Is the function determinant $A \rightarrow \det(A)$ a non-convex fuction?

Question

Is the function $$ \det: A\in \mathbb{M}^{n \times n}(\mathbb{R}) \rightarrow \det (A)$$ a convex function?

I think the answer is no, but I cannot prove it directly using the definition of convex function. How can I do?

Thanks for the help!

Answer 1

If $n=1$, $\det A=a_{1,1}$, which is trivially a convex function of $A$.

If $n>1$, take two diagonal matrices, with $a_{i,i}=1$ if $i\leq n/2$ and $0$ otherwise, and $b_{i,i}=0$ if $i\leq n/2$ and $1$ otherwise.

Then $\det A=\det B=0$.

However, for $t\in]0,1[$, $\det [tA+(1-t)B] > 0$, hence the function is not convex.