Linear Algebra and Its Applications, Exercise 3.1.5

Exercise 3.1.5. Of the following vectors

$v_1 = \begin{bmatrix} 1 \\ 2 \\ -2 \\ 1 \end{bmatrix} \qquad v_2 = \begin{bmatrix} 4 \\ 0 \\ 4 \\ 0 \end{bmatrix} \qquad v_3 = \begin{bmatrix} 1 \\ -1 \\ -1 \\ -1 \end{bmatrix}$

which are orthogonal to one another?

Answer: We have the following inner products among the vectors:

$v_1^Tv_2 = 1 \cdot 4 + 2 \cdot 0 + (-2) \cdot 4 + 1 \cdot 0$

$= 4 + 0 - 8 + 0 = -4$

$v_1^Tv_3 = 1 \cdot 1 + 2 \cdot (-1) + (-2) \cdot (-1) + 1 \cdot (-1)$

$= 1 - 2 + 2 - 1 = 0$

$v_2^Tv_3 = 4 \cdot 1 + 0 \cdot (-1) + 4 \cdot (-1) + 0 \cdot (-1)$

$= 4 + 0 - 4 + 0 = 0$

So $v_1$ and $v_2$ are orthogonal to $v_3$ but not to each other.

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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