or fewer vectors that span V , a vector space of dimension t, with m < t. However, this contradicts Theorem G, so our assumption is false and \mathop{ dim}\nolimits \left (U\right ) ≥\mathop{ dim}\nolimits \left (V \right ).

Example NSDAT
Not surjective by dimension, Archetype T
The linear transformation in Archetype T is

 T : {P}_{4} → {P}_{5},\quad T\left (p(x)\right ) = (x − 2)p(x)

Since \mathop{ dim}\nolimits \left ({P}_{4}\right ) = 5 < 6 =\mathop{ dim}\nolimits \left ({P}_{5}\right ), T cannot be surjective for then it would violate Theorem SLTD.

Notice that the previous example made no use of the actual formula defining the function. Merely a comparison of the dimensions of the domain and codomain are enough to conclude that the linear transformation is not surjective. Archetype O and Archetype P are two more examples of linear transformations that have “small” domains and “big” codomains, resulting in an inability to create all possible outputs and thus they are non-surjective linear transformations.

#### Subsection CSLT: Composition of Surjective Linear Transformations

In Subsection LT.NLTFO we saw how to combine linear transformations to build new linear transformations, specifically, how to build the composition of two linear transformations (Definition LTC). It will be useful later to know that the composition of surjective linear transformations is again surjective, so we prove that here.

Theorem CSLTS
Composition of Surjective Linear Transformations is Surjective
Suppose that T : U → V and S : V → W are surjective linear transformations. Then (S ∘ T): U → W is a surjective linear transformation.

Proof   That the composition is a linear transformation was established in Theorem CLTLT, so we need only establish that the composition is surjective. Applying Definition SLT, choose w ∈ W.

Because S is surjective, there must be a vector v ∈ V , such that S\left (v\right ) = w. With the existence of v established, that T is surjective guarantees a vector u ∈ U such that T\left (u\right ) = v. Now,

\eqalignno{ \left (S ∘ T\right )\left (u\right ) & = S\left (T\left (u\right )\right ) & &\text{@(a href="fcla-jsmath-2.23li51.html#definition.LTC")Definition LTC@(/a)} & & & & \cr & = S\left (v\right ) & &\text{Definition of $u$} & & & & \cr & = w & &\text{Definition of $v$} & & & & }

This establishes that any element of the codomain (w) can be created by evaluating S ∘ T with the right input (u). Thus, by Definition SLT, S ∘ T is surjective.

1. Suppose T : {ℂ}^{5} → {ℂ}^{8} is a linear transformation. Why can’t T be surjective?
2. What is the relationship between a surjective linear transformation and its range?
3. Compare and contrast injective and surjective linear transformations.

#### Subsection EXC: Exercises

C10 Each archetype below is a linear transformation. Compute the range for each.
Archetype M
Archetype N
Archetype O
Archetype P
Archetype Q
Archetype R
Archetype S
Archetype T
Archetype U
Archetype V
Archetype W
Archetype X

Contributed by Robert Beezer

C20 Example SAR concludes with an expression for a vector u ∈ {ℂ}^{5} that we believe will create the vector v ∈ {ℂ}^{5} when used to evaluate T. That is, T\left (u\right ) = v. Verify this assertion by actually evaluating T with u. If you don’t have the patience to push around all these symbols, try choosing a numerical instance of v, compute u, and then compute T\left (u\right ), which should result in v.
Contributed by Robert Beezer

C22 The linear transformation S : {ℂ}^{4} → {ℂ}^{3} is not surjective. Find an output w ∈ {ℂ}^{3} that has an empty pre-image (that is {S}^{−1}\left (w\right ) = ∅.)

 S\left (\left [\array{ {x}_{1} \cr {x}_{2} \cr {x}_{3} \cr {x}_{4} } \right ]\right ) = \left [\array{ 2{x}_{1} + {x}_{2} + 3{x}_{3} − 4{x}_{4} \cr {x}_{1} + 3{x}_{2} + 4{x}_{3} + 3{x}_{4} \cr −{x}_{1} + 2{x}_{2} + {x}_{3} + 7{x}_{4} } \right ]

Contributed by Robert Beezer Solution [1552]

C23 Determine whether or not the following linear transformation T : {ℂ}^{5} → {P}_{ 3} is surjective:

\eqalignno{ T\left (\left [\array{ a\cr b \cr c\cr d \cr e } \right ]\right ) & = a + (b + c)x + (c + d){x}^{2} + (d + e){x}^{3} & & }

Contributed by Chris Black Solution [1554]

C24 Determine whether or not the linear transformation T : {P}_{3} → {ℂ}^{5} below is surjective:

\eqalignno{ T\left (a + bx + c{x}^{2} + d{x}^{3}\right ) & = \left [\array{ a + b \cr b + c \cr c + d\cr a + c \cr b + d } \right ]. & & }

Contributed by Chris Black Solution [1556]

C25 Define the linear transformation

 T : {ℂ}^{3} → {ℂ}^{2},\quad T\left (\left [\array{ {x}_{1}\cr {x}_{2} \cr {x}_{3} } \right ]\right ) = \left [\array{ 2{x}_{1} − {x}_{2} + 5{x}_{3} \cr −4{x}_{1} + 2{x}_{2} − 10{x}_{3} } \right ]

Find a basis for the range of T, ℛ\kern -1.95872pt \left (T\right ). Is T surjective?
Contributed by Robert Beezer Solution [1556]

C26 Let T : {ℂ}^{3} → {ℂ}^{3} be given by T\left (\left [\array{ a\cr b \cr c } \right ]\right ) = \left [\array{ a + b + 2c\cr 2c \cr a + b + c } \right ]. Find a basis of ℛ\kern -1.95872pt \left (T\right ). Is T surjective?
Contributed by Chris Black Solution [1557]

C27 Let T : {ℂ}^{3} → {ℂ}^{4} be given by T\left (\left [\array{ a\cr b \cr c } \right ]\right ) = \left [\array{ a + b − c \cr a − b + c \cr −a + b + c \cr a + b + c } \right ]. Find a basis of ℛ\kern -1.95872pt \left (T\right ). Is T surjective?
Contributed by Chris Black Solution [1558]

C28 Let T : {ℂ}^{4} → {M}_{ 2,2} be given by T\left (\left [\array{ a\cr b \cr c\cr d } \right ]\right ) = \left [\array{ a + b &a + b + c\cr a + b + c & a + d } \right ]. Find a basis of ℛ\kern -1.95872pt \left (T\right ). Is T surjective?
Contributed by Chris Black Solution [1559]

C29 Let T : {P}_{2} → {P}_{4} be given by T\left (p(x)\right ) = {x}^{2}p(x). Find a basis of ℛ\kern -1.95872pt \left (T\right ). Is T surjective?
Contributed by Chris Black Solution [1560]

C30 Let T : {P}_{4} → {P}_{3} be given by T\left (p(x)\right ) = {p}^{′}(x), where {p}^{′}(x) is the derivative. Find a basis of ℛ\kern -1.95872pt \left (T\right ). Is T surjective?
Contributed by Chris Black Solution [1560]

C40 Show that the linear transformation T is not surjective by finding an element of the codomain, v, such that there is no vector u with T\left (u\right ) = v.

 T : {ℂ}^{3} → {ℂ}^{3},\quad T\left (\left [\array{ a\cr b \cr c } \right ]\right ) = \left [\array{ 2a + 3b − c \cr 2b − 2c \cr a − b + 2c } \right ]

Contributed by Robert Beezer Solution [1561]

M60 Suppose U and V are vector spaces. Define the function Z : U → V by T\left (u\right ) = {0}_{V } for every u ∈ U. Then by Exercise LT.M60, Z is a linear transformation. Formulate a condition on V that is equivalent to Z being an surjective linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof): Z is surjective if and only if V is __________________. (See Exercise ILT.M60, Exercise IVLT.M60.)
Contributed by Robert Beezer

T15 Suppose that that T : U → V and S : V → W are linear transformations. Prove the following relationship between ranges.

