A First Course in Linear Algebra: (Beta Version)

In the vector space of polynomials with degree 4 or less, \(P_4\) (Example VSP) consider the set \(S\) below.

Is this set of vectors linearly independent or dependent? Consider that

This is a nontrivial relation of linear dependence (Definition RLD) on the set \(S\) and so convinces us that \(S\) is linearly dependent (Definition LI).

Now, I hear you say, “Where did those scalars come from?” Do not worry about that right now, just be sure you understand why the above explanation is sufficient to prove that \(S\) is linearly dependent. The remainder of the example will demonstrate how we might find these scalars if they had not been provided so readily.

Let us look at another set of vectors (polynomials) from \(P_4\text{.}\) Let

Suppose we have a relation of linear dependence on this set,

Using our definitions of vector addition and scalar multiplication in \(P_4\) (Example VSP), we arrive at,

Equating coefficients, we arrive at the homogeneous system of equations,

We form the coefficient matrix of this homogeneous system of equations and row-reduce to find

We expected the system to be consistent (Theorem HSC) and so can compute \(n-r=4-4=0\) and Theorem CSRN tells us that the solution is unique. Since this is a homogeneous system, this unique solution is the trivial solution (Definition TSHSE), \(\alpha_1=0\text{,}\) \(\alpha_2=0\text{,}\) \(\alpha_3=0\text{,}\) \(\alpha_4=0\text{.}\) So by Definition LI the set \(T\) is linearly independent.

A few observations. If we had discovered infinitely many solutions, then we could have used one of the nontrivial solutions to provide a linear combination in the manner we used to show that \(S\) was linearly dependent. It is important to realize that it is not interesting that we can create a relation of linear dependence with zero scalars — we can always do that — but for \(T\text{,}\) this is the only way to create a relation of linear dependence. It was no accident that we arrived at a homogeneous system of equations in this example, it is related to our use of the zero vector in defining a relation of linear dependence. It is easy to present a convincing statement that a set is linearly dependent (just exhibit a nontrivial relation of linear dependence) but a convincing statement of linear independence requires demonstrating that there is no relation of linear dependence other than the trivial one. Notice how we relied on theorems from Chapter SLE to provide this demonstration. Whew! There is a lot going on in this example. Spend some time with it, we will be waiting patiently right here when you get back.

Consider the two sets of vectors \(R\) and \(S\) from the vector space of all \(3\times 2\) matrices, \(M_{32}\) (Example VSM).

One set is linearly independent, the other is not. Which is which? Let us examine \(R\) first. Build a generic relation of linear dependence (Definition RLD),

Massaging the left-hand side with our definitions of vector addition and scalar multiplication in \(M_{32}\) (Example VSM) we obtain,

Using our definition of matrix equality (Definition ME) to equate entries we get the homogeneous system of six equations in four variables,

Form the coefficient matrix of this homogeneous system and row-reduce to obtain

Analyzing this matrix we are led to conclude that \(\alpha_1=0\text{,}\) \(\alpha_2=0\text{,}\) \(\alpha_3=0\text{,}\) \(\alpha_4=0\text{.}\) This means there is only a trivial relation of linear dependence on the vectors of \(R\) and so we call \(R\) a linearly independent set (Definition LI).

So it must be that \(S\) is linearly dependent. Let us see if we can find a nontrivial relation of linear dependence on \(S\text{.}\) We will begin as with \(R\text{,}\) by constructing a relation of linear dependence (Definition RLD) with unknown scalars,

Massaging the left-hand side with our definitions of vector addition and scalar multiplication in \(M_{32}\) (Example VSM) we obtain,

Using our definition of matrix equality (Definition ME) to equate entries we get the homogeneous system of six equations in four variables,

Form the coefficient matrix of this homogeneous system and row-reduce to obtain

Analyzing this we see that the system is consistent (we expected this since the system is homogeneous, Theorem HSC) and has \(n-r=4-2=2\) free variables, namely \(\alpha_2\) and \(\alpha_4\text{.}\) This means there are infinitely many solutions, and in particular, we can find a nontrivial solution, so long as we do not pick all of our free variables to be zero. The mere presence of a nontrivial solution for these scalars is enough to conclude that \(S\) is a linearly dependent set (Definition LI). But let us go ahead and explicitly construct a nontrivial relation of linear dependence.

