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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