0
$\begingroup$

I came across the following claim:

The multiplication of an orthogonal matrix $A \in \Bbb R^{m\times n}$ by a vector $x$ with $||x||_2=1$ must have a sum of squared elements (i.e. Frobenius norm) at most $1$.

This can be cast into a clumsy-looking inequality of real numbers: We assume $x_1+...+x_n=1$ and $\sum_i a_{ij}^2=1$ and $\sum _ka_{ij}a_{kj}=0$ for all $k,j,i$. Then we claim $\sum _i (\sum_j a_{ij}x_j)^2 \leq 1$.

Cauchy's inequality only gives me a weak bound (of $m$ instead of $1$), but I saw this version used and it seems true. How can I show this?

  • 0
    An orthogonal matrix respects norms, so $\;\left\|Ax\right\|=\left\|x\right\|\;$ and we're done – 2017-02-19

2 Answers 2

1

It's because $$\|Ax\|_2^2 = x^T A^T A x = x^T x = \|x\|_2^2 = 1.$$ The result always has sum of squared elements exactly $1$.

1

The result you are looking for is a classical property of orthogonal matrices:

An orthogonal matrix gives an isometry, i.e., preserves the norm:

$$\forall X, \ \ \|AX\|=\|X\|$$