Norm inequality for sum and difference of positive-definite matrices

Question

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.

Answer 1

This answer only answers the first question. It does so in a way that may be unnecessarily complicated for your taste.

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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 $\max\{\|X_1\|,\|X_2\|\}\leq \|X_1+X_2\|$.