## Linear Algebra and Its Applications, Exercise 3.1.1

Exercise 3.1.1. For $x = (1, 4, 0, 2)$ and $y = (2, -2, 1, 3)$ what is the length of each vector and their inner product? $\|x\|^2 = 1 \cdot 1 + 4 \cdot 4 + 0 \cdot 0 + 2 \cdot 2$ $= 1 + 16 + 0 + 4 = 21$

and $\|y\|^2 = 2 \cdot 2 + (-2) \cdot (-2) + 1 \cdot 1 + 3 \cdot 3$ $= 4 + 4 + 1 + 9 = 18$

so that $\|x\| = \sqrt{21}$ and $\|y\| = \sqrt{18}$.

The inner product of $x$ and $y$ is then $x^Ty = 1 \cdot 2 + 4 \cdot (-2) + 0 \cdot 1 + 2 \cdot 3$ $= 2 - 8 + 0 + 6 = 0$

Note that $x$ and $y$ are thus orthogonal.

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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