0
$\begingroup$

Consider the matrix

$$ B = A + \begin{bmatrix} R_{1} & R_{2} \\ R_{1} & R_{2} \end{bmatrix} = A + \begin{bmatrix} I \\ I\end{bmatrix} \begin{bmatrix} R_{1} & R_{2} \end{bmatrix} $$ where $A$ is an invertible square matrix, and $R_1$ and $R_2$ are symmetric matrices. Notice that the block matrix $R$ is akin to a rank-one matrix, in the sense that the two rows are the same.

If $R_1$ and $R_2$ were scalar, we could use the Sherman–Morrison formula to compute $B^{-1}$ in a simple way. Is there a version of that formula that would apply when $R_1$ and $R_2$ are matrices?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

I think you are looking for something like the Woodbury matrix identity. In the link, choose: $U,V^{-1}$ to be eigenvectors corresponding to non-zero eigenvalues of the given matrix

$$UDU^{-1} = \begin{pmatrix} R_1 & R_2 \\ R_1 & R_2 \end{pmatrix}$$

$V = U^{-1}$, $C = D$ (in the link). Use only those columns in $U$ with non-zero eigenvalues in $D$.

$\endgroup$
2
  • $\begingroup$ Thanks for your help! I'll give this a try, but I suspect that doing the spectral decomposition of $R$ won't be simple in my case. $\endgroup$
    – user_lambda
    May 26 at 13:42
  • 1
    $\begingroup$ You can also use/try $U = [I,I]^T$, $C = I$, $V = [R_1,R_2]$. $\endgroup$
    – Balaji sb
    May 26 at 14:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .