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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.

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  • $\begingroup$ You are already given a solution. What additional help do you need? $\endgroup$ – user1551 Jul 22 '20 at 18:28
  • $\begingroup$ @user1551 I would appreciate any help on outlining why my solution yields just $\lambda > 1$ not $|\lambda| > 1$ as the problem asks $\endgroup$ – VIVID Jul 22 '20 at 18:30
  • $\begingroup$ 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 $A$. $\endgroup$ – user1551 Jul 22 '20 at 18:46
  • $\begingroup$ @user1551 Thanks for catching that. Now I know my mistake, so will try to proceed that solution in a different direction. $\endgroup$ – VIVID Jul 22 '20 at 18:53
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$AB$ 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 $1$.

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

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