SubsectionLISVLinearly Independent Sets of Vectors¶ permalink
Theorem SLSLC tells us that a solution to a homogeneous system of equations is a linear combination of the columns of the coefficient matrix that equals the zero vector. We used just this situation to our advantage (twice!) in Example SCAD where we reduced the set of vectors used in a span construction from four down to two, by declaring certain vectors as surplus. The next two definitions will allow us to formalize this situation.
DefinitionRLDCVRelation of Linear Dependence for Column Vectors
Given a set of vectors \(S=\set{\vectorlist{u}{n}}\text{,}\) a true statement of the form
\begin{equation*}
\lincombo{\alpha}{u}{n}=\zerovector
\end{equation*}
is a relation of linear dependence on \(S\text{.}\) If this statement is formed in a trivial fashion, i.e. \(\alpha_i=0\text{,}\) \(1\leq i\leq n\text{,}\) then we say it is the trivial relation of linear dependence on \(S\text{.}\)
DefinitionLICVLinear Independence of Column Vectors
The set of vectors \(S=\set{\vectorlist{u}{n}}\) is linearly dependent if there is a relation of linear dependence on \(S\) that is not trivial. In the case where the only relation of linear dependence on \(S\) is the trivial one, then \(S\) is a linearly independent set of vectors.
Notice that a relation of linear dependence is an equation. Though most of it is a linear combination, it is not a linear combination (that would be a vector). Linear independence is a property of a set of vectors. It is easy to take a set of vectors, and an equal number of scalars, all zero, and form a linear combination that equals the zero vector. When the easy way is the only way, then we say the set is linearly independent. Here are a couple of examples.
Consider the set of \(n=4\) vectors from \(\complex{5}\)
\begin{equation*}
S=\set{
\colvector{2\\1\\3\\1\\2},\,
\colvector{1\\2\\1\\5\\2},\,
\colvector{2\\1\\3\\6\\1},\,
\colvector{6\\7\\1\\0\\1}
}\text{.}
\end{equation*}
To determine linear independence we first form a relation of linear dependence,
\begin{equation*}
\alpha_1\colvector{2\\1\\3\\1\\2}+
\alpha_2\colvector{1\\2\\1\\5\\2}+
\alpha_3\colvector{2\\1\\3\\6\\1}+
\alpha_4\colvector{6\\7\\1\\0\\1}
=\zerovector\text{.}
\end{equation*}
We know that \(\alpha_1=\alpha_2=\alpha_3=\alpha_4=0\) is a solution to this equation, but that is of no interest whatsoever. That is always the case, no matter what four vectors we might have chosen. We are curious to know if there are other, nontrivial, solutions. Theorem SLSLC tells us that we can find such solutions as solutions to the homogeneous system \(\homosystem{A}\) where the coefficient matrix has these four vectors as columns, which we then rowreduce
\begin{equation*}
A=
\begin{bmatrix}
2&1&2&6\\
1&2&1&7\\
3&1&3&1\\
1&5&6&0\\
2&2&1&1
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1}&0&0&2\\
0&\leading{1}&0&4\\
0&0&\leading{1}&3\\
0&0&0&0\\
0&0&0&0
\end{bmatrix}\text{.}
\end{equation*}
We could solve this homogeneous system completely, but for this example all we need is one nontrivial solution. Setting the lone free variable to any nonzero value, such as \(x_4=1\text{,}\) yields the nontrivial solution \(\vect{x}\) which we use to complete our application of Theorem SLSLC
\begin{align*}
\vect{x}&=\colvector{2\\4\\3\\1}
&
2\colvector{2\\1\\3\\1\\2}+
(4)\colvector{1\\2\\1\\5\\2}+
3\colvector{2\\1\\3\\6\\1}+
1\colvector{6\\7\\1\\0\\1}
&=\zerovector\text{.}
\end{align*}
This is a relation of linear dependence on \(S\) that is not trivial, so we conclude that \(S\) is linearly dependent.
Consider the set of \(n=4\) vectors from \(\complex{5}\text{,}\)
\begin{equation*}
T=\set{
\colvector{2\\1\\3\\1\\2},\,
\colvector{1\\2\\1\\5\\2},\,
\colvector{2\\1\\3\\6\\1},\,
\colvector{6\\7\\1\\1\\1}
}\text{.}
\end{equation*}
To determine linear independence we first form a relation of linear dependence,
\begin{equation*}
\alpha_1\colvector{2\\1\\3\\1\\2}+
\alpha_2\colvector{1\\2\\1\\5\\2}+
\alpha_3\colvector{2\\1\\3\\6\\1}+
\alpha_4\colvector{6\\7\\1\\1\\1}
=\zerovector\text{.}
\end{equation*}
We know that \(\alpha_1=\alpha_2=\alpha_3=\alpha_4=0\) is a solution to this equation, but that is of no interest whatsoever. That is always the case, no matter what four vectors we might have chosen. We are curious to know if there are other, nontrivial, solutions. Theorem SLSLC tells us that we can find such solutions as solution to the homogeneous system \(\linearsystem{B}{\zerovector}\) where the coefficient matrix has these four vectors as columns. Rowreducing this coefficient matrix yields
\begin{align*}
B=
\begin{bmatrix}
2&1&2&6\\
1&2&1&7\\
3&1&3&1\\
1&5&6&1\\
2&2&1&1
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1}&0&0&0\\
0&\leading{1}&0&0\\
0&0&\leading{1}&0\\
0&0&0&\leading{1}\\
0&0&0&0
\end{bmatrix}\text{.}
\end{align*}
From the form of this matrix, we see that there are no free variables, so the solution is unique, and because the system is homogeneous, this unique solution is the trivial solution. So we now know that there is but one way to combine the four vectors of \(T\) into a relation of linear dependence, and that one way is the easy and obvious way. In this situation we say that the set, \(T\text{,}\) is linearly independent.
Example LDS and Example LIS relied on solving a homogeneous system of equations to determine linear independence. We can codify this process in a timesaving theorem.
TheoremLIVHSLinearly Independent Vectors and Homogeneous Systems
Suppose that \(S=\set{\vectorlist{v}{n}}\subseteq\complex{m}\) is a set of vectors and \(A\) is the \(m\times n\) matrix whose columns are the vectors in \(S\text{.}\) Then \(S\) is a linearly independent set if and only if the homogeneous system \(\homosystem{A}\) has a unique solution.
Proof
(⇐)
Suppose that \(\homosystem{A}\) has a unique solution. Since it is a homogeneous system, this solution must be the trivial solution \(\vect{x}=\zerovector\text{.}\) By Theorem SLSLC, this means that the only relation of linear dependence on \(S\) is the trivial one. So \(S\) is linearly independent.
