3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Hello and welcome to math.stackexchange!! The answer is yes, sine $I + AA^\ast$ is always a non-singular matrix. $\endgroup$ Aug 21, 2016 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, 2016 at 14:51
  • $\begingroup$ @Ennar, hang on sorry I thought he asked if $det(I + AA^*) = 0$ $\endgroup$
    – IAmNoOne
    Aug 21, 2016 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, 2016 at 14:55

3 Answers 3

10
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks, that's actually what I want. $\endgroup$
    – QDE TRY
    Aug 21, 2016 at 15:13
1
$\begingroup$

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 $\lambda\ge0$.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .