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.
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$\begingroup$ Hello and welcome to math.stackexchange!! The answer is yes, sine $I + AA^\ast$ is always a non-singular matrix. $\endgroup$– Hans EnglerAug 21 '16 at 14:48
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$\begingroup$ @Nameless, it's true, why do you think it is not? Zero matrix is not a counterexample. $\endgroup$– EnnarAug 21 '16 at 14:51
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$\begingroup$ @Ennar, hang on sorry I thought he asked if $det(I + AA^*) = 0$ $\endgroup$– IAmNoOneAug 21 '16 at 14:52
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$\begingroup$ @HansEngler In fact, I want to ask whether it is always a non-singular, so I ask for the det. $\endgroup$– QDE TRYAug 21 '16 at 14:55
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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.
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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 $\lambda\ge0$.
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Another way (sketch)
If $P>0$ and $Q\ge 0 $ then $P +Q>0$ (here $>0$ means "strictly positive definite" and $\ge 0$ means "(semi) positive definite".
$P>0 \implies |P| \ne 0$
$A A^* \ge 0$
For any $\epsilon>0$, $\epsilon I>0$. Hence, $\epsilon I + A A^* > 0$
Therefore, $|\epsilon I + A A^*| \ne 0$