(⇒)
We will prove the contrapositive. Suppose that \(\linearsystem{A}{\zerovector}\) does not have a unique solution. Since it is a homogeneous system, it is consistent (Theorem HSC), and so must have infinitely many solutions (Theorem PSSLS). One of these infinitely many solutions must be nontrivial (in fact, almost all of them are), so choose one. By Theorem SLSLC this nontrivial solution will give a nontrivial relation of linear dependence on \(S\text{,}\) so we can conclude that \(S\) is a linearly dependent set.
Since Theorem LIVHS is an equivalence, we can use it to determine the linear independence or dependence of any set of column vectors, just by creating a matrix and analyzing the rowreduced form. Let us illustrate this with two more examples.
Is the set of vectors
\begin{equation*}
S=\set{
\colvector{2\\1\\3\\4\\2},\,
\colvector{6\\2\\1\\3\\4},\,
\colvector{4\\3\\4\\5\\1}
}
\end{equation*}
linearly independent or linearly dependent?
Theorem LIVHS suggests we study the matrix, \(A\text{,}\) whose columns are the vectors in \(S\text{.}\) Specifically, we are interested in the size of the solution set for the homogeneous system \(\homosystem{A}\text{,}\) so we rowreduce \(A\text{.}\)
\begin{equation*}
A=
\begin{bmatrix}
2 & 6 & 4\\
1 & 2 & 3\\
3 & 1 & 4\\
4 & 3 & 5\\
2 & 4 & 1
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 0\\
0 & \leading{1} & 0\\
0 & 0 & \leading{1}\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
Now, \(r=3\text{,}\) so there are \(nr=33=0\) free variables and we see that \(\homosystem{A}\) has a unique solution (Theorem HSC, Theorem FVCS). By Theorem LIVHS, the set \(S\) is linearly independent.
Is the set of vectors
\begin{equation*}
S=\set{
\colvector{2\\1\\3\\4\\2},\,
\colvector{6\\2\\1\\3\\4},\,
\colvector{4\\3\\4\\1\\2}
}
\end{equation*}
linearly independent or linearly dependent?
Theorem LIVHS suggests we study the matrix, \(A\text{,}\) whose columns are the vectors in \(S\text{.}\) Specifically, we are interested in the size of the solution set for the homogeneous system \(\homosystem{A}\text{,}\) so we rowreduce \(A\text{.}\)
\begin{equation*}
A=
\begin{bmatrix}
2 & 6 & 4\\
1 & 2 & 3\\
3 & 1 & 4\\
4 & 3 & 1\\
2 & 4 & 2
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 1\\
0 & \leading{1} & 1\\
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}
\end{equation*}
Now, \(r=2\text{,}\) so there are \(nr=32=1\) free variables and we see that \(\homosystem{A}\) has infinitely many solutions (Theorem HSC, Theorem FVCS). By Theorem LIVHS, the set \(S\) is linearly dependent.
As an equivalence, Theorem LIVHS gives us a straightforward way to determine if a set of vectors is linearly independent or dependent.
Review Example LIHS and Example LDHS. They are very similar, differing only in the last two slots of the third vector. This resulted in slightly different matrices when rowreduced, and slightly different values of \(r\text{,}\) the number of nonzero rows. Notice, too, that we are less interested in the actual solution set, and more interested in its form or size. These observations allow us to make a slight improvement in Theorem LIVHS.
TheoremLIVRNLinearly Independent Vectors, \(r\) and \(n\)
Suppose that \(S=\set{\vectorlist{v}{n}}\subseteq\complex{m}\) is a set of vectors and \(A\) is the \(m\times n\) matrix whose columns are the vectors in \(S\text{.}\) Let \(B\) be a matrix in reduced rowechelon form that is rowequivalent to \(A\) and let \(r\) denote the number of pivot columns in \(B\text{.}\) Then \(S\) is linearly independent if and only if \(n=r\text{.}\)
Proof
Theorem LIVHS says the linear independence of \(S\) is equivalent to the homogeneous linear system \(\homosystem{A}\) having a unique solution. Since \(\homosystem{A}\) is consistent (Theorem HSC) we can apply Theorem CSRN to see that the solution is unique exactly when \(n=r\text{.}\)
So now here is an example of the most straightforward way to determine if a set of column vectors is linearly independent or linearly dependent. While this method can be quick and easy, do not forget the logical progression from the definition of linear independence through homogeneous system of equations which makes it possible.
Is the set of vectors
\begin{equation*}
S=\set{
\colvector{2\\1\\3\\1\\0\\3},\,
\colvector{9\\6\\2\\3\\2\\1},\,
\colvector{1\\1\\1\\0\\0\\1},\,
\colvector{3\\1\\4\\2\\1\\2},\,
\colvector{6\\2\\1\\4\\3\\2}
}
\end{equation*}
linearly independent or linearly dependent?
Theorem LIVRN suggests we place these vectors into a matrix as columns and analyze the rowreduced version of the matrix
\begin{equation*}
\begin{bmatrix}
2 & 9 & 1 & 3 & 6\\
1 & 6 & 1 & 1 & 2\\
3 & 2 & 1 & 4 & 1\\
1 & 3 & 0 & 2 & 4\\
0 & 2 & 0 & 1 & 3\\
3 & 1 & 1 & 2 & 2
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 0 & 0 & 1\\
0 & \leading{1} & 0 & 0 & 1\\
0 & 0 & \leading{1} & 0 & 2\\
0 & 0 & 0 & \leading{1} & 1\\
0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
Now we need only compute that \(r=4\lt 5=n\) to recognize, via Theorem LIVRN that \(S\) is a linearly dependent set. Boom!
Consider the set of \(n=9\) vectors from \(\complex{4}\text{,}\)
\begin{equation*}
R=\set{
\colvector{1\\3\\1\\2},\,
\colvector{7\\1\\3\\6},\,
\colvector{1\\2\\1\\2},\,
\colvector{0\\4\\2\\9},\,
\colvector{5\\2\\4\\3},\,
\colvector{2\\1\\6\\4},\,
\colvector{3\\0\\3\\1},\,
\colvector{1\\1\\5\\3},\,
\colvector{6\\1\\1\\1}
}\text{.}
\end{equation*}
To employ Theorem LIVHS, we form a \(4\times 9\) matrix, \(C\text{,}\) whose columns are the vectors in \(R\)
\begin{equation*}
C=
\begin{bmatrix}
1&7&1&0&5&2&3&1&6\\
3&1&2&4&2&1&0&1&1\\
1&3&1&2&4&6&3&5&1\\
2&6&2&9&3&4&1&3&1
\end{bmatrix}\text{.}
\end{equation*}
To determine if the homogeneous system \(\homosystem{C}\) has a unique solution or not, we would normally rowreduce this matrix. But in this particular example, we can do better. Theorem HMVEI tells us that since the system is homogeneous with \(n=9\) variables in \(m=4\) equations, and \(n\gt m\text{,}\) there must be infinitely many solutions. Since there is not a unique solution, Theorem LIVHS says the set is linearly dependent.
The situation in Example LLDS is slick enough to warrant formulating as a theorem.
