The following theorem is a classical result, cf. Fischer  and Courant & Hilbert  (Vol. 1, Ch. I, §4). Our treatment follows Haemers  (Ch. I).
Proof: Let be an orthonormal basis of eigenvectors of such that for all , and let be such a basis for , with for all . Write and .
By expanding the vector on the basis one sees that if , then . Similarly it follows that if , then . In both cases, equality implies that we have an eigenvector.
(i) Let be a nonzero vector in
(this is a subspace of dimension at least one). Then
(iii) Induction on . We may suppose that . Now so that , and the above applies.
(iv) We have for all . Since the form a basis, it follows that .
Remark This theorem generalizes directly to complex Hermitean matrices instead of real symmetric matrices (with conjugate transpose instead of transpose) with virtually the same proof.
If the assumptions of (iv) hold, we say that the interlacing is tight. Below we formulate some corollaries in terms of graphs, since that is what we are mainly interested in. Of course analogous statements hold for arbitrary real symmetric matrices (and we shall use such an interlacing result to derive the Krein conditions).
Remark As an application we can show that if is a strongly regular graph that is not complete multipartite, then for each vertex the subgraph is connected. Indeed, if not, then its largest eigenvalue would have multiplicity at least two, and hence would be not larger than the second largest eigenvalue of . Then for , i.e., , contradiction.
For more detailed eigenvalue inequalities, see Haemers .