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Suppose that $a$ and $b$ below are scalars, $F$ a square matrix, $v$ a column vector. I'm trying to invert the matrix $M$ of the form $$ M=\begin{pmatrix} a & v' & 0\\ v & F & 0\\ 0 & 0 & b \end{pmatrix} $$ When $v$ and $F$ are actually scalars, we can use $M^{-1}=\frac{1}{\det M}\text{adj}(M)$ but how does one handle the general case above please?

yurnero
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    Why include $b$? If there is an inverse it will have a block decomposition $\begin{bmatrix}A&0\0&b^{-1}\end{bmatrix}$. Also, is $v'$ the transpose of $v$? Or some arbitrary row vector? This is probably helpful. – 2'5 9'2 Aug 10 '15 at 07:32
  • @alex.jordan I included $b$ because it was there in the original setup. Your link is helpful. Thanks. – yurnero Aug 10 '15 at 12:44

2 Answers2

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Let us define the matrix $$ X = \begin{pmatrix} c & w^T & 0\\ w & G & 0 \\ 0 & 0 &b^{-1}\end{pmatrix}, $$ that will be our candidate for $M^{-1}$.

Then, $$ M X = \begin{pmatrix} ac + v^Tw & aw^T + v^TG & 0\\ cv + Fw & vw^T + FG & 0 \\ 0 & 0 & 1. \end{pmatrix} $$ In order to have $MX = I$, we solve the linear system $$ \tag{1} \begin{cases} ac + v^Tw = 1,\\ aw^T + v^TG = 0,\\ cv + Fw = 0,\\ vw^T + FG = I. \end{cases} $$ This can be solved by substitution, provided $F$ invertible and $a-v^TF^{-1}v \neq 0$. Indeed we get $$ \begin{cases} c = (a-v^TF^{-1}v)^{-1},\\ w = -c F^{-1}v,\\ G = F^{-1}(I-vw^T), \end{cases} $$ using only the $1^{st}$, $3^{rd}$ and $4^{th}$ equations in $(1)$. If you plug the obtained expressions of $b,w$ and $G$ in the $2^{nd}$ equation, you get an additional hypothesis for the invertibility of $M$.

Paglia
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  • From the first you obtain $ac = 1-v^T w$, and from the third you have $w = -c F^{-1}v$. Plug the second relation into the first and you obtain $ac = 1+c v^TF^{-1}v$. Hence, $c(a-v^TF^{-1}v)=1$ – Paglia Aug 10 '15 at 13:09
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    You are right. I made a mistake. :) – yurnero Aug 10 '15 at 13:12
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    I unticked (temporarily) because I couldn't verify the 2nd equation using the $c,w,G$ as you have given. I have also used the pseudo block inversion formulas to obtain an answer that looks different from yours. – yurnero Aug 11 '15 at 00:26
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By a block inversion formula, we can write $$ M^{-1}=\begin{pmatrix}N & 0 \\ 0 & b^{-1} \end{pmatrix} \quad\text{with}\quad N=\begin{pmatrix}a & v' \\ v & F\end{pmatrix}^{-1}. $$ Using the block inversion formula once more, we have $$ N=\begin{pmatrix} (a-v'F^{-1}v)^{-1} & -v'(aF-vv')^{-1}\\ -F^{-1}v(a-v'F^{-1}v)^{-1} & a(aF-vv')^{-1} \end{pmatrix}\cdot $$ In sum, $$ M^{-1}=\begin{pmatrix} (a-v'F^{-1}v)^{-1} & -v'(aF-vv')^{-1} & 0\\ -F^{-1}v(a-v'F^{-1}v)^{-1} & a(aF-vv')^{-1} & 0\\ 0 & 0 & b^{-1} \end{pmatrix}\cdot $$ One can verify that $$ MM^{-1}=\begin{pmatrix}1&0&0\\0&I&0\\0&0&1\end{pmatrix}\cdot $$ And because $M$ and $M^{-1}$ as defined above are square matrices, we infer that $MM^{-1}=I$ implies $M^{-1}M=I$. The verification that $M^{-1}$ is symmetric can be done along the line shown here.

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yurnero
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