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Peter Denton, Stephen Parke, Xining Zhang, and I have just uploaded to the arXiv the short unpublished note “Eigenvectors from eigenvalues“. This note gives two proofs of a general eigenvector identity observed recently by Denton, Parke and Zhang in the course of some quantum mechanical calculations. The identity is as follows:

Theorem 1 Let {A} be an {n \times n} Hermitian matrix, with eigenvalues {\lambda_1(A),\dots,\lambda_n(A)}. Let {v_i} be a unit eigenvector corresponding to the eigenvalue {\lambda_i(A)}, and let {v_{i,j}} be the {j^{th}} component of {v_i}. Then

\displaystyle  |v_{i,j}|^2 \prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A)) = \prod_{k=1}^{n-1} (\lambda_i(A) - \lambda_k(M_j))

where {M_j} is the {n-1 \times n-1} Hermitian matrix formed by deleting the {j^{th}} row and column from {A}.

For instance, if we have

\displaystyle  A = \begin{pmatrix} a & X^* \\ X & M \end{pmatrix}

for some real number {a}, {n-1}-dimensional vector {X}, and {n-1 \times n-1} Hermitian matrix {M}, then we have

\displaystyle  |v_{i,1}|^2 = \frac{\prod_{k=1}^{n-1} (\lambda_i(A) - \lambda_k(M))}{\prod_{k=1; k \neq i}^n (\lambda_i(A) - \lambda_k(A))} \ \ \ \ \ (1)

assuming that the denominator is non-zero.

Once one is aware of the identity, it is not so difficult to prove it; we give two proofs, each about half a page long, one of which is based on a variant of the Cauchy-Binet formula, and the other based on properties of the adjugate matrix. But perhaps it is surprising that such a formula exists at all; one does not normally expect to learn much information about eigenvectors purely from knowledge of eigenvalues. In the random matrix theory literature, for instance in this paper of Erdos, Schlein, and Yau, or this later paper of Van Vu and myself, a related identity has been used, namely

\displaystyle  |v_{i,1}|^2 = \frac{1}{1 + \| (M-\lambda_i(A))^{-1} X \|^2}, \ \ \ \ \ (2)

but it is not immediately obvious that one can derive the former identity from the latter. (I do so below the fold; we ended up not putting this proof in the note as it was longer than the two other proofs we found. I also give two other proofs below the fold, one from a more geometric perspective and one proceeding via Cramer’s rule.) It was certainly something of a surprise to me that there is no explicit appearance of the {a,X} components of {A} in the formula (1) (though they do indirectly appear through their effect on the eigenvalues {\lambda_k(A)}; for instance from taking traces one sees that {a = \sum_{k=1}^n \lambda_k(A) - \sum_{k=1}^{n-1} \lambda_k(M)}).

One can get some feeling of the identity (1) by considering some special cases. Suppose for instance that {A} is a diagonal matrix with all distinct entries. The upper left entry {a} of {A} is one of the eigenvalues of {A}. If it is equal to {\lambda_i(A)}, then the eigenvalues of {M} are the other {n-1} eigenvalues of {A}, and now the left and right-hand sides of (1) are equal to {1}. At the other extreme, if {a} is equal to a different eigenvalue of {A}, then {\lambda_i(A)} now appears as an eigenvalue of {M}, and both sides of (1) now vanish. More generally, if we order the eigenvalues {\lambda_1(A) \leq \dots \leq \lambda_n(A)} and {\lambda_1(M) \leq \dots \leq \lambda_{n-1}(M)}, then the Cauchy interlacing inequalities tell us that

\displaystyle  0 \leq \lambda_i(A) - \lambda_k(M) \leq \lambda_i(A) - \lambda_k(A)

for {1 \leq k < i}, and

\displaystyle  \lambda_i(A) - \lambda_{k+1}(A) \leq \lambda_i(A) - \lambda_k(M) < 0

for {i \leq k \leq n-1}, so that the right-hand side of (1) lies between {0} and {1}, which is of course consistent with (1) as {v_i} is a unit vector. Thus the identity relates the coefficient sizes of an eigenvector with the extent to which the Cauchy interlacing inequalities are sharp.

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