# Let $\lambda$ be a real eigenvalue of matrix $AB$. Prove that $|\lambda| > 1$.

Let $$A$$ and $$B$$ be real symmetric matrices with all eigenvalues strictly greater than 1. Let $$\lambda$$ be a real eigenvalue of matrix $$AB$$. Prove that $$|\lambda| > 1$$.

My solution:

Let $$a$$ and $$b$$ be eigenvalues of $$A$$ and $$B$$ corresponding the eigenvectors $$y$$ and $$x$$, respectively.

Looking at the following dot product: $$\langle ABx,y \rangle = \langle Bx,A^Ty \rangle=\langle Bx,Ay \rangle = \langle bx,ay \rangle=ab\langle x,y \rangle=\langle abx,y \rangle$$ we get $$(AB)x=(ab)x$$ Therefore, $$\lambda := ab$$ is an eigenvalue of $$AB$$. Since $$a>1$$ and $$b>1$$, it follows that $$\lambda > 1$$

However, it doesn't seem ok as the problem was actually asking to prove $$|\lambda|>1$$. Indeed, $$\lambda > 1 \implies |\lambda|>1$$, but then the problem wouldn't write $$|\lambda|$$ in my opinion.

The given solution:

The transforms given by $$A$$ and $$B$$ strictly increase the length of every nonzero vector, this can be seen easily on a basis where the matrix is diagonal with entries greater than $$1$$ in the diagonal. Hence their product $$AB$$ also strictly increases the length of any nonzero vector, and therefore its real eigenvalues are all greater than $$1$$ or less than $$-1$$.

Any help is appreciated.

• You are already given a solution. What additional help do you need? – user1551 Jul 22 '20 at 18:28
• @user1551 I would appreciate any help on outlining why my solution yields just $\lambda > 1$ not $|\lambda| > 1$ as the problem asks – VIVID Jul 22 '20 at 18:30
• Your answer doesn't sound right. You've correctly shown that $\langle ABx,y\rangle=\langle abx,y\rangle$ when $(a,y)$ is an eigenpair of $A$ and $(b,x)$ is an eigenpair of $B$. However, the inference $ABx=abx$ is incorrect because it wrongly implies that $A(bx)=abx$, i.e. $x$ is an eigenvector of $A$. – user1551 Jul 22 '20 at 18:46
• @user1551 Thanks for catching that. Now I know my mistake, so will try to proceed that solution in a different direction. – VIVID Jul 22 '20 at 18:53

$$AB$$ is similar to $$A^{1/2}BA^{1/2}$$. Hence its eigenvalues are positive.

Full solution:

Since $$A$$ and $$B$$ are positive definite, $$AB$$ is similar to the positive definite matrix $$A^{1/2}BA^{1/2}$$. Hence $$AB$$ has a positive spectrum.

Furthermore, since $$A$$ and $$B$$ are unitarily diagonalisable and their eigenvalues are greater than $$1$$, we have $$\|Ax\|_2,\|Bx\|_2>\|x\|_2$$ for all nonzero vector $$x$$. It follows that $$\|ABx\|_2=\|A(Bx)\|_2>\|Bx\|_2>\|x\|_2$$ for all nonzero $$x$$. As $$AB$$ has a positive spectrum, the eigenvalues of $$AB$$ must be positive numbers greater than $$1$$.

Alternative solution (that uses the induced $$2$$-norm for matrices). Since $$A,B\succ I$$, we have $$0\prec A^{-1},B^{-1}\prec I$$ and $$\|A^{-1/2}B^{-1}A^{-1/2}\|_2\le\|A^{-1/2}\|_2^2\|B^{-1}\|_2=\|A^{-1}\|_2\|B^{-1}\|_2<1$$. Hence $$0\prec A^{-1/2}B^{-1}A^{-1/2}\prec I$$ and $$A^{1/2}BA^{1/2}\succ I$$. Since $$A^{1/2}BA^{1/2}$$ is similar to $$AB$$, all eigenvalues of $$AB$$ are positive numbers greater than $$1$$.

• Then the problem is not as strict as possible about the range of $\lambda$? – VIVID Jul 22 '20 at 18:33
• @VIVID The conditions allow a stronger result ($\lambda>1$), but the problem statement only asks for a weaker one ($|\lambda|>1$). – user1551 Jul 22 '20 at 18:36
• Thanks for the update! But it seems I'm not quite familiar with the notation $\|Ax\|_2$. – VIVID Jul 22 '20 at 18:46
• @VIVID It's the Euclidean norm of a vector, i.e. $\|(x_1,\ldots,x_n)^T\|_2=\sqrt{x_1^2+\cdots+x_n^2}$. – user1551 Jul 22 '20 at 18:49
• Ok got it! Your answer is like a more formal version of the original solution and also more clearer for me. Thanks again! – VIVID Jul 22 '20 at 18:55