Finding a matrix and its inverse in terms of another one.

Question

I have a symmetric matrix $A$. I wanted to modify its first row (thus it is first column). Let's call the new row $c^T$. How do I write the modified matrix $B$ in terms of $A$ and $c$. What is the inverse of $B$ in terms of $A$ and $c$?

Answer 1

Let $\alpha$ denote the first column of $A$, and let $e_1$ denote the vector $e_1 = (1,0,\dots,0)$ (i.e. the first standard basis vector). Let $a_{11}$ denote the top-left entry of $A$, and let $c_1$ denote the first entry of $c$. Denote $\beta = (c - \alpha) - \frac 12 (c_1 - a_{11})e_1$.

If you prefer to work with fewer symbols, $\alpha$ can be expressed as $\alpha = Ae_1$, $a_{11}$ can be expressed as $a_{11} = e_1^TAe_1$, and $c_1$ can be expressed as $c_1 = e_1^Tc$.

The matrix $B$ can be expressed in the following forms: $$ B = A + \beta e_1^T + e_1 \beta ^T\\ B = A + UCV, \text{ where} \quad U = V^T = \pmatrix{e_1 & \beta}, \quad C = \pmatrix{0&1\\1&0}. $$ In order to find the find the inverse of $A$ (assuming that this inverse exists), one could apply the Woodbury matrix identity to $B = A + UCV$ to find that $$ B^{-1} = \left(A + UCV \right)^{-1} = A^{-1} - A^{-1}U \left(C^{-1} + VA^{-1}U \right)^{-1} VA^{-1}. $$ Note that obtaining the inverse of the matrix $\left(C^{-1} + VA^{-1}U \right)$ requires knowing the inverse of $A$, but $C^{-1} = C$ and the matrix $\left(C^{-1} + VA^{-1}U \right)$ is only of size $2 \times 2$.