A determinant identity

Topic: Estimation and inference for the multivariate normal · more questions on this topic
‹ Two factors of one covariance A small T^2 test ›

Verify numerically that for

A=\begin{pmatrix}4&1\\1&3\end{pmatrix} \qquad\text{and}\qquad \mathbf{d}=\sqrt{n}(\bar{\mathbf{x}}-\boldsymbol{\mu}_0)=(1,2)^\top,

|A+\mathbf{d}\mathbf{d}^\top|=|A|\,(1+\mathbf{d}^\top A^{-1}\mathbf{d}).

Worked answer (click to show)

Compute the rank-one update directly, then use the inverse formula as an independent check.

|A|=12-1=11.

Here

\mathbf{d}\mathbf{d}^\top=\begin{pmatrix}1&2\\2&4\end{pmatrix}, \qquad A+\mathbf{d}\mathbf{d}^\top=\begin{pmatrix}5&3\\3&7\end{pmatrix},

with determinant 35-9=26. Also

A^{-1}=\frac{1}{11}\begin{pmatrix}3&-1\\-1&4\end{pmatrix},

so

\mathbf{d}^\top A^{-1}\mathbf{d}=\frac{1}{11}(3-4+16)=\frac{15}{11}

and

|A|\bigl(1+\mathbf{d}^\top A^{-1}\mathbf{d}\bigr)=11\bigl(1+15/11\bigr)=11+15=26.

The two sides agree.