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I've been struggling with a question. It goes as follows: Given a matrix $A$ $$ \text{s.t. } \rho(A^TA)=1.$$ Prove: $$\rho((I-A^TA)(A^TA)^v)\le \frac{v^v}{(v+1)^{v+1}}$$

I've started thinking on a direction, but haven't been successful:

$A^TA$ 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 $A^TA$.

One of these values is zero, due to the known spectral radius, but how can I move towards that bound?

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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.

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