Throughout, matrices are real and all orthogonality statements refer to the Frobenius inner product
Theorem (Eckart–Young–Mirsky)
Among all matrices of rank at most , is a best approximation to for every unitarily invariant norm. In particular,
and
The general result follows from Mirsky's singular-value inequality. The two common cases admit short direct proofs.
Frobenius-norm proof
Take any with , and let be the orthogonal projector onto the column space of . Since , the two terms in
are orthogonal in the Frobenius inner product. Hence
Also, and are orthogonal because
where . Therefore
Using the SVD of ,
For any and ,
The diagonal identity again uses . Thus the summands are mutually orthogonal, and
Set . Since an orthogonal projector is positive semidefinite and contractive,
Extend to an orthonormal basis of , and let
Since is orthogonal, cyclicity of the trace gives
Since is a symmetric idempotent matrix, it is orthogonally diagonalizable with eigenvalues in . Its rank counts the eigenvalues equal to , while its trace sums them. Hence
Therefore
Under these constraints, moving weight from a later index to an earlier one cannot decrease . Since , the maximum is attained at
and therefore
Combining this estimate with gives
Equality holds for , which proves the Frobenius-norm result.
If , this Frobenius-norm minimizer is unique. If , rotations inside the corresponding singular subspaces give other minimizers.
Spectral-norm proof
Let
Since , the intersection contains a nonzero vector. Choose a unit vector in this intersection. Then and
It follows that . On the other hand,
so . This gives the spectral-norm result.
Unlike the Frobenius-norm minimizer when , the spectral-norm minimizer need not be unique. Indeed, replacing the retained singular values in by nonzero values satisfying
still gives a rank- approximation with spectral-norm error .