Frobenius norm of these two matrices is the same

Question

Suppose $B$ is a real symmetric matrix. Then it has spectral decomposition $Z\Lambda Z^T$, where $\Lambda$ is a diagonal matrix with $i$th diagonal entry equal to an eigenvalue $\lambda_i(B)$, $Z$ is orthogonal (i.e. $ZZ^T=I$) with $k$th column equal to the eigenvector corresponding to $\lambda_i(B)$. I'd like to show that for any positive semidefinite matrix $X$,

$$||Z\Lambda Z^T-X||_F=||\Lambda-Z^TXZ||_F,$$ where $||\cdot||_F$ is the Frobenius norm.

I've tried writing out each of the matrices' entries and computing the norm explicitly but wasn't able show that they're equal.

It looks almost like we can do the following computation, which is unfortunately nonsensical: $$||Z\Lambda Z^T-X||_F=Z^TZ||Z\Lambda Z^T-X||_F=Z^T||Z\Lambda Z^T-X||_FZ$$

$$=||Z^T(Z\Lambda Z^T-X)Z||_F=||\Lambda-Z^TXZ||_F.$$

The first equality (for instance) is nonsensical since $Z^TZ$ is the identity matrix and $||Z\Lambda Z^T-X||_F$ is a scalar.

But maybe there's some similar trick. Any guidance greatly appreciated.

Context: The claimed equality appears in Higham's paper Computing a Nearest Symmetric Positive Semidefinite Matrix.

Answer 1

Recall that if $Z$ is an $n\times n$ orthogonal matrix then $||ZA||_F=||A||_F=||AZ||_F$ for any $n\times n$ matrix $A$. Hence $$ ||Z\Lambda Z^T-X||_F=||Z(\Lambda-Z^TXZ)Z^T||_F=||\Lambda-Z^TXZ||_F$$ using the above property twice.

Answer 2

Try this one out: $$ \|Z \Lambda Z^T - X\|_F^2 = \operatorname{tr}([Z \Lambda Z^T - X]^T[Z \Lambda Z^T - X])\\ = \operatorname{tr}([Z (\Lambda - Z^TXZ)Z^T]^T[Z (\Lambda - Z^TXZ)Z^T]) $$ perhaps you can figure out the rest.