This was the first problem of the IMC 2014. Let $A$ and $B$ be two $n\times n$ symmetric matrices with real entries which have all their eigenvalues strictly larger than $1$. Prove all the eigenvalues of $AB$ have absolute value larger than $1$. I would like for people to try the problem and tell me about their thought process.

The solution is rather straightforward, if you know the right information. The solution is the following:

the matrix $A$,being symmetric is not only diagonalizable, but is diagonalizable by a matrix of orthogonal columns, in other words there is a matrix $C$ of pairwise orthogonal columns $C$ so that $CAC^{-1}$ is diagonal. This practically finishes the problem. Why? because the linear operation induced by $A$ increases the norm of every vector, this is because all the eigenvalues are greater than $1$, so it increases the magnitude in each of the orthogonal projections.

Hence the linear map induced by $AB$ also increases the norms, (since it is the composition of both linear transforms), this implies every eigenvalue is larger than $1$.

What I need is a proof of the fact that an appropriate $C$ exists (is this theorem well known?), and some insight as to how the thought process of solving the problem may work.

Thank you very much in advance, regards.

  • $\begingroup$ You neglected to mention that $A$ was symmetric $\endgroup$ – Omnomnomnom Jul 6 '15 at 15:25
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    $\begingroup$ If $A$ is symmetric, then the theorem you're looking for is the spectral theorem (in its finite dimensional case). $\endgroup$ – Omnomnomnom Jul 6 '15 at 15:27
  • $\begingroup$ Oh yes, thank you very much! Is this theorem well known? :) $\endgroup$ – Jorge Fernández Hidalgo Jul 6 '15 at 15:28
  • $\begingroup$ Yes, it is. ${}{}$ $\endgroup$ – Omnomnomnom Jul 6 '15 at 15:29
  • $\begingroup$ Where can I find an exposition of this level in linear algebra? $\endgroup$ – Jorge Fernández Hidalgo Jul 6 '15 at 15:30

See the comments. The missing theorem here is the spectral theorem.


I am amazed by the official (if I correctly understand the dream's post) solution. If a candidate failed to solve this exercise, how can he understand the proposed answer?

Point 1. We must diagonalize $A^2$ and not $A$ and the "good" conditions about the eigenvalues are $|\lambda_i|\geq 1$. $||Ax||^2=x^TA^2x=x^TPD^2P^Tx=y^TD^2y$ where $P=diag((\lambda_i)_i)$, $y=P^Tx$ and $||y||=||x||$. Then $||Ax||^2=\sum \lambda_i^2y_i^2\geq \sum_i y_i^2=||x||^2$.

Point 2. By composition of $A,B$, we obtain $||ABx||\geq ||x||$. And then ?? Of course, it is well-known that $AB$ is diagonalizable with $>0$ eigenvalues. Yet, this property is not used in the "official" solution. Moreover, the statement implies that the eigenvalues are real.

Anyway, if $\lambda$ is a real eigenvalue of $AB$, then clearly $|\lambda|\geq 1$; how to obtain $\lambda\geq 1$ ?

EDIT. Fortunately the answer given by dream is not the official one !

dream, your edit is still wrong; the correct question is "Prove all the REAL eigenvalues of AB have absolute value larger than 1.


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