## 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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