Suppose $x \in X$ is a unit vector such that $||A|| = ||Ax||$; show that $A^TAx = ||A||^2 x$.

Question

Let $X$ be a linear space with real Euclidean structure. Let $A : X \to X$ be a linear map. Suppose $x \in X$ is a unit vector such that $||A|| = ||Ax||$; show that $A^TAx = ||A||^2 x$.

I tried in this way, considering $g(t)=||\frac{A(x+ty)}{||x+ty||}||$. It has a maximum at $t=0$ because of the definition of the norm of the matrix. Hence we get $g'(0)=0$. Now can I use this data to work with transpose?

Answer 1

Take $g(t) = \frac12\frac{ \|A(x+ty)\|^2 } { \|x+ty\|^2 }$, i.e., take the square of your original function. Then $g'(t)=0$, where $$ g'(t) = \frac{ y^TA^TA(x+ty) } { \|x+ty\|^2 } -\frac{ \|A(x+ty)\|^2 } { \|x+ty\|^4 } y^T(x+ty). $$ Hence $$ g'(0) = y^TA^TAx - \|Ax\|^2 y^Tx. $$ This is zero for all $y$, hence $$ A^TAx = \|Ax\|^2 x. $$

Answer 2

$$\|A^{T} Ax-\|A||^{2}x\|^{2}$$ $$=\|A^{T} Ax\|^{2}-2\|A||^{2} \langle A^{T}Ax, x \rangle+\|A\|^{4}$$ $$=\|A^{T} Ax\|^{2}-\|A\|^{4}.$$ Now $\|A^{T} Ax\| \leq \|A^{T}\|\|A\|=\|A\|^{2}$ so we get $0 \leq \|A^{T} Ax-\|A||^{2}x\|^{2} \leq 0$