Let $A$ be an Hermitian matrix in a vector space $V$, and let $U\le V$ be a subspace of $V$.

If $U$ is invariant under $A$, then the maximum of $\langle x,Ax\rangle$ over all unit vectors $x\in U$ equals the largest eigenvalue of $A$ whose eigenvector is in $U$ (and we know $U$ is spanned by eigenvectors of $A$, as otherwise it wouldn't be invariant under its action). Similar ideas are used for example to prove the min-max principle.

What about subspaces $U$ that are not invariant under $A$? More precisely, is there a way to find $$\max_{x\in U}\frac{\langle x,Ax\rangle}{\|x\|^2}$$ for arbitrary subspaces $U$? Of course, feel free to remove the Hermitianity constraint if the problem is better posed in a more general setting.

Decomposing an arbitrary $x\in U$ as $x=\sum_k c_k x_k=\sum_j d_j u_j$ where $x_k$ are a basis of eigenvectors of $A$ and $u_k$ an orthonormal basis for $U$, we have $$\langle x,Ax\rangle = \sum_k \lambda_k |c_k|^2,$$ but the problem is that the maximisation is constrained to those coefficients $(c_k)$ such that $\sum_k c_k x_k\in U$,

  • $\begingroup$ If $P$ is the projector onto the subspace $U$ then for all $ x\in U$ we have $\langle x, Ax\rangle = \langle P x, A P x \rangle = \langle x, PAP x\rangle$. Then $PAP$ Is a Hermitian matrix which leaves the subspace $U$ invariant and so we can solve it using the same method presented in the question. Would such a method be sufficient? $\endgroup$
    – Rammus
    Jun 1, 2020 at 21:25
  • $\begingroup$ @Rammus , you may be waiting for someone to copy your comment and write it as an answer. $\endgroup$
    – user91684
    Jun 1, 2020 at 22:06
  • $\begingroup$ @Rammus that sounds like a good approach. Maybe complemented by an observation about the relation between the largest eigenvalues of $A$ and $PAP$ $\endgroup$
    – glS
    Jun 1, 2020 at 23:22
  • $\begingroup$ @glS I'm not sure if I know of any relation that is not just trivial, i.e., $\|PAP\| \leq \|P\|\|A\|\|P\| = \|A\|$. This can be equality if the eigenvector corresponding to the largest absolute eigenvalue is contained within $U$. I'll update my answer if I think of something though. $\endgroup$
    – Rammus
    Jun 2, 2020 at 8:06
  • 1
    $\begingroup$ @glS It's not quite equivalent to the unconstrained maximization of $PAP$. To see this imagine $P$ projects onto the eigenspace of the strictly negative eigenvalues of $A$ (assuming it has some). Then, maximizing over U would give something strictly negative whereas maximizing over $V$ would give $0$. If you add an absolute value around the inner product then these two optimizations would be the same. $\endgroup$
    – Rammus
    Jun 3, 2020 at 7:50

1 Answer 1


Let P be the projector onto the subspace $U$. Then for all $x \in U$ we have $$ \begin{aligned} \langle x, A x \rangle &= \langle P x, A P x \rangle \\ &= \langle x, PAP x\rangle. \end{aligned} $$ Therefore, we have $$ \max_{x\in U}\frac{\langle x,Ax\rangle}{\|x\|^2} = \max_{x\in U}\frac{\langle x,PAPx\rangle}{\|x\|^2}. $$ Now $PAP$ is an operator that leaves the subspace $U$ invariant and so as noted in the question the maximum is given by the largest eigenvalue of $PAP$ whose eigenvector is in $U$. Note that we have an upper bound of $\lambda_{\max}(A)$ (largest eigenvalue of A) which can be seen be noticing the original problem is upper bounded by the same problem but with a maximization over the whole space $V$.


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