Describe $\det(A^*)$ in terms of $\det(A)$

Question

Let $A=[a_{i,j}]\in \mbox{Mat}_n \mathbb{C}$. We define $A^* =[\bar{a_{ji}}]$. Describe $\det(A^*)$ in terms of $\det(A)$.

My work:

Note $A^*$ is hermitic matrix of $n\times n$. this implies elements of the main diagonal of $A^*$ are reals.

Moreover, this matrix $A^*$ is the transposed conjugated of $A$. for example for $A_{2\times 2}$ matrix

$A=\begin{pmatrix} q & a+ib\\ a-ib & k \end{pmatrix}$ with $a,b,q,k\in \mathbb{R}^2$

Then

$A^*=\begin{pmatrix} q & a-ib\\ a+ib & k \end{pmatrix}$

This implies $det(A)=qk-(a+ib)(a-ib)=qk-(a^2-aib+aib-i^2b^2)=qk-a^2+i^2b^2=qk-a^2-b^2$
and
$det(A^*)=qk-(a-ib)(a+ib)=qk-a^2-b^2$

then, $det(A)=det(A^*)$

Here my question:

I think this happen for $A_{n\times n}$ but i don't know why... Can someone help me to solve this?

Answer 1

A matrix $A$ is hermitian if and only if its conjugate transpose $A^*$ is the same as $A$. In this case, obviously, $\det(A)=\det(A^*)$, but this is not generally true.

You can decompose taking the conjugate transpose in two steps: first conjugate each entry, then transpose. Since transposition doesn't change the determinant, you are reduced to find the relationship between $A$ and $$ \overline{A}=\bigl[\,\overline{a_{ij}}\,\bigr] $$ By taking a Laplace expansion and using induction, it should be easy for you to prove that $$ \det(\,\overline{A}\,)=\overline{\det(A)} $$ and so to get the answer to your initial problem.

If you know that the determinant can be expressed as $$ \det(A)=\sum_{\sigma\in S_n} \operatorname{sgn}(\sigma) a_{1,\sigma(1)}a_{2,\sigma(2)}\dotsm a_{n,\sigma(n)} $$ where $S_n$ is the set of permutations on $\{1,2,\dots,n\}$ and $\operatorname{sgn}(\sigma)$ is $1$ or $-1$, according to $\sigma$ being even or odd, then it's just applying the rules of conjugation, because this makes essentially obvious that the determinant of the transpose is the same as the determinant of the matrix.

Answer 2

This is clear if we know that $\det(A^T)=\det(A)$, where $A^T$ is the transpose of $A$. $\newcommand{\sgn}{\text{sgn}}$ One can give a very simple proof of that, valid over any field, using the definition of $\det(A)$ in terms of permutations: If $A=(a_{j,k})$ then $$\det(A)=\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{j=1}^n a_{j,\sigma(j)},$$ where $\sgn(\sigma)$ is $1$ if $\sigma$ is even, $-1$ if $\sigma$ is odd:

Since $(A^T)_{jk}=a_{kj}$ we have $$\begin{aligned}\det(A^T)&=\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{j=1}^n a_{\sigma(j),j} \\&=\sum_{\sigma\in S_n}\sgn(\sigma)\prod_{j=1}^n a_{j,\sigma^{-1}(j)} \\&=\sum_{\sigma\in S_n}\sgn(\sigma^{-1})\prod_{j=1}^n a_{j,\sigma(j)} \\&=\det(A) \end{aligned}$$

(The second equality holds because $\sigma$ is a bijection of $\{1,2\dots,n\}$, the third holds beccause $\sigma\mapsto\sigma^{-1}$ is an idempotent bijection of $S_n$, and the last holds because $\sgn(\sigma^{-1})=\sgn(\sigma)$.)

That formula in terms of permutaions looks nasty but it's very useful in proving things about determinants.

Answer 3

In general, it's easier to keep things coordinate-free as possible. Writing out an arbitrary $2\times 2$ matrix and then slogging through a computation of $\det A$ and $\det A^\dagger$ is tedious, and it doesn't generalize to higher-dimensions. There are a couple of ways of deriving $\det(A^\dagger) = \overline{\det A}$ for arbitrary $A\in M_n(\mathbb{C})$:

• The result is clear if $A$ is diagonalizable. Such matrices are dense in $M_n(\mathbb{C})$, and both $A \to \det A$ and $A \to \det A^\dagger$ are continuous.
• By definition, the determinant acts on $\Lambda^n \mathbb{C}^n$ by $Av = (\det A)v$. By unwinding the definition of $A^\dagger$ (in terms of the inner product, not the coordinate-unfree version), it's not hard to show that $A^\dagger v = (\overline{\det A})v$ on the same space.
• For arbitrary $A = (A_{ij})$, \begin{align*} \overline{(\det A^\dagger)} &= \sum_{g\in S_n} (-1)^g \prod_{j = g(i)} \overline{(A^\dagger)_{ij}} \\ &= \sum_{g\in S_n} (-1)^g \prod_{j = g(i)} A_{ji} \\ &= \sum_{g\in S_n} (-1)^g \prod_{g(j) = i} A_{ij} \\ &= \sum_{g\in S_n} (-1)^g \prod_{j = g^{-1}(i)} A_{ij} \\ &= \sum_{g\in S_n} (-1)^{g^{-1}} \prod_{j = g^{-1}(i)} A_{ij} \\ &= \det A. \end{align*}