Daily Archives: June 12, 2012

The Matrix Determinant Lemma

Posted on by

I had never really heard of this result, sometimes called the Matrix Determinant Lemma, but it came up in the process of answering a relatively simple question. Suppose I have an M-dimensional jointly Gaussian vector \mathbf{X} with covariance matrix A. The differential entropy of \mathbf{X} is \frac{1}{2} \log ( (2 \pi e)^M \det(A) . Suppose now I consider some rank-1 perturbation B = A + u u^T. What choice of u maximizes the differential entropy?

On the face of it, this seems intuitively easy — diagonalize A and then pick u to be the eigenvector corresponding to the smallest singular value of A. But is there an simple way to see this analytically?

Matrix Determinant Lemma. Let A be an M \times M positive definite matrix and U and V be two M \times k matrices. Then

\det(A + U V^H) = \det(A) \det(I_k + V^H A^{-1} U).

To see this, note that

\left[ \begin{array}{cc} A & -U \\ V^{H} & I \end{array} \right] = \left[ \begin{array}{cc} A & 0 \\ V^{H} & I \end{array} \right] \cdot \left[ \begin{array}{cc} I & - A^{-1} U \\ 0 & I + V^H A^{-1} U \end{array} \right],

and take determinants on both sides.

So now applying this to our problem,
\det(A + u u^T) = \det(A) ( 1 + u^T A^{-1} u )
But the right side is clearly maximized by choosing u corresponding to the largest singular value of A^{-1}, which in this case is the smallest singular value of A. Ta-da!

Advertisement