## Linear Algebra and Its Applications, Exercise 2.3.4

Exercise 2.3.4. Suppose the vectors $v_1$, $v_2$, and $v_3$ are linearly independent. (The book says linearly dependent, but I believe this is a typo.) Are the vectors $w_1 = v_1 + v_2$, $w_2 = v_1 + v_3$, and $w_3 = v_2 + v_3$ also linearly independent?

Answer: Consider the linear combination of $w_1$, $w_2$, and $w_3$ with weights $c_1$, $c_2$, and $c_3$. We have

$c_1w_1 + c_2w_2 + c_3w_3 = c_1(v_1 + v_2) + c_2(v_1 + v_3) + c_3(v_2 + v_3)$

$= (c_1+c_2)v_1 + (c_1+c_3)v_2 + (c_2+c_3)v_1$

Since the vectors $v_1$, $v_2$, and $v_3$ are linearly independent the above expression can be zero only if $c_1 + c_2 = 0$, $c_1 + c_3 = 0$, and $c_2 + c_3 = 0$.

We can express this as the following system of equations

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}c_1&+&c_2&&&=&0 \\ c_1&&&+&c_3&=&0 \\ &&c_2&+&c_3&=&0 \end{array}$

and can solve it via elimination. We first subtract the first equation from the second to obtain the following system:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}c_1&+&c_2&&&=&0 \\ &&-c_2&+&c_3&=&0 \\ &&c_2&+&c_3&=&0 \end{array}$

and then add the second equation to the third to obtain the final system:

$\setlength\arraycolsep{0.2em}\begin{array}{rcrcrcl}c_1&+&c_2&&&=&0 \\ &&-c_2&+&c_3&=&0 \\ &&&&2c_3&=&0 \end{array}$

Solving for the weights we have $2c_3 = 0$ or $c_3 = 0$, $-c_2 + 0 = 0$ or $c_2 = 0$, and $c_1 + 0 = 0$ or $c_1 = 0$.

Since $c_1w_1 + c_2w_2 + c_3w_3 = 0$ only if $c_1 = c_2 = c_3 = 0$ we see that $w_1$, $w_2$, and $w_3$ are linearly independent.

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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