 ℛ\kern -1.95872pt \left (S ∘ T\right ) ⊆ℛ\kern -1.95872pt \left (S\right )

Contributed by Robert Beezer Solution [1562]

T20 Suppose that A is an m × n matrix. Define the linear transformation T by

 T : {ℂ}^{n} → {ℂ}^{m},\quad T\left (x\right ) = Ax

Prove that the range of T equals the column space of A, ℛ\kern -1.95872pt \left (T\right ) = C\kern -1.95872pt \left (A\right ).
Contributed by Andy Zimmer Solution [1563]

#### Subsection SOL: Solutions

C22 Contributed by Robert Beezer Statement [1546]
To find an element of {ℂ}^{3} with an empty pre-image, we will compute the range of the linear transformation ℛ\kern -1.95872pt \left (S\right ) and then find an element outside of this set.

By Theorem SSRLT we can evaluate S with the elements of a spanning set of the domain and create a spanning set for the range.

\eqalignno{ S\left (\left [\array{ 1\cr 0 \cr 0\cr 0 } \right ]\right ) & = \left [\array{ 2\cr 1 \cr −1 } \right ] &S\left (\left [\array{ 0\cr 1 \cr 0\cr 0 } \right ]\right ) & = \left [\array{ 1\cr 3 \cr 2 } \right ] &S\left (\left [\array{ 0\cr 0 \cr 1\cr 0 } \right ]\right ) & = \left [\array{ 3\cr 4 \cr 1 } \right ] &S\left (\left [\array{ 0\cr 0 \cr 0\cr 1 } \right ]\right ) & = \left [\array{ −4\cr 3 \cr 7 } \right ] & & & & & & & & }

So

 ℛ\kern -1.95872pt \left (S\right ) = \left \langle \left \{\left [\array{ 2\cr 1 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 1\cr 3 \cr 2 } \right ],\kern 1.95872pt \left [\array{ 3\cr 4 \cr 1 } \right ],\kern 1.95872pt \left [\array{ −4\cr 3 \cr 7 } \right ]\right \}\right \rangle

This spanning set is obviously linearly dependent, so we can reduce it to a basis for ℛ\kern -1.95872pt \left (S\right ) using Theorem BRS, where the elements of the spanning set are placed as the rows of a matrix. The result is that

 ℛ\kern -1.95872pt \left (S\right ) = \left \langle \left \{\left [\array{ 1\cr 0 \cr −1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 1 \cr 1 } \right ]\right \}\right \rangle

Therefore, the unique vector in ℛ\kern -1.95872pt \left (S\right ) with a first slot equal to 6 and a second slot equal to 15 will be the linear combination

 6\left [\array{ 1\cr 0 \cr −1 } \right ]+15\left [\array{ 0\cr 1 \cr 1 } \right ] = \left [\array{ 6\cr 15 \cr 9 } \right ]

So, any vector with first two components equal to 6 and 15, but with a third component different from 9, such as

 w = \left [\array{ 6\cr 15 \cr −63 } \right ]

will not be an element of the range of S and will therefore have an empty pre-image. Another strategy on this problem is to guess. Almost any vector will lie outside the range of T, you have to be unlucky to randomly choose an element of the range. This is because the codomain has dimension 3, while the range is “much smaller” at a dimension of 2. You still need to check that your guess lies outside of the range, which generally will involve solving a system of equations that turns out to be inconsistent.

C23 Contributed by Chris Black Statement [1547]
The linear transformation T is surjective if for any p(x) = α + βx + γ{x}^{2} + δ{x}^{3}, there is a vector u = \left [\array{ a\cr b \cr c\cr d \cr e } \right ] in {ℂ}^{5} so that T\left (u\right ) = p(x). We need to be able to solve the system

\eqalignno{ a & = α & & \cr b + c & = β & & \cr c + d & = γ & & \cr d + e & = δ & & }

This system has an infinite number of solutions, one of which is a = α, b = β, c = 0, d = γ and e = δ − γ, so that

\eqalignno{ T\left (\left [\array{ α\cr β \cr 0\cr γ \cr δ − γ } \right ]\right ) & = α + (β + 0)x + (0 + γ){x}^{2} + (γ + (δ − γ)){x}^{3} & & \cr & = α + βx + γ{x}^{2} + δ{x}^{3} & & \cr & = p(x). & & }

Thus, T is surjective, since for every vector v ∈ {P}_{3}, there exists a vector u ∈ {ℂ}^{5} so that T\left (u\right ) = v.