TheoremMVSLDMore Vectors than Size implies Linear Dependence
Suppose that \(S=\set{\vectorlist{u}{n}}\subseteq\complex{m}\) and \(n\gt m\text{.}\) Then \(S\) is a linearly dependent set.
Proof
Form the \(m\times n\) matrix \(A\) whose columns are \(\vect{u}_i\text{,}\) \(1\leq i\leq n\text{.}\) Consider the homogeneous system \(\homosystem{A}\text{.}\) By Theorem HMVEI this system has infinitely many solutions. Since the system does not have a unique solution, Theorem LIVHS says the columns of \(A\) form a linearly dependent set, as desired.
SubsectionNSSLINull Spaces, Spans, Linear Independence¶ permalink
In Subsection SS.SSNS we proved Theorem SSNS which provided \(nr\) vectors that could be used with the span construction to build the entire null space of a matrix. As we have hinted in Example SCAD, and as we will see again going forward, linearly dependent sets carry redundant vectors with them when used in building a set as a span. Our aim now is to show that the vectors provided by Theorem SSNS form a linearly independent set, so in one sense they are as efficient as possible a way to describe the null space. Notice that the vectors \(\vect{z}_j\text{,}\) \(1\leq j\leq nr\) first appear in the vector form of solutions to arbitrary linear systems (Theorem VFSLS). The exact same vectors appear again in the span construction in the conclusion of Theorem SSNS. Since this second theorem specializes to homogeneous systems the only real difference is that the vector \(\vect{c}\) in Theorem VFSLS is the zero vector for a homogeneous system. Finally, Theorem BNS will now show that these same vectors are a linearly independent set. We will set the stage for the proof of this theorem with a moderately large example. Study the example carefully, as it will make it easier to understand the proof.
Suppose that we are interested in the null space of a \(3\times 7\) matrix, \(A\text{,}\) which rowreduces to
\begin{equation*}
B=
\begin{bmatrix}
\leading{1} & 0 & 2 & 4 & 0 & 3 & 9\\
0 & \leading{1} & 5 & 6 & 0 & 7 & 1\\
0 & 0 & 0 & 0 & \leading{1} & 8 & 5
\end{bmatrix}\text{.}
\end{equation*}
The set \(F=\set{3,\,4,\,6,\,7}\) is the set of indices for our four free variables that would be used in a description of the solution set for the homogeneous system \(\homosystem{A}\text{.}\) Applying Theorem SSNS we can begin to construct a set of four vectors whose span is the null space of \(A\text{,}\) a set of vectors we will reference as \(T\text{.}\)
\begin{equation*}
\nsp{A}=\spn{T}=\spn{\set{\vect{z}_1,\,\vect{z}_2,\,\vect{z}_3,\,\vect{z}_4}}=\spn{\set{
\colvector{ \\ \\1\\0\\ \\0\\0},\,
\colvector{ \\ \\0\\1\\ \\0\\0},\,
\colvector{ \\ \\0\\0\\ \\1\\0},\,
\colvector{ \\ \\0\\0\\ \\0\\1}
}}
\end{equation*}
So far, we have constructed as much of these individual vectors as we can, based just on the knowledge of the contents of the set \(F\text{.}\) This has allowed us to determine the entries in slots 3, 4, 6 and 7, while we have left slots 1, 2 and 5 blank. Without doing any more, let us ask if \(T\) is linearly independent? Begin with a relation of linear dependence on \(T\text{,}\) and see what we can learn about the scalars,
\begin{align*}
\zerovector&=\alpha_1\vect{z}_1+\alpha_2\vect{z}_2+\alpha_3\vect{z}_3+\alpha_4\vect{z}_4\\
\colvector{0\\0\\0\\0\\0\\0\\0}
&=
\alpha_1\colvector{ \\ \\1\\0\\ \\0\\0}+
\alpha_2\colvector{ \\ \\0\\1\\ \\0\\0}+
\alpha_3\colvector{ \\ \\0\\0\\ \\1\\0}+
\alpha_4\colvector{ \\ \\0\\0\\ \\0\\1}\\
&=
\colvector{ \\ \\\alpha_1\\0\\ \\0\\0}+
\colvector{ \\ \\0\\\alpha_2\\ \\0\\0}+
\colvector{ \\ \\0\\0\\ \\\alpha_3\\0}+
\colvector{ \\ \\0\\0\\ \\0\\\alpha_4}
=
\colvector{ \\ \\\alpha_1\\\alpha_2\\ \\\alpha_3\\\alpha_4}
\end{align*}
Applying Definition CVE to the two ends of this chain of equalities, we see that \(\alpha_1=\alpha_2=\alpha_3=\alpha_4=0\text{.}\) So the only relation of linear dependence on the set \(T\) is a trivial one. By Definition LICV the set \(T\) is linearly independent. The important feature of this example is how the “pattern of zeros and ones” in the four vectors led to the conclusion of linear independence.
The proof of Theorem BNS is really quite straightforward, and relies on the “pattern of zeros and ones” that arise in the vectors \(\vect{z}_i\text{,}\) \(1\leq i\leq nr\) in the entries that arise with the locations of the nonpivot columns. Play along with Example LINSB as you study the proof. Also, take a look at Example VFSAD, Example VFSAI and Example VFSAL, especially at the conclusion of Step 2 (temporarily ignore the construction of the constant vector, \(\vect{c}\)). This proof is also a good first example of how to prove a conclusion that states a set is linearly independent.
TheoremBNSBasis for Null Spaces
Suppose that \(A\) is an \(m\times n\) matrix, and \(B\) is a rowequivalent matrix in reduced rowechelon form with \(r\) pivot columns. Let \(D=\{d_1,\,d_2,\,d_3,\,\ldots,\,d_r\}\) and \(F=\{f_1,\,f_2,\,f_3,\,\ldots,\,f_{nr}\}\) be the sets of column indices where \(B\) does and does not (respectively) have pivot columns. Construct the \(nr\) vectors \(\vect{z}_j\text{,}\) \(1\leq j\leq nr\) of size \(n\) as
\begin{equation*}
\vectorentry{\vect{z}_j}{i}=
\begin{cases}
1&\text{if }i\in F, i=f_j\\
0&\text{if }i\in F, i\neq f_j\\
\matrixentry{B}{k,f_j}&\text{if }i\in D, i=d_k
\end{cases}\text{.}
\end{equation*}
Define the set \(S=\set{\vectorlist{z}{nr}}\text{.}\) Then
 \(\nsp{A}=\spn{S}\text{.}\)
 \(S\) is a linearly independent set.
Proof
Notice first that the vectors \(\vect{z}_j\text{,}\) \(1\leq j\leq nr\) are exactly the same as the \(nr\) vectors defined in Theorem SSNS. Also, the hypotheses of Theorem SSNS are the same as the hypotheses of the theorem we are currently proving. So it is then simply the conclusion of Theorem SSNS that tells us that \(\nsp{A}=\spn{S}\text{.}\) That was the easy half, but the second part is not much harder. What is new here is the claim that \(S\) is a linearly independent set.
