Let $A, B$ be positive definite matrices. What can be said about the maximum eigenvalue of $B^{1/2}AB^{1/2}$ in terms of the eigenvalues of $A$ and $B$?
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Welcome to Math.SE! Can you provide some context for your question? What motivates it? Did you get it out of a textbook? What did you try so far? – fonini Jan 24 '16 at 02:26
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What can be said: as Omnomnomnom mentioned, it is less than or equal the product of the largest eigenvalues of $A$ and $B$. Other than that, not a lot can be said.
For instance, you could have $$ A=\begin{bmatrix}2&0\\0&1\end{bmatrix},\ \ \ B=\begin{bmatrix}1&0\\0&3\end{bmatrix}. $$ Here $\lambda_1(A)=2$, $\lambda_2(B)=3$, $\lambda_1(AB)=3$.
But if $$ A=\begin{bmatrix}2&0\\0&1\end{bmatrix},\ \ \ B=\begin{bmatrix}3&0\\0&1\end{bmatrix}, $$ now $\lambda_1(A)=2$, $\lambda_2(B)=3$, $\lambda_1(AB)=6$.
Or if $$ A=\begin{bmatrix}2&0\\0&1/10\end{bmatrix},\ \ \ B=\begin{bmatrix}1/10&0\\0&3\end{bmatrix}, $$ now $\lambda_1(A)=2$, $\lambda_2(B)=3$, $\lambda_1(AB)=0.3$. Pushing this idea one can make $\lambda_1(AB)$ as close to $0$ as desired, without changing $\lambda_1(A)$ nor $\lambda_1(B)$.
Martin Argerami
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Let $\|A\|$ denote the spectral norm of $A$ (which gives the largest eigenvalue for positive definite matrices). We have $$ \|B^{1/2}AB^{1/2}\| \leq \|B^{1/2}\| \cdot \|A\| \cdot \|B^{1/2}\| = \|A\|\|B\| $$ that is, it is at most the product of the largest eigenvalues of $A$ and $B$.
Ben Grossmann
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See also Bhatia's matrix analysis for some more elaborate inequalities – Ben Grossmann Jan 24 '16 at 04:37
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Are you sure that doesn't hold with equality? This post seems to suggest that the eigenvalues of $B^{1/2}AB^{1/2}$ are equal to those of $AB$: http://math.stackexchange.com/questions/326944/evaluating-eigenvalues-of-a-product-of-two-positive-definite-matrices?rq=1 – user307522 Jan 24 '16 at 05:50
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That post is correct. However, the largest eigenvalue of $AB$ is not necessarily $|A||B|$. – Ben Grossmann Jan 24 '16 at 13:10