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Let $<$ denote the matrix order ([[Loewner order]]). Most the following inequalities apply to more general linear operators.
As you'd expect, there are matrix versions of the Markov and Chebyshev inequalities.
A good overview is given in Appendix C here: https://arxiv.org/pdf/quant-ph/0012127
For more sophisticated matrix inequalities (which often use the following inequalities in the background) see [[matrix inequalities]].
Markov
For matrix $X$ and PSD $A$,
$$
\Pr(X \not\le A) \leq \Tr(\E[X]A^{-1}).
$$
Of course, this reduces to usual Markov inequality ([[basic inequalities#Markov's inequality|basic inequalities:Markov's inequality]]).
Chebyshev
Markov's inequality extends to Chebyshev's inequality in the same way as in the scalar case:
$$
\Pr(|X-\E X| \not\leq A) \leq \Tr(\E|X-\E X|^2 A^{-2}).
$$
A Chernoff-like inequality
For matrix $Y$, symmetric matrix $B$ and matrix $T$ such that $T^* T >0$ where $T^$ is the [[conjugate transpose]] of $T$, we have
$$
\Pr(Y\not\leq B)\leq \Tr(\E\exp(TYT^ - TBT^)).
$$
We can prove this easily using Markov's inequality:
$$
\begin{align}
\Pr(Y\not\leq B) &= \Pr(Y - B\not\leq 0) \
&= \Pr(T YT^ - TBT^* \not\leq 0) \
&= \Pr(\exp(TYT^* - TBT^)\not\leq I) \
&\leq \Tr(\E\exp(TYT^ - TBT^)I^{-1}).
\end{align}
$$
Here we've used that the exponential of the zero matrix is the identity. Note also that since the trace is a linear operator, so
$$
\Tr(\E\exp(TYT^ - TBT^)) = \E\Tr(\exp(TYT^ - TBT^*)).
$$