To prove the linear independence of a set, we need to start with a relation of linear dependence and somehow conclude that the scalars involved must all be zero, i.e. that the relation of linear dependence only happens in the trivial fashion. So to establish the linear independence of \(S\text{,}\) we start with
\begin{equation*}
\lincombo{\alpha}{z}{nr}=\zerovector\text{.}
\end{equation*}
For each \(j\text{,}\) \(1\leq j\leq nr\text{,}\) consider the equality of the individual entries of the vectors on both sides of this equality in position \(f_j\)
\begin{align*}
0
&=\vectorentry{\zerovector}{f_j}\\
&=\vectorentry{\lincombo{\alpha}{z}{nr}}{f_j}&&
\knowl{./knowl/definitionCVE.html}{\text{Definition CVE}}\\
&=
\vectorentry{\alpha_1\vect{z}_1}{f_j}+
\vectorentry{\alpha_2\vect{z}_2}{f_j}+
\vectorentry{\alpha_3\vect{z}_3}{f_j}+
\cdots+
\vectorentry{\alpha_{nr}\vect{z}_{nr}}{f_j}&&
\knowl{./knowl/definitionCVA.html}{\text{Definition CVA}}\\
&=
\alpha_1\vectorentry{\vect{z}_1}{f_j}+
\alpha_2\vectorentry{\vect{z}_2}{f_j}+
\alpha_3\vectorentry{\vect{z}_3}{f_j}+
\cdots+\\
&\quad\quad
\alpha_{j1}\vectorentry{\vect{z}_{j1}}{f_j}+
\alpha_{j}\vectorentry{\vect{z}_j}{f_j}+
\alpha_{j+1}\vectorentry{\vect{z}_{j+1}}{f_j}+
\cdots+\\
&\quad\quad
\alpha_{nr}\vectorentry{\vect{z}_{nr}}{f_j}&&
\knowl{./knowl/definitionCVSM.html}{\text{Definition CVSM}}\\
&=\alpha_1(0)+
\alpha_2(0)+
\alpha_3(0)+
\cdots+\\
&\quad\quad
\alpha_{j1}(0)+
\alpha_{j}(1)+
\alpha_{j+1}(0)+
\cdots+\alpha_{nr}(0)&&
\text{Definition of }\vect{z}_j\\
&=\alpha_{j}\text{.}
\end{align*}
So for all \(j\text{,}\) \(1\leq j\leq nr\text{,}\) we have \(\alpha_j=0\text{,}\) which is the conclusion that tells us that the only relation of linear dependence on \(S=\set{\vectorlist{z}{nr}}\) is the trivial one. Hence, by Definition LICV the set is linearly independent, as desired.
In Example VFSAL we previewed Theorem SSNS by finding a set of two vectors such that their span was the null space for the matrix in Archetype L. Writing the matrix as \(L\text{,}\) we have
\begin{equation*}
\nsp{L}=
\spn{\set{\colvector{1\\2\\2\\1\\0},\,\colvector{2\\2\\1\\0\\1}}
}\text{.}
\end{equation*}
Solving the homogeneous system \(\homosystem{L}\) resulted in recognizing \(x_4\) and \(x_5\) as the free variables. So look in entries 4 and 5 of the two vectors above and notice the pattern of zeros and ones that provides the linear independence of the set.
Determine if the sets of vectors in Exercises C20–C25 are linearly independent or linearly dependent. When the set is linearly dependent, exhibit a nontrivial relation of linear dependence.
C20
\(\set{\colvector{1\\2\\1},\,\colvector{2\\1\\3},\,\colvector{1\\5\\0}}\)
SolutionWith three vectors from \(\complex{3}\text{,}\) we can form a square matrix by making these three vectors the columns of a matrix. We do so, and rowreduce to obtain
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0\\
0 & \leading{1} & 0\\
0 & 0 & \leading{1}
\end{bmatrix}
\end{equation*}
the \(3\times 3\) identity matrix. So by Theorem NME2 the original matrix is nonsingular and its columns are therefore a linearly independent set.
C21
\(\set{\colvector{1\\2\\4\\2},\,\colvector{3\\3\\1\\3},\,\colvector{7\\3\\6\\4}}\)
SolutionTheorem LIVRN says we can answer this question by putting theses vectors into a matrix as columns and rowreducing. Doing this we obtain
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0\\
0 & \leading{1} & 0\\
0 & 0 & \leading{1}\\
0 & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
With \(n=3\) (3 vectors, 3 columns) and \(r=3\) (3 pivot columns) we have \(n=r\) and the theorem says the vectors are linearly independent.
C22
\(\set{
\colvector{2\\1\\1},\,
\colvector{1\\0\\1},\,
\colvector{3\\3\\6},\,
\colvector{5\\4\\6},\,
\colvector{4\\4\\7}
}\)
SolutionFive vectors from \(\complex{3}\text{.}\) Theorem MVSLD says the set is linearly dependent. Boom.
C23
\(\set{
\colvector{1\\2\\2\\5\\3},\,
\colvector{3\\3\\1\\2\\4},\,
\colvector{2\\1\\2\\1\\1},\,
\colvector{1\\0\\1\\2\\2}
}\)
SolutionTheorem LIVRN suggests we analyze a matrix whose columns are the vectors of \(S\)
\begin{equation*}
A=\begin{bmatrix}
1 & 3 & 2 & 1\\
2 & 3 & 1 & 0\\
2 & 1 & 2 & 1\\
5 & 2 & 1 & 2\\
3 & 4 & 1 & 2
\end{bmatrix}\text{.}
\end{equation*}
Rowreducing the matrix \(A\) yields
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0 & 0\\
0 & \leading{1} & 0 & 0\\
0 & 0 & \leading{1} & 0\\
0 & 0 & 0 & \leading{1}\\
0 & 0 & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
We see that \(r=4=n\text{,}\) where \(r\) is the number of nonzero rows and \(n\) is the number of columns. By Theorem LIVRN, the set \(S\) is linearly independent.
C24
\(\set{
\colvector{1\\2\\1\\0\\1},\,
\colvector{3\\2\\1\\2\\2},\,
\colvector{4\\4\\2\\2\\3},\,
\colvector{1\\2\\1\\2\\0}
}\)
SolutionTheorem LIVRN suggests we analyze a matrix whose columns are the vectors from the set
\begin{equation*}
A=\begin{bmatrix}
1 & 3 & 4 & 1\\
2 & 2 & 4 & 2\\
1 & 1 & 2 & 1\\
0 & 2 & 2 & 2\\
1 & 2 & 3 & 0
\end{bmatrix}\text{.}
\end{equation*}
Rowreducing the matrix \(A\) yields
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 1 & 2\\
0 & \leading{1} & 1 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
We see that \(r=2\neq 4=n\text{,}\) where \(r\) is the number of nonzero rows and \(n\) is the number of columns. By Theorem LIVRN, the set \(S\) is linearly dependent.
