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?