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I have matrices $A, B, C, D \in \mathbb R^{n\times n}$ which are all symmetric positive (semi-)definite.

If I have that

\begin{align} \rho(AB) &< \gamma^2, \\ \rho(CD) &< \gamma^2, \quad \text{and}\\ \rho(BD) &< \gamma^2, \end{align}

can I conclude that $$ \rho(AC) < \gamma^2 \tag{?} $$

Here, $\rho$ denotes the spectral radius, i.e., the largest in modulo eigenvalue, and $\gamma \in \mathbb R$ is a constant.

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1 Answer 1

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No. If $n=1$ and $X \ge 0$, then $ \rho(X)=X.$

Now let $A=1, B=3, C=5, D=1/2$ and $ \gamma =2.$

Then we have

$\rho(AB) < \gamma^2,$ $\rho(CD) <\gamma^2$ and $\rho(BD) < \gamma^2$,

but $\rho(AC) > \gamma^2 .$

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