C25
\(\set{
\colvector{2\\1\\3\\1\\2},\,
\colvector{4\\2\\1\\3\\2},\,
\colvector{10\\7\\0\\10\\4}
}\)
SolutionTheorem LIVRN suggests we analyze a matrix whose columns are the vectors from the set
\begin{equation*}
A=\begin{bmatrix}
2 & 4 & 10 \\
1 & 2 & 7 \\
3 & 1 & 0 \\
1 & 3 & 10 \\
2 & 2 & 4
\end{bmatrix}\text{.}
\end{equation*}
Rowreducing the matrix \(A\) yields
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 1 \\
0 & \leading{1} & 3 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
We see that \(r=2\neq 3=n\text{,}\) where \(r\) is the number of nonzero rows and \(n\) is the number of columns. By Theorem LIVRN, the set \(S\) is linearly dependent.
C30
For the matrix \(B\) below, find a set \(S\) that is linearly independent and spans the null space of \(B\text{,}\) that is, \(\nsp{B}=\spn{S}\text{.}\)
\begin{equation*}
B=
\begin{bmatrix}
3 & 1 & 2 & 7\\
1 & 2 & 1 & 4\\
1 & 1 & 2 & 1
\end{bmatrix}
\end{equation*}
SolutionThe requested set is described by Theorem BNS. It is easiest to find by using the procedure of Example VFSAL. Begin by rowreducing the matrix, viewing it as the coefficient matrix of a homogeneous system of equations. We obtain
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 1 & 2\\
0 & \leading{1} & 1 & 1\\
0 & 0 & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
Now build the vector form of the solutions to this homogeneous system (Theorem VFSLS). The free variables are \(x_3\) and \(x_4\text{,}\) due to the locations of the nonpivot columns
\begin{equation*}
\colvector{x_1\\x_2\\x_3\\x_4}=
x_3\colvector{1\\1\\1\\0}+
x_4\colvector{2\\1\\0\\1}\text{.}
\end{equation*}
The desired set \(S\) is simply the constant vectors in this expression, and these are the vectors \(\vect{z}_1\) and \(\vect{z}_2\) described by Theorem BNS.
\begin{equation*}
S=\set{
\colvector{1\\1\\1\\0},\,
\colvector{2\\1\\0\\1}
}
\end{equation*}
C31
For the matrix \(A\) below, find a linearly independent set \(S\) so that the null space of \(A\) is spanned by \(S\text{,}\) that is, \(\nsp{A}=\spn{S}\text{.}\)
\begin{equation*}
A=
\begin{bmatrix}
1 & 2 & 2 & 1 & 5 \\
1 & 2 & 1 & 1 & 5 \\
3 & 6 & 1 & 2 & 7 \\
2 & 4 & 0 & 1 & 2
\end{bmatrix}
\end{equation*}
SolutionTheorem BNS provides formulas for \(nr\) vectors that will meet the requirements of this question. These vectors are the same ones listed in Theorem VFSLS when we solve the homogeneous system \(\homosystem{A}\text{,}\) whose solution set is the null space (Definition NSM).
To apply Theorem BNS or Theorem VFSLS we first rowreduce the matrix, resulting in
\begin{equation*}
B=
\begin{bmatrix}
\leading{1} & 2 & 0 & 0 & 3 \\
0 & 0 & \leading{1} & 0 & 6 \\
0 & 0 & 0 & \leading{1} & 4 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
So we see that \(nr=53=2\) and \(F=\set{2,5}\text{,}\) so the vector form of a generic solution vector is
\begin{equation*}
\colvector{x_1\\x_2\\x_3\\x_4\\x_5}
=
x_2\colvector{2\\1\\ 0\\ 0\\0}
+
x_5\colvector{3\\0\\6 \\4 \\1}\text{.}
\end{equation*}
So we have
\begin{equation*}
\nsp{A}=\spn{\set{
\colvector{2\\1\\ 0\\ 0\\0},\,
\colvector{3\\0\\6 \\4 \\1}
}}\text{.}
\end{equation*}
C32
Find a set of column vectors, \(T\text{,}\) such that (1) the span of \(T\) is the null space of \(B\text{,}\) \(\spn{T}=\nsp{B}\) and (2) \(T\) is a linearly independent set.
\begin{align*}
B&=
\begin{bmatrix}
2 & 1 & 1 & 1 \\
4 & 3 & 1 & 7 \\
1 & 1 & 1 & 3
\end{bmatrix}
\end{align*}
SolutionThe conclusion of Theorem BNS gives us everything this question asks for. We need the reduced rowechelon form of the matrix so we can determine the number of vectors in \(T\text{,}\) and their entries.
\begin{align*}
\begin{bmatrix}
2 & 1 & 1 & 1 \\
4 & 3 & 1 & 7 \\
1 & 1 & 1 & 3
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1} & 0 & 2 & 2 \\
0 & \leading{1} & 3 & 5 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align*}
We can build the set \(T\) in immediately via Theorem BNS, but we will illustrate its construction in two steps. Since \(F=\set{3,\,4}\text{,}\) we will have two vectors and can distribute strategically placed ones, and many zeros. Then we distribute the negatives of the appropriate entries of the nonpivot columns of the reduced rowechelon matrix.
\begin{align*}
T&=\set{
\colvector{\\\\1\\0},\,
\colvector{\\\\0\\1}
}
&
T&=\set{
\colvector{2\\3\\1\\0},\,
\colvector{2\\5\\0\\1}
}
\end{align*}
C33
Find a set \(S\) so that \(S\) is linearly independent and \(\nsp{A}=\spn{S}\text{,}\) where \(\nsp{A}\) is the null space of the matrix \(A\) below.
\begin{align*}
A&=
\begin{bmatrix}
2 & 3 & 3 & 1 & 4 \\
1 & 1 & 1 & 1 & 3 \\
3 & 2 & 8 & 1 & 1
\end{bmatrix}
\end{align*}
SolutionA direct application of Theorem BNS will provide the desired set. We require the reduced rowechelon form of \(A\text{.}\)
\begin{align*}
\begin{bmatrix}
2 & 3 & 3 & 1 & 4 \\
1 & 1 & 1 & 1 & 3 \\
3 & 2 & 8 & 1 & 1
\end{bmatrix}
&\rref
\begin{bmatrix}
\leading{1} & 0 & 6 & 0 & 3 \\
0 & \leading{1} & 5 & 0 & 2 \\
0 & 0 & 0 & \leading{1} & 4\end{bmatrix}
\end{align*}
The nonpivot columns have indices \(F=\set{3,\,5}\text{.}\) We build the desired set in two steps, first placing the requisite zeros and ones in locations based on \(F\text{,}\) then placing the negatives of the entries of columns 3 and 5 in the proper locations. This is all specified in Theorem BNS.
\begin{align*}
S&=
\set{
\colvector{ \\ \\ 1\\ \\ 0},\,
\colvector{ \\ \\ 0\\ \\ 1}
}
=
\set{
\colvector{ 6\\ 5\\ 1\\ 0\\ 0},\,
\colvector{ 3\\ 2\\ 0\\ 4\\ 1}
}
\end{align*}
C50
Consider each archetype that is a system of equations and consider the solutions listed for the homogeneous version of the archetype. (If only the trivial solution is listed, then assume this is the only solution to the system.) From the solution set, determine if the columns of the coefficient matrix form a linearly independent or linearly dependent set. In the case of a linearly dependent set, use one of the sample solutions to provide a nontrivial relation of linear dependence on the set of columns of the coefficient matrix (Definition RLD). Indicate when Theorem MVSLD applies and connect this with the number of variables and equations in the system of equations.
