Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Of course $\overline M$ is the complex conjugate of an $n\times n$ matrix $M$.

Someone gave me advice to use the definition of determinant, then it means I have to use cofactor expasion here?

share|cite|improve this question
It is easier with the closed expression of $\det(A)$ as a sum over all permutation in $n$ : $\det(A)=\sum_{\sigma \in S_n}sgn(\sigma)\cdots$ – Bebop Apr 28 '13 at 15:37
Antoine's answer is great and transparent. As an alternative, you can triangularize $M=PUP^{-1}$ and observe that this reduces to the upper triangular case where the determinant is just the product of the diagonal terms. The only trick is to observe that $\overline{P^{-1}}=\overline{P}^{-1}$. But I would still go with Antoine's answer. – 1015 Apr 28 '13 at 16:06
up vote 2 down vote accepted

If you want to use the Laplace development (cofactors), then you can do it by induction.

The case for $1\times1$ matrices is obvious, so let $n>1$ and assume the result for $(n-1)\times(n-1)$ matrices.

If $A=[a_{ij}]$ is an $n\times n$ matrix, denote by $A_{ij}$ the matrix obtained by removing the $i$-th row and the $j$-th column. To not complicate notations, let $B=\bar{A}=[b_{ij}]$ with $b_{ij}=\overline{a_{ij}}$.

The expansion of $\det B$ along its first line is

$$\det B=(-1)^{1+1}b_{11}\det B_{11}+(-1)^{1+2}b_{12}\det B_{12}+\dots+ (-1)^{1+n}b_{1n}\det B_{1n}$$

By induction hypothesis, $\det B_{ij}=\overline{\det A_{ij}}$, so you can write

\begin{align} \det B&= (-1)^{1+1}\overline{a_{11}}\,\,\overline{\det A_{11}}+ (-1)^{1+2}\overline{a_{12}}\,\,\overline{\det A_{12}}+\dots+ (-1)^{1+n}\overline{a_{1n}}\,\,\overline{\det A_{1n}} \\[4pt] &= (-1)^{1+1}\overline{a_{11}\det A_{11}}+ (-1)^{1+2}\overline{a_{12}\det A_{12}}+\dots+ (-1)^{1+n}\overline{a_{1n}\det A_{1n}} \\[4pt] &=\overline{ (-1)^{1+1}a_{11}\det A_{12}+ (-1)^{1+2}a_{12}\det A_{12}+\dots+ (-1)^{1+n}a_{11}\det A_{1n} } \\[4pt] &=\overline{\det A} \end{align}

share|cite|improve this answer

You have the following definition of the determinant:

$$\det(M) = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}m_{i,\sigma(i)}.$$

Then one has $$\det(\bar M) = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}\bar m_{i,\sigma(i)} = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\overline{\prod_{i=1}^{n}m_{i,\sigma(i)}} = \overline{\sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}m_{i,\sigma(i)}} = \overline{\det M},$$ all this equalities holding because conjugation is a $\mathbb R$-algebra homomorphism.

share|cite|improve this answer
Sorry, I can't understand some of your notations. Can you explain it more simply just appropriate to linear algebra level? – postman Apr 28 '13 at 15:45
@postman you can read wikipedia for the permutation definition of determinant. – mez Apr 28 '13 at 16:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.