when linear mapping keeps monotonicity of $L_2$ norm

Question

Consider an arbitrary vector $\alpha$ from vector space $R^p$, a linear mapping $A: \alpha\rightarrow A\alpha$ transforms $\alpha$ to $A\alpha$ in space $R^q$. What condition should $A$ satisfy so that $\alpha_1'\alpha_1 \geq \alpha_2'\alpha_2 \Rightarrow$ $\alpha_1'A'A\alpha_1 \geq \alpha_2'A'A\alpha_2$ for any $\alpha_1, \alpha_2$ in $R^p$? I know that when $A'A=kI_p$ this implication holds, but any more situations?

Answer 1

Let's note $B=A'A$ - it's a square symetric matrix, it's eigenvalues $\lambda_j$ are therefore real and it has an orthogonal basis of eigenvectors $v_j$. We can suppose that these vectors are all of norm $\|v_j\|=1$.

Let's take two eigenvectors $v_1$ and $v_2$. $\|v_1\|\ge \|v_2\|$, so $(Bv_1,v_1)\ge (Bv_2,v_2)$, in other words, $\lambda_1\ge \lambda_2$. By a similar argument we obtain $\lambda_2\ge \lambda_1$ and conclude that $\lambda_1= \lambda_2$. Obviously, this means that all eigenvalues of $B$ are equal (among others, equal to the norm $\|B\|$).

This implies that for an arbitrary vector $x$ we have $(Bx,x)=\|B\|\|x\|^2$ and thus the desire inequality will always hold.

Theferore, the final answer is all eigenvalues of $A'A$are equal.