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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