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If $A$ and $B$ are symmetric positive semi-definite matrices such that $A \succeq B$ (i.e., $A-B$ is p.s.d.), does the following hold: $\Vert A\Vert_2 \geq \Vert B \Vert_2$?

Here $\Vert \cdot \Vert_2$ is the induced/operator 2-norm defined as

$\| A \|_2 = \sup \limits _{x \ne 0} \frac{\| A x\| _2}{\|x\|_2}$

iMath
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  • If positive semidefinitness includes symmetry, then yes. Check the eigen values of $A$ and $B$. – user251257 May 01 '18 at 18:02
  • Yes they are symmetric, but I don't know how the eigenvalues of $A-B$ are related to the eigenvalues of $A$ and $B$.. – iMath May 01 '18 at 18:10
  • $|A|_2$ is the largest eigenvalue of $A$... – user251257 May 01 '18 at 18:12
  • Yes, but how do we know that's larger than the largest eigenvalue of $B$? All we know is that the eigenvalues of $A-B$ are all non-negative, but there is no obvious way to relate these sets of eigenvalues to each other except when, e.g., $A$ and $B$ are both diagonal. – iMath May 01 '18 at 18:18

1 Answers1

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Assuming $A$ and $B$ are symmetric:

Let $u$ be an unit eigen vector to eigen value $\|B\|_2$ of $B$. Then, we have $$ 0 \le u^T(A-B) u = u^T A u - u^T B u \le \|A\|_2 - \|B\|_2. $$

user251257
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