Archetype A,
Archetype B,
Archetype C,
Archetype D/Archetype E,
Archetype F,
Archetype G/Archetype H,
Archetype I,
Archetype J
C51
For each archetype that is a system of equations consider the homogeneous version. Write elements of the solution set in vector form (Theorem VFSLS) and from this extract the vectors \(\vect{z}_j\) described in Theorem BNS. These vectors are used in a span construction to describe the null space of the coefficient matrix for each archetype. What does it mean when we write a null space as \(\spn{\set{\ }}\text{?}\)
Archetype A,
Archetype B,
Archetype C,
Archetype D/Archetype E,
Archetype F,
Archetype G/Archetype H,
Archetype I,
Archetype J
C52
For each archetype that is a system of equations consider the homogeneous version. Sample solutions are given and a linearly independent spanning set is given for the null space of the coefficient matrix. Write each of the sample solutions individually as a linear combination of the vectors in the spanning set for the null space of the coefficient matrix.
Archetype A,
Archetype B,
Archetype C,
Archetype D/Archetype E,
Archetype F,
Archetype G/Archetype H,
Archetype I,
Archetype J
C60
For the matrix \(A\) below, find a set of vectors \(S\) so that (1) \(S\) is linearly independent, and (2) the span of \(S\) equals the null space of \(A\text{,}\) \(\spn{S}=\nsp{A}\text{.}\) (See Exercise SS.C60.)
\begin{equation*}
A=
\begin{bmatrix}
1 & 1 & 6 & 8\\
1 & 2 & 0 & 1\\
2 & 1 & 6 & 7
\end{bmatrix}
\end{equation*}
SolutionTheorem BNS says that if we find the vector form of the solutions to the homogeneous system \(\homosystem{A}\text{,}\) then the fixed vectors (one per free variable) will have the desired properties. Rowreduce \(A\text{,}\) viewing it as the augmented matrix of a homogeneous system with an invisible columns of zeros as the last column.
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 4 & 5\\
0 & \leading{1} & 2 & 3\\
0 & 0 & 0 & 0
\end{bmatrix}
\end{equation*}
Moving to the vector form of the solutions (Theorem VFSLS), with free variables \(x_3\) and \(x_4\text{,}\) solutions to the consistent system (it is homogeneous, Theorem HSC) can be expressed as
\begin{equation*}
\colvector{x_1\\x_2\\x_3\\x_4}
=
x_3\colvector{4\\2\\1\\0}+
x_4\colvector{5\\3\\0\\1}\text{.}
\end{equation*}
Then with \(S\) given by
\begin{equation*}
S=\set{\colvector{4\\2\\1\\0},\,\colvector{5\\3\\0\\1}}
\end{equation*}
Theorem BNS guarantees the set has the desired properties.
M20
Suppose that \(S=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}\) is a set of three vectors from \(\complex{873}\text{.}\) Prove that the set
\begin{align*}
T&=
\set{
2\vect{v}_1+3\vect{v}_2+\vect{v}_3,\,
\vect{v}_1\vect{v}_22\vect{v}_3,\,
2\vect{v}_1+\vect{v}_2\vect{v}_3
}
\end{align*}
is linearly dependent.
SolutionBy Definition LICV, we can complete this problem by finding scalars, \(\alpha_1,\,\alpha_2,\,\alpha_3\text{,}\) not all zero, such that
\begin{align*}
\alpha_1\left(2\vect{v}_1+3\vect{v}_2+\vect{v}_3\right)+
\alpha_2\left(\vect{v}_1\vect{v}_22\vect{v}_3\right)+
\alpha_3\left(2\vect{v}_1+\vect{v}_2\vect{v}_3\right)
&=
\zerovector\text{.}
\end{align*}
Using various properties in Theorem VSPCV, we can rearrange this vector equation to
\begin{align*}
\left(2\alpha_1+\alpha_2+2\alpha_3\right)\vect{v}_1+
\left(3\alpha_1\alpha_2+\alpha_3\right)\vect{v}_2+
\left(\alpha_12\alpha_2\alpha_3\right)\vect{v}_3
&=
\zerovector\text{.}
\end{align*}
We can certainly make this vector equation true if we can determine values for the \(\alpha\)'s such that
\begin{align*}
2\alpha_1+\alpha_2+2\alpha_3&=0\\
3\alpha_1\alpha_2+\alpha_3&=0\\
\alpha_12\alpha_2\alpha_3&=0
\end{align*}
Aah, a homogeneous system of equations. And it has infinitely many nonzero solutions. By the now familiar techniques, one such solution is \(\alpha_1=3\text{,}\) \(\alpha_2=4\text{,}\) \(\alpha_3=5\text{,}\) which you can check in the original relation of linear dependence on \(T\) above.
Note that simply writing down the three scalars, and demonstrating that they provide a nontrivial relation of linear dependence on \(T\text{,}\) could be considered an ironclad solution. But it would not have been very informative for you if we had only done just that here. Compare this solution very carefully with Solution M21.1.
M21
Suppose that \(S=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}\) is a linearly independent set of three vectors from \(\complex{873}\text{.}\) Prove that the set
\begin{align*}
T&=
\set{
2\vect{v}_1+3\vect{v}_2+\vect{v}_3,\,
\vect{v}_1\vect{v}_2+2\vect{v}_3,\,
2\vect{v}_1+\vect{v}_2\vect{v}_3
}
\end{align*}
is linearly independent.
