I have a problem in which, in principle I can apply twice Sherman-Morrison formula but it seems to me that for this case, there should be a simpler solution so my question is "May the process described below simplified thanks to the fact that the matrix is Hermitian?". Moreover, invertibility of the matrices is known (for the particular matrices I use) never to be a problem (it is a consequence of some existence and uniqueness theorem). Suppose I have an infinite hermitian matrix $A=(a_{i,j})_{i,j=0}^{\infty}$ and I want to find the inverses of the right-up corner diagonal matrices $A_n= (a_{i,j})_{i,j=0}^n$ for each value of $n$. I could decompose $A_{n+1}$ as an $n \times n$ block containing $A_n$, $u:= ((a_{i,n+1})_{i=0}^n,0)$, $u^*$, the conjugate transpose of $u$, and the corner element $a_{n+1,n+1}$. Call $B$ to the matrix containing the blocks $A_{n}$ and $a_{n+1,n+1}$ but zeros on the other positions. Then $B^{-1}$ is a matrix with blocks $A_n^{-1}$, a vector of zeros, and its transposed and $a_{n+1,n+1}^{-1}$. Define $C$ the matrix equal to $B$ except that we update the last column by summing $u$ (so $C$ is also equal to $A_{n+1}$ substituting $u^*$ by zeros). From that, Sherman-Morrison formula, gives me the inverse of the matrix $C$, $C^{-1}$, depending on $B^{-1}$ and $u$ (the summing of a vector process can be considered as summing the matrix which is result of multiplying pointwise a vector $(0,...,0,1)$ with $((a_{i,n+1})_{i=1}^n,0)$). Once I have the inverse of $C$, I can repeat the process to update with the transpose conjugate vector $u^*$ and form a matrix $D$ and apply Sherman-Morrison formula to obtain $D^{-1}$ from $C^{-1}$, and the elements of $u^*$.

It seems like the fact that everything is hermitian would be useful to reduce the number of steps here, right?

If I can work in abstract with this inversion process, this would have important applications in complex analysis.


I'm not sure the fact that the matrix is Hermitian simplifies things much. Rather than the Sherman-Morrison formula, use the more general Woodbury matrix identity, essentially reducing the problem to the solution of a 2x2 matrix.

A simplified version of it should work for you:



$$B=\pmatrix{A_n&0\\0&a_{n+1,n+1}}$$ $$C=\pmatrix{u&e_{n+1}}$$ $$R=\pmatrix{e_{n+1}^*\\u^*}$$

And $e_{n+1}$ is the component unit vector (all $0s$ except the last coefficient which is $1$):


Let us define $v$ as $u$ with the last coefficient ($0$) dropped, so:

$$u=\pmatrix{v\\0}$$ $$C=\pmatrix{v&0\\0&1}$$ $$R=\pmatrix{0&1\\v^*&0}$$ $$I+RB^{-1}C=I+\pmatrix{0&1\\v^*&0}\pmatrix{A_n^{-1}&0\\0&a_{n+1,n+1}^{-1}}\pmatrix{v&0\\0&1}=\pmatrix{1&a_{n+1,n+1}^{-1}\\v^*A_n^{-1}v&1}$$ $$\left(I+RB^{-1}C\right)^{-1}=\frac{1}{1-a_{n+1,n+1}^{-1}v^*A_n^{-1}v}\pmatrix{1&-a_{n+1,n+1}^{-1}\\-v^*A_n^{-1}v&1}$$ $$C\left(I+RB^{-1}C\right)^{-1}R=\frac{1}{1-a_{n+1,n+1}^{-1}v^*A_n^{-1}v}\pmatrix{-a_{n+1,n+1}^{-1}vv^*&v\\v^*&-v^*A_n^{-1}v}$$ $$B^{-1}C\left(I+RB^{-1}C\right)^{-1}RB^{-1}=\frac{\pmatrix{-a_{n+1,n+1}^{-1}A_n^{-1}vv^*A_n^{-1}&a_{n+1,n+1}^{-1}A_n^{-1}v\\a_{n+1,n+1}^{-1}v^*A_n^{-1}&-a_{n+1,n+1}^{-2}v^*A_n^{-1}v}}{1-a_{n+1,n+1}^{-1}v^*A_n^{-1}v}$$

Or slightly cleaner:


And there you have your correction matrix which you subtract from $B^{-1}$ to get $A_{n+1}^{-1}$.

$$A_{n+1}^{-1}=\pmatrix{A_n^{-1}&0\\0&a_{n+1,n+1}^{-1}}-\frac{\pmatrix{-A_n^{-1}vv^*A_n^{-1}&A_n^{-1}v\\v^*A_n^{-1}&-a_{n+1,n+1}^{-1}v^*A_n^{-1}v}}{a_{n+1,n+1}-v^*A_n^{-1}v}$$ $$A_{n+1}^{-1}=\pmatrix{A_n^{-1}+\frac{A_n^{-1}vv^*A_n^{-1}}{a_{n+1,n+1}-v^*A_n^{-1}v}&\frac{-A_n^{-1}v}{a_{n+1,n+1}-v^*A_n^{-1}v}\\\frac{-v^*A_n^{-1}}{a_{n+1,n+1}-v^*A_n^{-1}v}&\frac{1}{a_{n+1,n+1}-v^*A_n^{-1}v}}$$

Computationally, the fact that $A$ is Hermitian simplifies things since $v^*A_n^{-1}$ is the Hermitian of $A_n^{-1}v$, but it may not simplify your theoretical work.

  • $\begingroup$ A better derivation would have been to use the blockwise matrix inverse formula. The derivation above counts on $a_{n+1,n+1}$ being non-zero where the derivation of the blockwise matrix inverse formula does not. In any case the formula above only requires that A be non-singular and $a_{n+1,n+1}-v^*A_n^{-1}v$ be non-zero. $\endgroup$ – Paul Hanson Mar 6 '15 at 4:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.