Choose \(\alpha_2=1\) and \(\alpha_4=-1\text{.}\) There is nothing special about this choice, there are infinitely many possibilities, some “easier” than this one, just avoid picking both variables to be zero. (Why not?) Then we find the dependent variables to have values \(\alpha_1=-2\) and \(\alpha_3=3\text{.}\) So the relation of linear dependence,

is an iron-clad demonstration that \(S\) is linearly dependent. Can you construct another such demonstration?

Is the set \(R=\set{(1,\,0),\,(6,\,3)}\) linearly independent in the crazy vector space \(C\) (Example CVS)?

We begin with an arbitrary relation of linear dependence on \(R\)

and then massage it to a point where we can apply the definition of equality in \(C\text{.}\) Recall the definitions of vector addition and scalar multiplication in \(C\) are not what you would expect.

Equality in \(C\) (Example CVS) then yields the two equations,

which becomes the homogeneous system

Since the coefficient matrix of this system is nonsingular (check this!) the system has only the trivial solution \(a_1=a_2=0\text{.}\) By Definition LI the set \(R\) is linearly independent. Notice that even though the zero vector of \(C\) is not what we might have first suspected, a question about linear independence still concludes with a question about a homogeneous system of equations. Hmmm.

In Example SP4 we showed that

is a subspace of \(P_4\text{,}\) the vector space of polynomials with degree at most \(4\) (Example VSP). In this example, we will show that the set

is a spanning set for \(W\text{.}\) To do this, we require that \(W=\spn{S}\text{.}\) This is an equality of sets. We can check that every polynomial in \(S\) has \(x=2\) as a root and therefore \(S\subseteq W\text{.}\) Since \(W\) is closed under addition and scalar multiplication, \(\spn{S}\subseteq W\) also.

So it remains to show that \(W\subseteq \spn{S}\) (Definition SE). To do this, begin by choosing an arbitrary polynomial in \(W\text{,}\) say \(r(x)=ax^4+bx^3+cx^2+dx+e\in W\text{.}\) This polynomial is not as arbitrary as it would appear, since we also know it must have \(x=2\) as a root. This translates to

as a condition on \(r\text{.}\)

We wish to show that \(r\) is a polynomial in \(\spn{S}\text{,}\) that is, we want to show that \(r\) can be written as a linear combination of the vectors (polynomials) in \(S\text{.}\) So let us try.

Equating coefficients (vector equality in \(P_4\)) gives the system of five equations in four variables,

Any solution to this system of equations will provide the linear combination we need to determine if \(r\in\spn{S}\text{,}\) but we need to be convinced there is a solution for any values of \(a,\,b,\,c,\,d,\,e\) that qualify \(r\) to be a member of \(W\text{.}\) So the question is: is this system of equations consistent? We will form the augmented matrix, and row-reduce. (We probably need to do this by hand, since the matrix is symbolic — reversing the order of the first four rows is the best way to start). We obtain a matrix in reduced row-echelon form

For your results to match our first matrix, you may find it necessary to multiply the final row of your row-reduced matrix by the appropriate scalar, and/or add multiples of this row to some of the other rows. To obtain the second version of the matrix, the last entry of the last column has been simplified to zero according to the one condition we were able to impose on an arbitrary polynomial from \(W\text{.}\) Since the last column is not a pivot column, Theorem RCLS tells us this system is consistent. Therefore, any polynomial from \(W\) can be written as a linear combination of the polynomials in \(S\text{,}\) so \(W\subseteq\spn{S}\text{.}\) Therefore, \(W=\spn{S}\) and \(S\) is a spanning set for \(W\) by Definition SSVS.

Notice that an alternative to row-reducing the augmented matrix by hand would be to appeal to Theorem FS by expressing the column space of the coefficient matrix as a null space, and then verifying that the condition on \(r\) guarantees that \(r\) is in the column space, thus implying that the system is always consistent. Give it a try, we will wait. This has been a complicated example, but worth studying carefully.

