## Linear Algebra and Its Applications, Exercise 3.1.14

Exercise 3.1.14. Given two vectors $x$ and $y$ in $\mathbb{R}^n$, show that their difference $x-y$ is orthogonal to their sum $x+y$ if and only if their lengths $\|x\|$ and $\|y\|$ are the same.

Answer: First we assume that $x-y$ is orthogonal to $x+y$. This means that their inner product  must be zero, or $0 = \sum_{i=1}^n (x_i-y_i)(x_i+y_i)$ $= \sum_{i=1}^n (x_i^2 - y_ix_i + x_iy_i - y_i^2)$ $= \sum_{i=1}^n (x_i^2 - y_i^2)$ $= \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2$ $= \|x\|^2 - \|y\|^2$

So we have $\|x\|^2 - \|y\|^2 = 0$ which implies that $\|x\|^2 = \|y\|^2$ or $\|x\| = \|y\|$ (keeping in mind that the length of a vector must be a positive quantity).

So if $x-y$ is orthogonal to $x+y$ then we must have $\|x\| = \|y\|$.

Now suppose that $\|x\| = \|y\|$. This implies that $\|x\|^2 = \|y\|^2$ or $\|x\|^2 - \|y\|^2 = 0$. We then use the definitions of $\|x\|$ and $\|y\|$ to run the argument above in reverse: $0 = \sum_{i=1}^n x_i^2 - \sum_{i=1}^n y_i^2$ $= \sum_{i=1}^n (x_i^2 - y_i^2)$ $= \sum_{i=1}^n (x_i^2 - y_ix_i + x_iy_i - y_i^2)$ $= \sum_{i=1}^n (x_i-y_i)(x_i+y_i)$

Since $\sum_{i=1}^n (x_i-y_i)(x_i+y_i) = 0$ we see that $x-y$ is orthogonal to $x+y$.

So if $\|x\| = \|y\|$ then $x-y$ must be orthogonal to $x+y$.

Combining the two proofs above, we have shown that for any two vectors $x$ and $y$ in in $\mathbb{R}^n$, $x-y$ is orthogonal to $x+y$ if and only if $\|x\| = \|y\|$

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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