# Showing whether or not a matrix is a hermitian positive definite matrix

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 $\alpha > 1$.

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.

$$A^{*}A - \alpha \lambda_{min} I= Q\Sigma Q^{*} - Q\alpha \lambda_{min} I Q^{*} = Q(\Sigma -\alpha \lambda_{min} I) Q^{*}$$

$$(Q(\Sigma -\alpha \lambda_{min} I) Q^{*})^{*}= Q(\Sigma^{*} -\alpha \lambda_{min} I^{*}) Q^{*} = Q(\Sigma -\alpha \lambda_{min} I) Q^{*}$$
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 $\alpha < 1$.