C24 Contributed by Chris Black Statement [1547]
According to Theorem SLTD, if a linear transformation T : U → V is surjective, then \mathop{ dim}\nolimits \left (U\right ) ≥\mathop{ dim}\nolimits \left (V \right ). In this example, U = {P}_{3} has dimension 4, and V = {ℂ}^{5} has dimension 5, so T cannot be surjective. (There is no way T can “expand” the domain {P}_{3} to fill the codomain {ℂ}^{5}.)

C25 Contributed by Robert Beezer Statement [1548]
To find the range of T, apply T to the elements of a spanning set for {ℂ}^{3} as suggested in Theorem SSRLT. We will use the standard basis vectors (Theorem SUVB).

 ℛ\kern -1.95872pt \left (T\right ) = \left \langle \left \{T\left ({e}_{1}\right ),\kern 1.95872pt T\left ({e}_{2}\right ),\kern 1.95872pt T\left ({e}_{3}\right )\right \}\right \rangle = \left \langle \left \{\left [\array{ 2\cr −4 } \right ],\kern 1.95872pt \left [\array{ −1\cr 2 } \right ],\kern 1.95872pt \left [\array{ 5\cr −10 } \right ]\right \}\right \rangle

Each of these vectors is a scalar multiple of the others, so we can toss two of them in reducing the spanning set to a linearly independent set (or be more careful and apply Theorem BCS on a matrix with these three vectors as columns). The result is the basis of the range,

 \left \{\left [\array{ 1\cr −2 } \right ]\right \}

With r\left (T\right )\mathrel{≠}2, ℛ\kern -1.95872pt \left (T\right )\mathrel{≠}{ℂ}^{2}, so Theorem RSLT says T is not surjective.

C26 Contributed by Chris Black Statement [1548]
The range of T is

\eqalignno{ ℛ\kern -1.95872pt \left (T\right ) & = \left \{\left .\left [\array{ a + b + 2c\cr 2c \cr a + b + c } \right ]\right \vert a,b,c ∈ ℂ\right \} & & \cr & = \left \{\left .a\left [\array{ 1\cr 0 \cr 1 } \right ] + b\left [\array{ 1\cr 0 \cr 1 } \right ] + c\left [\array{ 2\cr 2 \cr 1 } \right ]\right \vert a,b,c ∈ ℂ\right \} & & \cr & = \left \langle \left [\array{ 1\cr 0 \cr 1 } \right ],\left [\array{ 2\cr 2 \cr 1 } \right ]\right \rangle & & }

Since the vectors \left [\array{ 1\cr 0 \cr 1 } \right ] and \left [\array{ 2\cr 2 \cr 1 } \right ]are linearly independent (why?), a basis of ℛ\kern -1.95872pt \left (T\right ) is \left \{\left [\array{ 1\cr 0 \cr 1 } \right ],\left [\array{ 2\cr 2 \cr 1 } \right ]\right \}. Since the dimension of the range is 2 and the dimension of the codomain is 3, T is not surjective.

