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Let $A$ be 3x3 hermitian matrix with eigenvalues $\lambda_1\leq\lambda_2\leq\lambda_3$. Let $B$ be the 2x2 matrix which is the upper left-hand corner of $A$. Let $\mu_1\leq\mu_2$ be the eigenvalues of $B$. Prove that $\lambda_1\leq\mu_1\leq\lambda_2\leq\mu_2\leq\lambda_3$.

I know $B$ is hermitian since $A$ is, and I'm almost positive you want to use the minmax theorem for this, but I'm having difficulty making progress. Does anyone have any suggestions how you might show at least the first inequality $\lambda_1\leq\mu_1$?

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Sure, here's a demonstration of that first inequality. We have $$ \lambda_1 = \min \{x^*Ax : x = (x_1,x_2,x_3)^T \in \Bbb C^3, \|x\| = 1\}\\ \leq \min\left\{ x^*Ax : x = (x_1,x_2,x_3)^T \in \Bbb C^3, \|x\| = 1, x_3 = 0\right\}\\ = \min\left\{ \pmatrix{\bar x_1 & \bar x_2 & 0}A\pmatrix{x_1\\x_2\\0} : x \in \Bbb C^3,|x_1|^2 + |x_2|^2 = 1 \right\}\\ = \min\left\{ x^*Bx : x \in \Bbb C^2,\|x\| = 1 \right\} = \mu_1 $$ The last eigenvalue has the same trick but with a max. For the middle eigenvalue, we have to do this with an actual min-max, as in the theorem.

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