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I was asked the following question:

Let $A,B \in \mbox{Mat}_n(\mathbb R)$ be symmetric positive definite matrices. Show that $AB$ does not have negative eigenvalues.

My answer

I'm getting something very weird. According to my answer, $AB$ has to be positive definite. which is weird because we don't know that $AB$ is symmetric.

$B$ is symmetric so there is a basis $v_1,...,v_n$ of eigenvectors such that $Bv_i=\lambda_i$ and $\lambda_i>0$ for all $i$, and $<v_i,v_j>=0$ when $i\neq j$

let's look at $<ABx,x>$ where $x$ is some vector in $\mathbb R^n$

$<ABx,x>=<AB\sum_{i=1}^n \alpha_i v_i,\sum_{i=1}^n \alpha_i v_i> = <AB\alpha_1 v_1,\alpha_1 v_1>+<AB\alpha_2 v_2,\alpha_2 v_2>+...+<AB\alpha_n v_n,\alpha_n v_n> = \alpha_1 ^2 \lambda_1<Av_1,v_1>+\alpha_2 ^2\lambda_2<Av_2,v_2>+...+\alpha_n ^2 \lambda_n<Av_n,v_n>$

obviously for all $i$ $\alpha_i ^2$ is positive.

Because $B$ is positive definite, for all $i : \lambda_i >0$.

Because $A$ is positive definite, for all $i: <Av_i,v_i>$ is positive.

so overall, for all $i: \alpha_i ^2 \lambda_i <Av_i,v_i>$ is positive.

So we got that $<ABx,x>$ is larger than zero for any vector $x$, and so I infer that $AB$ is positive definite. And so it doesn't have negative eigenvalues.

But how can we say that it's positive definite if it's not symmetric? I did something wrong I'm sure.

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The matrix $AB$ is similar to a symmetric positive definite matrix $A^{-1/2}(AB)A^{1/2}=A^{1/2}BA^{1/2}$. So, its eigenvalues are real and positive.

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    $\begingroup$ Just in case the OP doesn't know how to find the square root of a matrix, I would like to add that if $U$ is an invertible matrix such that $UAU^{-1} = D$ is the diagonalization of $A$, we write $A^{1/2} = U^{-1} D^{1/2} U$ where $D^{1/2}$ is the diagonal matrix of the square roots of the diagonal entries of $D$. $\endgroup$ – Joel Feb 10 '14 at 14:49

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