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If $X_{1}$ and $X_{2}$ are positive definite matrices, how to show that $\left\Vert X_{1}-X_{2}\right\Vert \le\left\Vert X_{1}+X_{2}\right\Vert$ for the spectral norm? and how about for the nuclear norm? Thanks in advance.

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BTW, the matrices are symmetric. –  hi9879 Jan 2 '13 at 6:50
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2 Answers 2

The spectral norm of a Hermitian matrix $X$ can be rewritten as $\|X\|=\max_{\|u\|=1}|u^\ast Xu|$ (exercise). Now for any $\|u\|=1$, we have $-u^\ast (X_1+X_2)u < u^\ast (X_1-X_2)u < u^\ast (X_1+X_2)u$ because $X_1,X_2$ are positive definite. Therefore your inequality holds.

Edit: The OP has edited the question and ask further that if the equality holds for the nuclear norm (trace norm). The answer is yes. This follows from the polarization identity $$ \|X_1+X_2\|^2-\|X_1-X_2\|^2 = 4\operatorname{tr}\,\left\{X_1X_2\right\} = 4\operatorname{tr}\,\left\{X_1^{1/2}X_2X_1^{1/2}\right\}\ge0. $$

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user1551, thank you for your answer, but how about for the nuclear norm? –  hi9879 Jan 2 '13 at 7:06
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This answer only answers the first question. It does so in a way that may be unnecessarily complicated for your taste.


If $A$ and $B$ are Hermitian matrices, let $A\leq B$ mean that $B-A$ is positive semidefinite. A Hermitian matrix is positive semidefinite if and only if all of its eigenvalues are nonnegative, and the spectral norm of a Hermitian matrix is the maximum of the absolute values of its eigenvalues.

Then $$-\|X_2\|I\leq -X_2\leq X_1-X_2\leq X_1\leq \|X_1\|I.$$ This implies that all of the eigenvalues of $X_1-X_2$ lie in the interval $[-\|X_2\|,\|X_1\|]$, which implies that $\|X_1-X_2\|\leq \max\{\|X_1\|,\|X_2\|\}$.

On the other hand, $0\leq X_1\leq X_1+X_2$ implies that $\|X_1\|\leq\|X_1+X_2\|$, and similarly $\|X_2\|\leq \|X_1+X_2\|$, so $\max\{\|X_1\|,\|X_2\|\}\leq \|X_1+X_2\|$.

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+1 Actually I think your proof is simpler than mine. –  user1551 Jan 4 '13 at 12:11
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