$\det A \neq 0$. Prove that $\det A^* \neq 0$.

Question

$A$ is matrix representing operator $\mathcal{A}$. $*$ is such operator that respects following equality: $(\mathcal{A}x,y)=(x, \mathcal{A}^*y)$; (I don't know what term is used in English).

$A^*=\overline{A}^T$. I know that $\det A=\det A^T$, but $A^*$ isn't just $A^T$, it's actually $\overline{A}^T$.

How to prove $\det A^* \neq 0$ if $\det A \neq 0$?

Answer 1

Here's the way I'd prefer to do it that doesn't refer to the expression $A^* = \overline{A^T}$.

Suppose we have a $y$ such that $A^*y = 0$. If we show that $y=0$ necessarily then $A^*$ is invertible and has a nonzero determinant. Well $(Ax,y)=(x,A^*y)=(x,0) = 0$ for all $x$. It follows $y$ is perpendicular to the range of $A$. But the range of $A$ includes all elements in the vector space because it has a nonzero determinant. In particular, it includes $y$ (choose $x=A^{-1}y$). Thus, $(y,y)=0$. By the properties of inner products, we conclude that $y=0$ and hence that $A^*$ has a nonzero determinant.

Answer 2

The term you're looking for is "adjoint."

The thing to notice here is that $(A^{*})^{-1} = (A^{-1})^*$. Thus if $A$ is invertible, so is $A^*$.

Answer 3

$$\det A^*=\det \overline A^t=\det\overline A=\overline{\det A}$$

The last equality following from the well-known properties of the complex conjugate