C27 Contributed by Chris Black Statement [1549]
The range of T is

\eqalignno{ ℛ\kern -1.95872pt \left (T\right ) & = \left \{\left .\left [\array{ a + b − c \cr a − b + c \cr −a + b + c \cr a + b + c } \right ]\right \vert a,b,c ∈ ℂ\right \} & & \cr & = \left \{\left .a\left [\array{ 1\cr 1 \cr −1\cr 1 } \right ] + b\left [\array{ 1\cr −1 \cr 1\cr 1 } \right ] + c\left [\array{ −1\cr 1 \cr 1\cr 1 } \right ]\right \vert a,b,c ∈ ℂ\right \} & & \cr & = \left \langle \left [\array{ 1\cr 1 \cr −1\cr 1 } \right ],\left [\array{ 1\cr −1 \cr 1\cr 1 } \right ],\left [\array{ −1\cr 1 \cr 1\cr 1 } \right ]\right \rangle & & }

By row reduction (not shown), we can see that the set

\eqalignno{ \left \{\left [\array{ 1\cr 1 \cr −1\cr 1 } \right ],\kern 1.95872pt \left [\array{ 1\cr −1 \cr 1\cr 1 } \right ],\kern 1.95872pt \left [\array{ −1\cr 1 \cr 1\cr 1 } \right ]\right \} & & }

are linearly independent, so is a basis of ℛ\kern -1.95872pt \left (T\right ). Since the dimension of the range is 3 and the dimension of the codomain is 4, T is not surjective. (We should have anticipated that T was not surjective since the dimension of the domain is smaller than the dimension of the codomain.)

C28 Contributed by Chris Black Statement [1549]
The range of T is

\eqalignno{ ℛ\kern -1.95872pt \left (T\right ) & = \left \{\left .\left [\array{ a + b &a + b + c\cr a + b + c & a + d } \right ]\right \vert a,b,c,d ∈ ℂ\right \} & & \cr & = \left \{\left .a\left [\array{ 1&1\cr 1&1 } \right ] + b\left [\array{ 1&1\cr 1&0 } \right ] + c\left [\array{ 0&1\cr 1&0 } \right ] + d\left [\array{ 0&0\cr 0&1 } \right ]\right \vert a,b,c,d ∈ ℂ\right \} & & \cr & = \left \langle \left [\array{ 1&1\cr 1&1 } \right ],\left [\array{ 1&1\cr 1&0 } \right ],\left [\array{ 0&1\cr 1&0 } \right ],\left [\array{ 0&0\cr 0&1 } \right ]\right \rangle & & \cr & = \left \langle \left [\array{ 1&1\cr 1&0 } \right ],\left [\array{ 0&1\cr 1&0 } \right ],\left [\array{ 0&0\cr 0&1 } \right ]\right \rangle . & & }

Can you explain the last equality above?

These three matrices are linearly independent, so a basis of ℛ\kern -1.95872pt \left (T\right ) is \left \{\left [\array{ 1&1\cr 1&0 } \right ],\left [\array{ 0&1\cr 1&0 } \right ],\left [\array{ 0&0\cr 0&1 } \right ]\right \}. Thus, T is not surjective, since the range has dimension 3 which is shy of \mathop{ dim}\nolimits \left ({M}_{2,2}\right ) = 4. (Notice that the range is actually the subspace of symmetric 2 × 2 matrices in {M}_{2,2}.)

C29 Contributed by Chris Black Statement [1549]
If we transform the basis of {P}_{2}, then Theorem SSRLT guarantees we will have a spanning set of ℛ\kern -1.95872pt \left (T\right ). A basis of {P}_{2} is \left \{1,x,{x}^{2}\right \}. If we transform the elements of this set, we get the set \left \{{x}^{2},{x}^{3},{x}^{4}\right \} which is a spanning set for ℛ\kern -1.95872pt \left (T\right ). These three vectors are linearly independent, so \left \{{x}^{2},{x}^{3},{x}^{4}\right \} is a basis of ℛ\kern -1.95872pt \left (T\right ).

C30 Contributed by Chris Black Statement [1549]
If we transform the basis of {P}_{4}, then Theorem SSRLT guarantees we will have a spanning set of ℛ\kern -1.95872pt \left (T\right ). A basis of {P}_{4} is \left \{1,x,{x}^{2},{x}^{3},{x}^{4}\right \}. If we transform the elements of this set, we get the set \left \{0, 1, 2x, 3{x}^{2}, 4{x}^{3}\right \} which is a spanning set for ℛ\kern -1.95872pt \left (T\right ). Reducing this to a linearly independent set, we find that \{1, 2x, 3{x}^{2}, 4{x}^{3}\} is a basis of ℛ\kern -1.95872pt \left (T\right ). Since ℛ\kern -1.95872pt \left (T\right ) and {P}_{3} both have dimension 4, T is surjective.

C40 Contributed by Robert Beezer Statement [1549]
We wish to find an output vector v that has no associated input. This is the same as requiring that there is no solution to the equality

 v = T\left (\left [\array{ a\cr b \cr c } \right ]\right ) = \left [\array{ 2a + 3b − c \cr 2b − 2c \cr a − b + 2c } \right ] = a\left [\array{ 2\cr 0 \cr 1 } \right ]+b\left [\array{ 3\cr 2 \cr −1 } \right ]+c\left [\array{ −1\cr −2 \cr 2 } \right ]

In other words, we would like to find an element of {ℂ}^{3} not in the set

 Y = \left \langle \left \{\left [\array{ 2\cr 0 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 3\cr 2 \cr −1 } \right ],\kern 1.95872pt \left [\array{ −1\cr −2 \cr 2 } \right ]\right \}\right \rangle

If we make these vectors the rows of a matrix, and row-reduce, Theorem BRS provides an alternate description of Y ,

 Y = \left \langle \left \{\left [\array{ 2\cr 0 \cr 1 } \right ],\kern 1.95872pt \left [\array{ 0\cr 4 \cr −5 } \right ]\right \}\right \rangle

If we add these vectors together, and then change the third component of the result, we will create a vector that lies outside of Y , say v = \left [\array{ 2\cr 4 \cr 9 } \right ].

T15 Contributed by Robert Beezer Statement [1550]
This question asks us to establish that one set (ℛ\kern -1.95872pt \left (S ∘ T\right )) is a subset of another (ℛ\kern -1.95872pt \left (S\right )). Choose an element in the “smaller” set, say w ∈ℛ\kern -1.95872pt \left (S ∘ T\right ). Then we know that there is a vector u ∈ U such that

 w = \left (S ∘ T\right )\left (u\right ) = S\left (T\left (u\right )\right )

Now define v = T\left (u\right ), so that then

 S\left (v\right ) = S\left (T\left (u\right )\right ) = w

This statement is sufficient to show that w ∈ℛ\kern -1.95872pt \left (S\right ), so w is an element of the “larger” set, and ℛ\kern -1.95872pt \left (S ∘ T\right ) ⊆ℛ\kern -1.95872pt \left (S\right ).

T20 Contributed by Andy Zimmer Statement [1551]
This is an equality of sets, so we want to establish two subset conditions (Definition SE).

First, show C\kern -1.95872pt \left (A\right ) ⊆ℛ\kern -1.95872pt \left (T\right ). Choose y ∈C\kern -1.95872pt \left (A\right ). Then by Definition CSM and Definition MVP there is a vector x ∈ {ℂ}^{n} such that Ax = y. Then

\eqalignno{ T\left (x\right ) & = Ax & &\text{Definition of $T$} & & & & \cr & = y & & & & }

This statement qualifies y as a member of ℛ\kern -1.95872pt \left (T\right ) (Definition RLT), so C\kern -1.95872pt \left (A\right ) ⊆ℛ\kern -1.95872pt \left (T\right ).

Now, show ℛ\kern -1.95872pt \left (T\right ) ⊆C\kern -1.95872pt \left (A\right ). Choose y ∈ℛ\kern -1.95872pt \left (T\right ). Then by Definition RLT, there is a vector x in {ℂ}^{n} such that T\left (x\right ) = y. Then

\eqalignno{ Ax & = T\left (x\right ) & &\text{Definition of $T$} & & & & \cr & = y & & & & }

So by Definition CSM and Definition MVP, y qualifies for membership in C\kern -1.95872pt \left (A\right ) and so ℛ\kern -1.95872pt \left (T\right ) ⊆C\kern -1.95872pt \left (A\right ).