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This question may seem a little dumb, but I really googled around with no success.

Let $A$ and $B$ be two square matrices of size $n$ whose (real) eigenvalues are denoted by

$$\lambda_1(A) \leq \lambda_2(A) \leq \cdots \leq \lambda_n(A)$$

$$\lambda_1(B) \leq \lambda_2(B) \leq \cdots \leq \lambda_n(B)$$

What can I say about $\lambda_i (AB)$?

Rodrigo de Azevedo
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the_candyman
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    My first hit using google was http://mathoverflow.net/questions/106191/eigenvalues-of-product-of-two-symmetric-matrices. – Dietrich Burde Nov 19 '13 at 19:28
  • And for special cases I found http://math.stackexchange.com/questions/326944/evaluating-eigenvalues-of-a-product-of-two-positive-definite-matrices, http://math.stackexchange.com/questions/174737/are-there-any-bounds-on-the-eigenvalues-of-products-of-positive-semidefinite-mat – Dietrich Burde Nov 19 '13 at 19:31

1 Answers1

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Here are two results (this is exercise III.6.14 in Matrix Analysis by R. Bhatia):

Let $\lambda^\uparrow(A)$ be the vector of eigenvalues of $A$ in increasing order and $\lambda^\downarrow(A)$ be the vector of eigenvalues of $A$ in decreasing order.

Then, if $A$ and $B$ are positive definite: $$\lambda^\downarrow(A)\lambda^\uparrow(B) \prec \lambda(AB) \prec \lambda^\downarrow(A)\lambda^\downarrow(B).$$ The $\prec$ symbol means majorized; For $n$-dimensional vectors $x,y$, $x \prec y$ means that for $1 \leq k < n$: $$\sum_{i = 1}^k x_i^\downarrow \leq \sum_{i = 1}^k y_i^\downarrow$$ and $$\sum_{i = 1}^n x_i = \sum_{i =1}^n y_i$$ i.e., if we sum the $k$ largest elements in each vector, the sum of elements from $x$ will be smaller than or equal to the sum of elements from $y$, but if we sum all the elements in each vector, the sums are equal.

More generally, for any Hermitian (symmetric, if $A$ and $B$ are real) matrices $A,B$: $$\langle \lambda^\downarrow(A), \lambda^\uparrow(B) \rangle \leq \operatorname{tr} AB \leq \langle \lambda^\downarrow(A), \lambda^\downarrow(B) \rangle$$ and we have in general, for any square matrix $X$, that $$\operatorname{tr} X = \sum_{i=1}^n \lambda_i(X)$$.

Special case if $AB$ is symmetric

An interesting case occurs if the product $AB$ is also symmetric. This happens if and only if $AB = BA$. Since $A$ and $B$ are diagonalizable (since they are symmetric), this means that $A$ and $B$ are simultaneously diagonalizable, i.e. there exists $T$ such that $$\begin{align*} A &= TD_AT^{-1} \\ B &= TD_BT^{-1} \end{align*}$$ and hence $$AB = TD_AT^{-1}TD_BT^{-1} = TD_AD_BT^{-1}$$ and the product $D_AD_B$ will be a diagonal matrix, containing the eigenvalues of $AB$. Hence an eigenvalue of $AB$ will be a product of one eigenvalue from $A$ and one eigenvalue from $B$.

Calle
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    How can the majorizations in your first displayed equation hold simultaneously? (Since they both are equalities for $k=n$) – Bananach Aug 19 '21 at 08:24