Is it always true for $\det(I+AA^*)\neq 0$ ?

Here, $A^*$ is the adjoint of $A$.

I have found that $\det(I+BC)=\det(I+CB)$, but it seems no use here.

  • $\begingroup$ Hello and welcome to math.stackexchange!! The answer is yes, sine $I + AA^\ast$ is always a non-singular matrix. $\endgroup$ – Hans Engler Aug 21 '16 at 14:48
  • $\begingroup$ @Nameless, it's true, why do you think it is not? Zero matrix is not a counterexample. $\endgroup$ – Ennar Aug 21 '16 at 14:51
  • $\begingroup$ @Ennar, hang on sorry I thought he asked if $det(I + AA^*) = 0$ $\endgroup$ – IAmNoOne Aug 21 '16 at 14:52
  • $\begingroup$ @HansEngler In fact, I want to ask whether it is always a non-singular, so I ask for the det. $\endgroup$ – QDE TRY Aug 21 '16 at 14:55

Suppose $\det(I + A A^*) = 0$.

That means the quadratic matrix $B = I + A A^*$ is singular. Thus there exists a nonzero vector $v$ such that $B v = 0$, and therefore $v^* B v = 0$.

But $0 = v^* B v = v^* v + (A^* v)^* (A^* v) = \|v\|^2 + \|A^* v\|^2$ implies $v = 0$.

This is a contradiction.

  • $\begingroup$ Thanks, that's actually what I want. $\endgroup$ – QDE TRY Aug 21 '16 at 15:13

Saying that $\det(AA^*+I)=0$ is the same as saying that $-1$ is an eigenvalue of $AA^*$; however, if $AA^*=\lambda v$ for some $v\ne0$, we have $$ (A^*v)^*(A^*v)=v^*AA^*v=\lambda(v^*v) $$ and, for any vector $x$, $x^*x\ge0$. Thus $\lambda\ge0$.


Another way (sketch)

  1. If $P>0$ and $Q\ge 0 $ then $P +Q>0$ (here $>0$ means "strictly positive definite" and $\ge 0$ means "(semi) positive definite".

  2. $P>0 \implies |P| \ne 0$

  3. $A A^* \ge 0$

  4. For any $\epsilon>0$, $\epsilon I>0$. Hence, $\epsilon I + A A^* > 0$

  5. Therefore, $|\epsilon I + A A^*| \ne 0$


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