I've been struggling with a question. It goes as follows: Given a matrix $$A$$ $$\text{s.t. } \rho(A^TA)=1.$$ Prove: $$\rho((I-A^TA)(A^TA)^v)\le \frac{v^v}{(v+1)^{v+1}}$$
$$A^TA$$ is symmetric and PSD, all eigenvalues are real. The eigenvalues in question are all of the form $$(1-\lambda_i)\lambda_i^v$$ where $$\lambda_i$$ is an eigenvalue of $$A^TA$$.
Presumably $$v>0$$. The condition $$\rho(A^TA)=1$$ and the fact that $$A^TA\succeq0$$ together imply that $$0\le\lambda\le1$$ for every eigenvalue $$\lambda$$ of $$A^TA$$. Now, let $$f(\lambda)=(1-\lambda)\lambda^v=\lambda^v-\lambda^{v+1}$$. Then $$f'(\lambda)=v\lambda^{v-1}-(v+1)\lambda^v$$. Therefore, when $$f'(\lambda)=0$$ only if $$\lambda=0$$ or $$\lambda=\frac{v}{v+1}$$. Since $$f(0)=0<\frac{v^v}{(v+1)^{v+1}}=f(\frac{v}{v+1})$$, the result follows.