This is an exercise in linear algebra.

Suppose $A,B$ are $n\times n$ unitary matrices. Show that $|\det(A+B)|\leq2^n$.


When $n=1$, this is trivially true since by the definition of unitary matrices, we must have $|A|=|B|=1$ and we can check case by case that $|\det(A+B)|\leq2$.

But I am not able to proceed with the general case when $n>1$. All I know is that $\det(A)=\det(B)=1$ when $A$ and $B$ are both unitary. But since they are both matrices, I don't know how to relate this to the quantity $|\det(A+B)|$.

Can anyone help?


1 Answer 1


$|\det(A+B)| = |\det(A(I+A^{-1}B))| = |\det(I+A^{-1}B)|$. Now, $A^{-1}B$ is unitary, so it's eigenvalues lie on the unit circle $B_1(0)$. Thus the eigenvalues of $I+A^{-1}B$ lie in $1+B_1(0)\subset B_2(0)$. I think, that's all you need.

  • $\begingroup$ Do we also have the following tighter bound for $n\ge 2$? Let $\mathbf{\lambda} = (\lambda_1, \cdots, \lambda_n)$ be the eigenvalues of $I + A^{-1}B$. We then have $|\det(I + A^{-1}B)| = \prod_{i=1}^{n}|\lambda_i| \le n^{-n}\|\mathbf{\lambda}\|_2^{2n} \le n^{-n} 4^n$ by the GM-QM inequality. $\endgroup$
    – Tom Chen
    Oct 16, 2019 at 15:03
  • $\begingroup$ @TomChen First, you forgot to take the square root. And why is $\|\lambda\|_2\le 2$? As the example $A = B$ shows, you cannot get any better bound than $2^n$. $\endgroup$
    – amsmath
    Oct 16, 2019 at 15:27
  • $\begingroup$ Ah, apologies, I thought that when you say that the eigenvalues lie within $B_2(0)$, you meant $\|\lambda\|_2 \le 2$, but you mean each individual eigenvalue is $\le 2$ in absolute value? $\endgroup$
    – Tom Chen
    Oct 16, 2019 at 15:45
  • $\begingroup$ @TomChen Of course. "The eigenvalues lie in $U$" means each eigenvalue lies in $U$. $\endgroup$
    – amsmath
    Oct 16, 2019 at 16:36

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