## 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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