SolutionBy Definition LICV we can complete this problem by proving that if we assume that
\begin{align*}
\alpha_1\left(2\vect{v}_1+3\vect{v}_2+\vect{v}_3\right)+
\alpha_2\left(\vect{v}_1\vect{v}_2+2\vect{v}_3\right)+
\alpha_3\left(2\vect{v}_1+\vect{v}_2\vect{v}_3\right)
&=
\zerovector
\end{align*}
then we must conclude that \(\alpha_1=\alpha_2=\alpha_3=0\text{.}\) Using various properties in Theorem VSPCV, we can rearrange this vector equation to
\begin{align*}
\left(2\alpha_1+\alpha_2+2\alpha_3\right)\vect{v}_1+
\left(3\alpha_1\alpha_2+\alpha_3\right)\vect{v}_2+
\left(\alpha_1+2\alpha_2\alpha_3\right)\vect{v}_3
&=
\zerovector\text{.}
\end{align*}
Because the set \(S=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}\) was assumed to be linearly independent, by Definition LICV we must conclude that
\begin{align*}
2\alpha_1+\alpha_2+2\alpha_3&=0\\
3\alpha_1\alpha_2+\alpha_3&=0\\
\alpha_1+2\alpha_2\alpha_3&=0
\end{align*}
Aah, a homogeneous system of equations. And it has a unique solution, the trivial solution. So, \(\alpha_1=\alpha_2=\alpha_3=0\text{,}\) as desired. It is an inescapable conclusion from our assumption of a relation of linear dependence above. Done.
Compare this solution very carefully with Solution M20.1, noting especially how this problem required (and used) the hypothesis that the original set be linearly independent, and how this solution feels more like a proof, while the previous problem could be solved with a fairly simple demonstration of any nontrivial relation of linear dependence.
M50
Consider the set of vectors from \(\complex{3}\text{,}\) \(W\text{,}\) given below. Find a set \(T\) that contains three vectors from \(W\) and such that \(W=\spn{T}\text{.}\)
\begin{equation*}
W=\spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5}}=\spn{\set{\colvector{2\\1\\1},\,\colvector{1\\1\\1},\,\colvector{1\\2\\3},\,\colvector{3\\1\\3},\,\colvector{0\\1\\3}}}
\end{equation*}
SolutionWe want to first find some relations of linear dependence on \(\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5}\) that will allow us to “kick out” some vectors, in the spirit of Example SCAD. To find relations of linear dependence, we formulate a matrix \(A\) whose columns are \(\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5\text{.}\) Then we consider the homogeneous system of equations \(\homosystem{A}\) by rowreducing its coefficient matrix (remember that if we formulated the augmented matrix we would just add a column of zeros). After rowreducing, we obtain
\begin{equation*}
\begin{bmatrix}
\leading{1} & 0 & 0 & 2 & 1\\
0 & \leading{1} & 0 & 1 & 2\\
0 & 0 & \leading{1} & 0 & 0
\end{bmatrix}\text{.}
\end{equation*}
From this we find that solutions can be obtained employing the free variables \(x_4\) and \(x_5\text{.}\) With appropriate choices we will be able to conclude that vectors \(\vect{v}_4\) and \(\vect{v}_5\) are unnecessary for creating \(W\) via a span. By Theorem SLSLC the choice of free variables below lead to solutions and linear combinations, which are then rearranged.
\begin{align*}
x_4=1, x_5=0&&\Rightarrow&&(2)\vect{v}_1+(1)\vect{v}_2+(0)\vect{v}_3+(1)\vect{v}_4+(0)\vect{v}_5=\zerovector&&\Rightarrow&&\vect{v}_4=2\vect{v}_1+\vect{v}_2\\
x_4=0, x_5=1&&\Rightarrow&&(1)\vect{v}_1+(2)\vect{v}_2+(0)\vect{v}_3+(0)\vect{v}_4+(1)\vect{v}_5=\zerovector&&\Rightarrow&&\vect{v}_5=\vect{v}_12\vect{v}_2
\end{align*}
Since \(\vect{v}_4\) and \(\vect{v}_5\) can be expressed as linear combinations of \(\vect{v}_1\) and \(\vect{v}_2\) we can say that \(\vect{v}_4\) and \(\vect{v}_5\) are not needed for the linear combinations used to build \(W\) (a claim that we could establish carefully with a pair of set equality arguments). Thus
\begin{equation*}
W=\spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}}=\spn{\set{\colvector{2\\1\\1},\,\colvector{1\\1\\1},\,\colvector{1\\2\\3}}}\text{.}
\end{equation*}
That the \(\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}\) is linearly independent set can be established quickly with Theorem LIVRN.
There are other answers to this question, but notice that any nontrivial linear combination of \(\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4,\,\vect{v}_5\) will have a zero coefficient on \(\vect{v}_3\text{,}\) so this vector can never be eliminated from the set used to build the span.
M51
Consider the subspace \(W=\spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}}\text{.}\) Find a set \(S\) so that (1) \(S\) is a subset of \(W\text{,}\) (2) \(S\) is linearly independent, and (3) \(W=\spn{S}\text{.}\) Write each vector not included in \(S\) as a linear combination of the vectors that are in \(S\text{.}\)
\begin{align*}
\vect{v}_1&=\colvector{1\\1\\2}
&
\vect{v}_2&=\colvector{4\\4\\8}
&
\vect{v}_3&=\colvector{3\\2\\7}
&
\vect{v}_4&=\colvector{2\\1\\7}
\end{align*}
SolutionThis problem can be solved using the approach in Solution M50.1. We will provide a solution here that is more adhoc, but note that we will have a more straightforward procedure given by the upcoming Theorem BS.
\(\vect{v}_1\) is a nonzero vector, so in a set all by itself we have a linearly independent set. As \(\vect{v}_2\) is a scalar multiple of \(\vect{v}_1\text{,}\) the equation \(4\vect{v}_1+\vect{v}_2=\zerovector\) is a relation of linear dependence on \(\set{\vect{v}_1,\,\vect{v}_2}\text{,}\) so we will pass on \(\vect{v}_2\text{.}\) No such relation of linear dependence exists on \(\set{\vect{v}_1,\,\vect{v}_3}\text{,}\) though on \(\set{\vect{v}_1,\,\vect{v}_3,\,\vect{v}_4}\) we have the relation of linear dependence \(7\vect{v}_1+3\vect{v}_3+\vect{v}_4=\zerovector\text{.}\) So take \(S=\set{\vect{v}_1,\,\vect{v}_3}\text{,}\) which is linearly independent.
Then
\begin{align*}
\vect{v}_2&=4\vect{v_1}+0\vect{v_3}
&
\vect{v}_4&=7\vect{v_1}3\vect{v_3}
\end{align*}
The two equations above are enough to justify the set equality
\begin{align*}
W
&=\spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}}
=\spn{\set{\vect{v}_1,\,\vect{v}_3}}
=\spn{S}
\end{align*}
There are other solutions (for example, swap the roles of \(\vect{v}_1\) and \(\vect{v}_2\text{,}\) but by upcoming theorems we can confidently claim that any solution will be a set \(S\) with exactly two vectors.
T10
Prove that if a set of vectors contains the zero vector, then the set is linearly dependent. (Ed. “The zero vector is death to linearly independent sets.”)
T12
Suppose that \(S\) is a linearly independent set of vectors, and \(T\) is a subset of \(S\text{,}\) \(T\subseteq S\) (Definition SSET). Prove that \(T\) is linearly independent.
