# How do I prove that $|\det(A+B)|\leq2^n$ where $A,B$ are $n\times n$ unitary matrices?

This is an exercise in linear algebra.

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

THOUGHTS:

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?

$$|\det(A+B)| = |\det(A(I+A^{-1}B))| = |\det(I+A^{-1}B)|$$. 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.
• 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. – Tom Chen Oct 16 '19 at 15:03
• @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 $2^n$. – amsmath Oct 16 '19 at 15:27
• 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? – Tom Chen Oct 16 '19 at 15:45
• @TomChen Of course. "The eigenvalues lie in $U$" means each eigenvalue lies in $U$. – amsmath Oct 16 '19 at 16:36