Eigenvalue Bounds

While grading homeworks today, I came across the following bound:

Theorem 1: If A and B are symmetric n\times n matrices with eigenvalues \lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n and \mu_1 \geq \mu_2 \geq \ldots \geq \mu_n respectively, then Trace(A^TB) \leq \sum_{i=1}^n \lambda_i \mu_i.

For such a natural-looking statement, this was surprisingly hard to prove. However, I finally came up with a proof, and it was cool enough that I felt the need to share. To prove this, we actually need two ingredients. The first is the Cauchy Interlacing Theorem:

Theorem 2: If A is an n\times n symmetric matrix and B is an (n-k) \times (n-k) principle submatrix of A, then \lambda_{i-k}(A) \leq \lambda_i(B) \leq \lambda_i(A), where \lambda_i(X) is the ith largest eigenvalue of X.

As a corollary we have:

Corollary 1: For any symmetric matrix X, \sum_{i=1}^k X_{ii} \leq \sum_{i=1}^k \lambda_i(X).

Proof: The left-hand-side is just the trace of the upper-left k\times k principle submatrix of X, whose eigenvalues are by Theorem 2 bounded by the k largest eigenvalues of X. \square

The final ingredient we will need is a sort of “majorization” inequality based on Abel summation:

Theorem 3: If x_1,\ldots,x_n and y_1,\ldots,y_n are such that \sum_{i=1}^k x_i \leq \sum_{i=1}^k y_i for all k (with equality when k=n), and c_1 \geq c_2 \geq \ldots \geq c_n, then \sum_{i=1}^n c_ix_i \leq \sum_{i=1}^n c_iy_i.

Proof: We have:

\sum_{i=1}^n c_ix_i = c_n(x_1+\cdots+x_n) + \sum_{i=1}^{n-1} (c_i-c_{i+1})(x_1+\cdots+x_i) \leq c_n(y_1+\cdots+y_n) + \sum_{i=1}^{n-1} (c_i-c_{i+1})(y_1+\cdots+y_i) = \sum_{i=1}^n c_iy_i

where the equalities come from the Abel summation method. \square

Now, we are finally ready to prove the original theorem:

Proof of Theorem 1: First note that since the trace is invariant under similarity transforms, we can without loss of generality assume that A is diagonal, in which case we want to prove that \sum_{i=1}^n \lambda_i B_{ii} \leq \sum_{i=1}^n \lambda_i \mu_i. But by Corollary 1, we also know that \sum_{i=1}^k B_{ii} \leq \sum_{i=1}^k \mu_i for all k. Since by assumption the \lambda_i are a decreasing sequence, Theorem 3 then implies that \sum_{i=1}^n \lambda_i B_{ii} \leq \sum_{i=1}^n \lambda_i \mu_i, which is what we wanted to show. \square

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2 Responses to Eigenvalue Bounds

  1. tmbtw says:

    Nice! However, I would like to correct a tiny thing in your proof. Theorem 3 needs $\sum_1^n x_i = \sum_1^n y_i$. Fortunately, this condition is also satisfied when you apply Theorem 3 in your proof as $\sum_1^n B_{ii} = Trace(B) = \sum_1^n \mu_i$.

    tmbtw

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