T13
Suppose that \(T\) is a linearly dependent set of vectors, and \(T\) is a subset of \(S\text{,}\) \(T\subseteq S\) (Definition SSET). Prove that \(S\) is linearly dependent.
T15
Suppose that \(\set{\vectorlist{v}{n}}\) is a set of vectors. Prove that
\begin{gather*}
\set{\vect{v}_1\vect{v}_2,\,\vect{v}_2\vect{v}_3,\,\vect{v}_3\vect{v}_4,\,\dots,\,\vect{v}_n\vect{v}_1}
\end{gather*}
is a linearly dependent set.
SolutionConsider the following linear combination
\begin{align*}
1\left(\vect{v}_1\vect{v}_2\right)+&
1\left(\vect{v}_2\vect{v}_3\right)+
1\left(\vect{v}_3\vect{v}_4\right)+\cdots+
1\left(\vect{v}_n\vect{v}_1\right)\\
&=
\vect{v}_1\vect{v}_2+
\vect{v}_2\vect{v}_3+
\vect{v}_3\vect{v}_4+\dots+
\vect{v}_n\vect{v}_1\\
&=\vect{v}_1+\zerovector+\zerovector+\cdots+\zerovector\vect{v}_1\\
&=\zerovector\text{.}
\end{align*}
This is a nontrivial relation of linear dependence (Definition RLDCV), so by Definition LICV the set is linearly dependent.
T20
Suppose that \(\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}\) is a linearly independent set in \(\complex{35}\text{.}\) Prove that
\begin{equation*}
\set{
\vect{v}_1,\,
\vect{v}_1+\vect{v}_2,\,
\vect{v}_1+\vect{v}_2+\vect{v}_3,\,
\vect{v}_1+\vect{v}_2+\vect{v}_3+\vect{v}_4
}
\end{equation*}
is a linearly independent set.
SolutionOur hypothesis and our conclusion use the term linear independence, so it will get a workout. To establish linear independence, we begin with the definition (Definition LICV) and write a relation of linear dependence (Definition RLDCV)
\begin{equation*}
\alpha_1\left(\vect{v}_1\right)+
\alpha_2\left(\vect{v}_1+\vect{v}_2\right)+
\alpha_3\left(\vect{v}_1+\vect{v}_2+\vect{v}_3\right)+
\alpha_4\left(\vect{v}_1+\vect{v}_2+\vect{v}_3+\vect{v}_4\right)
=\zerovector\text{.}
\end{equation*}
Using the distributive and commutative properties of vector addition and scalar multiplication (Theorem VSPCV) this equation can be rearranged as
\begin{equation*}
\left(\alpha_1+\alpha_2+\alpha_3+\alpha_4\right)\vect{v}_1+
\left(\alpha_2+\alpha_3+\alpha_4\right)\vect{v}_2+
\left(\alpha_3+\alpha_4\right)\vect{v}_3+
\left(\alpha_4\right)\vect{v}_4
=
\zerovector\text{.}
\end{equation*}
However, this is a relation of linear dependence (Definition RLDCV) on a linearly independent set, \(\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}\) (this was our lone hypothesis). By the definition of linear independence (Definition LICV) the scalars must all be zero. This is the homogeneous system of equations
\begin{align*}
\alpha_1+\alpha_2+\alpha_3+\alpha_4&=0\\
\alpha_2+\alpha_3+\alpha_4&=0\\
\alpha_3+\alpha_4&=0\\
\alpha_4&=0\text{.}
\end{align*}
Rowreducing the coefficient matrix of this system (or backsolving) gives the conclusion
\begin{align*}
\alpha_1=0&&\alpha_2=0&&\alpha_3=0&&\alpha_4=0\text{.}
\end{align*}
This means, by Definition LICV, that the original set
\begin{equation*}
\set{
\vect{v}_1,\,
\vect{v}_1+\vect{v}_2,\,
\vect{v}_1+\vect{v}_2+\vect{v}_3,\,
\vect{v}_1+\vect{v}_2+\vect{v}_3+\vect{v}_4
}
\end{equation*}
is linearly independent.
T50
Suppose that \(A\) is an \(m\times n\) matrix with linearly independent columns and the linear system \(\linearsystem{A}{\vect{b}}\) is consistent. Show that this system has a unique solution. (Notice that we are not requiring \(A\) to be square.)
SolutionLet \(A=\matrixcolumns{A}{n}\text{.}\) \(\linearsystem{A}{\vect{b}}\) is consistent, so we know the system has at least one solution (Definition CS). We would like to show that there are no more than one solution to the system. Employing Proof Technique U, suppose that \(\vect{x}\) and \(\vect{y}\) are two solution vectors for \(\linearsystem{A}{\vect{b}}\text{.}\) By Theorem SLSLC we know we can write
\begin{align*}
\vect{b}&=
\vectorentry{\vect{x}}{1}A_1+
\vectorentry{\vect{x}}{2}A_2+
\vectorentry{\vect{x}}{3}A_3+
\cdots+
\vectorentry{\vect{x}}{n}A_n\\
\vect{b}&=
\vectorentry{\vect{y}}{1}A_1+
\vectorentry{\vect{y}}{2}A_2+
\vectorentry{\vect{y}}{3}A_3+
\cdots+
\vectorentry{\vect{y}}{n}A_n
\end{align*}
Then
\begin{align*}
\zerovector
&=\vect{b}\vect{b}\\
&=
\left(
\vectorentry{\vect{x}}{1}A_1+
\vectorentry{\vect{x}}{2}A_2+
\cdots+
\vectorentry{\vect{x}}{n}A_n
\right)

\left(
\vectorentry{\vect{y}}{1}A_1+
\vectorentry{\vect{y}}{2}A_2+
\cdots+
\vectorentry{\vect{y}}{n}A_n
\right)\\
&=
\left(\vectorentry{\vect{x}}{1}\vectorentry{\vect{y}}{1}\right)A_1+
\left(\vectorentry{\vect{x}}{2}\vectorentry{\vect{y}}{2}\right)A_2+
\cdots+
\left(\vectorentry{\vect{x}}{n}\vectorentry{\vect{y}}{n}\right)A_n\text{.}
\end{align*}
This is a relation of linear dependence (Definition RLDCV) on a linearly independent set (the columns of \(A\)). So the scalars must all be zero
\begin{align*}
\vectorentry{\vect{x}}{1}\vectorentry{\vect{y}}{1}&=0
&
\vectorentry{\vect{x}}{2}\vectorentry{\vect{y}}{2}&=0
&
\dots&
&
\vectorentry{\vect{x}}{n}\vectorentry{\vect{y}}{n}&=0\text{.}
\end{align*}
Rearranging these equations yields the statement that \(\vectorentry{\vect{x}}{i}=\vectorentry{\vect{y}}{i}\text{,}\) for \(1\leq i\leq n\text{.}\) However, this is exactly how we define vector equality (Definition CVE), so \(\vect{x}=\vect{y}\) and the system has only one solution.