# Is $\det(I+AA^*)$ always be non-zero?

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.

• Hello and welcome to math.stackexchange!! The answer is yes, sine $I + AA^\ast$ is always a non-singular matrix. Aug 21, 2016 at 14:48
• @Nameless, it's true, why do you think it is not? Zero matrix is not a counterexample. Aug 21, 2016 at 14:51
• @Ennar, hang on sorry I thought he asked if $det(I + AA^*) = 0$ Aug 21, 2016 at 14:52
• @HansEngler In fact, I want to ask whether it is always a non-singular, so I ask for the det. Aug 21, 2016 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 $v = 0$.

• Thanks, that's actually what I want. Aug 21, 2016 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, $|\epsilon I + A A^*| \ne 0$