In the space of all \(2\times 2\) matrices, \(M_{22}\) consider the subspace

and find a spanning set for \(Z\text{.}\)

We need to construct a limited number of matrices in \(Z\) so that every matrix in \(Z\) can be expressed as a linear combination of this limited number of matrices. Suppose that \(B=\begin{bmatrix}a&b\\c&d\end{bmatrix}\) is a matrix in \(Z\text{.}\) Then we can form a column vector with the entries of \(B\) and write

Row-reducing this matrix and applying Theorem REMES we obtain the equivalent statement,

We can then express the subspace \(Z\) in the following equal forms,

So the set

spans \(Z\) by Definition SSVS.

In Example LIC we determined that the set \(R=\set{(1,\,0),\,(6,\,3)}\) is linearly independent in the crazy vector space \(C\) (Example CVS). We now show that \(R\) is a spanning set for \(C\text{.}\)

Given an arbitrary vector \((x,\,y)\in C\) we desire to show that it can be written as a linear combination of the elements of \(R\text{.}\) In other words, are there scalars \(a_1\) and \(a_2\) so that

We will act as if this equation is true and try to determine just what \(a_1\) and \(a_2\) would be (as functions of \(x\) and \(y\)). Recall that our vector space operations are unconventional and are defined in Example CVS.

Equality in \(C\) then yields the two equations,

which becomes the linear system with a matrix representation

The coefficient matrix of this system is nonsingular, hence invertible (Theorem NI), and we can employ its inverse to find a solution (Theorem TTMI, Theorem SNCM),

We could chase through the above implications backwards and take the existence of these solutions as sufficient evidence for \(R\) being a spanning set for \(C\text{.}\) Instead, let us view the above as simply scratchwork and now get serious with a simple direct proof that \(R\) is a spanning set. Ready? Suppose \((x,\,y)\) is any vector from \(C\text{,}\) then compute the following linear combination using the definitions of the operations in \(C\text{,}\)

This final sequence of computations in \(C\) is sufficient to demonstrate that any element of \(C\) can be written (or expressed) as a linear combination of the two vectors in \(R\text{,}\) so \(C\subseteq\spn{R}\text{.}\) Since the reverse inclusion \(\spn{R}\subseteq C\) is trivially true, \(C=\spn{R}\) and we say \(R\) spans \(C\) (Definition SSVS). Notice that this demonstration is no more or less valid if we hide from the reader our scratchwork that suggested \(a_1=4x-7y-3\) and \(a_2=-x+2y+1\text{.}\)

Consider the set

from the vector space \(\complex{3}\text{.}\) Let \(A\) be the matrix whose columns are the set \(S\text{,}\) and verify that \(A\) is nonsingular. By Theorem NMLIC the elements of \(S\) form a linearly independent set. Suppose that \(\vect{b}\in\complex{3}\text{.}\) Then \(\linearsystem{A}{\vect{b}}\) has a (unique) solution (Theorem NMUS) and hence is consistent. By Theorem SLSLC, \(\vect{b}\in\spn{S}\text{.}\) Since \(\vect{b}\) is arbitrary, this is enough to show that \(\spn{S}=\complex{3}\text{,}\) and therefore \(S\) is a spanning set for \(\complex{3}\) (Definition SSVS). (This set comes from the columns of the coefficient matrix of Archetype B.)

Now examine the situation for a particular choice of \(\vect{b}\text{,}\) say \(\vect{b}=\colvector{-33\\24\\5}\text{.}\) Because \(S\) is a spanning set for \(\complex{3}\text{,}\) we know we can write \(\vect{b}\) as a linear combination of the vectors in \(S\text{,}\)

The nonsingularity of the matrix \(A\) tells that the scalars in this linear combination are unique. More precisely, it is the linear independence of \(S\) that provides the uniqueness. We will refer to the scalars \(a_1=-3\text{,}\) \(a_2=5\text{,}\) \(a_3=2\) as a “representation of \(\vect{b}\) relative to \(S\text{.}\)” In other words, once we settle on \(S\) as a linearly independent set that spans \(\complex{3}\text{,}\) the vector \(\vect{b}\) is recoverable just by knowing the scalars \(a_1=-3\text{,}\) \(a_2=5\text{,}\) \(a_3=2\) (use these scalars in a linear combination of the vectors in \(S\)). This is all an illustration of the following important theorem, which we prove in the setting of a general vector space.