## 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.

This entry was posted in linear algebra and tagged , , . Bookmark the permalink.