Assume $A$ is a nonzero $n \times n $ matrix. We know $A^*A$ is hermitian positive definite matrix, so all its eigenvalues are real and positive. Now, consider the matrix $$A^*A-\alpha \lambda_{min} I $$ where $\lambda_{min}$ is the smallest eigenvalue of $A^*A$.
How can we show :
(1) $A^*A$ is hermitian positive definite if $0 < \alpha < 1$.
(2) $A^*A$ is not postive defenite if $\alpha > 1$.
Can anyone please help to prove this. Thank you.
We know A*A is hermitian positive definite so it can be diagonalized:
$$ A^{*}A = Q\Sigma Q^{*} $$
where $\Sigma$ is a diagonal matrix will all positive entries.
$$ A^{*}A - \alpha \lambda_{min} I= Q\Sigma Q^{*} - Q\alpha \lambda_{min} I Q^{*} = Q(\Sigma -\alpha \lambda_{min} I) Q^{*}$$
This matrix is self-adjoint:
$$ (Q(\Sigma -\alpha \lambda_{min} I) Q^{*})^{*}= Q(\Sigma^{*} -\alpha \lambda_{min} I^{*}) Q^{*} = Q(\Sigma -\alpha \lambda_{min} I) Q^{*} $$
Also, it is positive definite if and only if the diagonal entries are +ve. But the diagonal entries are $ \lambda_{i} - \alpha \lambda_{min}$ which is $>0$ if and only if $ \alpha < 1$.