Linear Algebra and Its Applications, Exercise 2.3.1

Exercise 2.3.1. State whether the following vectors are linearly independent or not

$v_1 = \begin{bmatrix} 1 \\ 1 \\ 0 \\ 0 \end{bmatrix} \qquad v_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} \qquad v_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 1 \end{bmatrix} \qquad v_4 = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 1 \end{bmatrix}$

by solving the equation $c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 = 0$. Also, solve $c_1v_1 + \cdots + c_4v_4 = (0, 0, 0, 1)$ to determine whether the vectors span $\mathbf{R}^4$.

Answer: The equation $c_1v_1 + c_2v_2 + c_3v_3 + c_4v_4 = 0$ is equivalent to

$\begin{bmatrix} 1&1&0&0 \\ 1&0&0&1 \\ 0&1&1&0 \\ 0&0&1&1 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

We can use elimination to solve this system

$\begin{bmatrix} 1&1&0&0 \\ 1&0&0&1 \\ 0&1&1&0 \\ 0&0&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&0&0 \\ 0&-1&0&1 \\ 0&1&1&0 \\ 0&0&1&1 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 1&1&0&0 \\ 0&-1&0&1 \\ 0&0&1&1 \\ 0&0&1&1 \end{bmatrix} \Rightarrow \begin{bmatrix} 1&1&0&0 \\ 0&-1&0&1 \\ 0&0&1&1 \\ 0&0&0&0 \end{bmatrix}$

producing the following system $Ux = c$

$\begin{bmatrix} 1&1&0&0 \\ 0&-1&0&1 \\ 0&0&1&1 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$

Note that $c_4$ is a free variable and $c_1$, $c_2$, and $c_4$ are basic. Setting the free variable $c_4 = 1$, from the third equation we have $c_3 + c_4 = 0$ or $c_3 = -c_4 = -1$. From the second equation we have $-c_2 + c_4 = 0$ or $c_2 = c_4 = 1$. Finally, from the first equation we have $c_1 + c_2 = 0$ or $c_1 = -c_2 = -1$. This means that we have $-v_1 + v_2 - v_3 + v_4 = 0$ and thus the vectors are not linearly independent. More specifically, we can express $v_4$ as a linear combination of the other vectors: $v_4 = v_1 - v_2 + v_3$.

We next attempt to solve $c_1v_1 + \cdots + c_4v_4 = (0, 0, 0, 1)$. As before we can use elimination to obtain the system $Ux = c$

$\begin{bmatrix} 1&1&0&0 \\ 0&-1&0&1 \\ 0&0&1&1 \\ 0&0&0&0 \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ c_3 \\ c_4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

In this case the fourth equation produces the contradictory result $0 = 1$. Thus this system has no solution, and we conclude that the vectors in question do not span $\mathbf{R}^4$.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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