# Sherman-Woodbery Formula for the Inverse Matrix

## Problem 250

Let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $\R^n$, and let $I$ be the $n \times n$ identity matrix. Suppose that the inner product of $\mathbf{u}$ and $\mathbf{v}$ satisfies

\[\mathbf{v}^{\trans}\mathbf{u}\neq -1.\]
Define the matrix

\[A=I+\mathbf{u}\mathbf{v}^{\trans}.\]

Prove that $A$ is invertible and the inverse matrix is given by the formula

\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans},\]
where

\[a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}.\]
This formula is called the **Sherman-Woodberry formula**.

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## Proof.

Let us put

\[B=I-a\mathbf{u}\mathbf{v}^{\trans},\]
the matrix given by the Sherman-Woodberry formula.

We compute $AB$ and $BA$ and show that they are equal to the identity matrix $I$.

Let us first compute the matrix product $AB$. We have

\begin{align*}

AB&=(I+\mathbf{u}\mathbf{v}^{\trans})(I-a\mathbf{u}\mathbf{v}^{\trans})\\

&=I-a\mathbf{u}\mathbf{v}^{\trans}+\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}\\

&=I+(1-a)\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans} \tag{*}

\end{align*}

By using the defining formula for $a=\frac{1}{1+\mathbf{v}^{\trans}\mathbf{u}}$, we have

\[a(1+\mathbf{v}^{\trans}\mathbf{u})=1,\]
and thus

\[1-a=a\mathbf{v}^{\trans}\mathbf{u}. \tag{**}\]

Note that the third term in (*) $-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}$ contains $\mathbf{v}^{\trans}\mathbf{u}$ in the middle, and $\mathbf{v}^{\trans}\mathbf{u}$ is just a number. Thus we can factor out this number and get

\begin{align*}

-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}&=-a\mathbf{u}(\mathbf{v}^{\trans}\mathbf{u})\mathbf{v}^{\trans}\\

&=-a(\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans} \tag{***}

\end{align*}

Inserting (**) and (***) into (*), it follows that we have

\begin{align*}

AB&=I+(a\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans}-a(\mathbf{v}^{\trans}\mathbf{u})\mathbf{u}\mathbf{v}^{\trans}\\

&=I.

\end{align*}

Thus we have proved $AB=I$.

Now we compute $BA$. We have

\begin{align*}

BA&=(I-a\mathbf{u}\mathbf{v}^{\trans})(I+\mathbf{u}\mathbf{v}^{\trans})\\

&=I+\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}\\

&=I+(1-a)\mathbf{u}\mathbf{v}^{\trans}-a\mathbf{u}\mathbf{v}^{\trans}\mathbf{u}\mathbf{v}^{\trans}

\end{align*}

and this is exactly the expression (*), hence $BA=AB=I$.

Therefore, we conclude that the matrix $A$ is invertible and the inverse matrix is $B$. Hence

\[A^{-1}=I-a\mathbf{u}\mathbf{v}^{\trans}\]
and we have proved the Sherman-Woodberry formula.

## Comment.

The invertible matrix theorem says that once we have $AB=I$, then we have automatically $BA=I$ and the inverse matrix of $A$ is $B$, that is, $A^{-1}=B$.

So in the above proof, after proving $AB=I$, you may conclude that $A$ is invertible and $A^{-1}=B$.

## Related Question.

**Problem**.

Let $A$ be a singular $2\times 2$ matrix such that $\tr(A)\neq -1$ and let $I$ be the $2\times 2$ identity matrix.

Then prove that the inverse matrix of the matrix $I+A$ is given by the following formula:

\[(I+A)^{-1}=I-\frac{1}{1+\tr(A)}A.\]

See the post ↴

The Formula for the Inverse Matrix of $I+A$ for a $2\times 2$ Singular Matrix $A$

for a proof of this problem.

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