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$.

  • $\begingroup$ Thanks for the answer. I have two questions. How do we know the diagonal is all positive? and why having a negative diagonal means it is not a hermitian positive definite matrix? $\endgroup$ – Crimson Oct 6 '16 at 14:31
  • $\begingroup$ I rewrote it, hope its clearer. $\endgroup$ – RSS Oct 6 '16 at 15:07
  • $\begingroup$ Thank you very much. I understood. $\endgroup$ – Crimson Oct 6 '16 at 15:13

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