Question

Let \(W\) be a linear space with an inner product and \(A: W \rightarrow W\) be a linear map whose image is one dimensional (so in the case of matrices, it has rank one). Let \(\vec{v} \neq 0\) be in the image of \(A\), so it is a basis for the image. If \(\langle\vec{v},(I+A) \vec{v}\rangle \neq 0\), show that \(I+A\) is invertible by finding a formula for the inverse.

Answer 1

The solution of \((I+A) \vec{x}=\vec{y}\) is \(\vec{x}=\vec{y}-\frac{\|\vec{v}\|^{2}}{\|\vec{v}\|^{2}+\langle\vec{v}, A \vec{v}\rangle} A \vec{y}\) so
\[
(I+A)^{-1}=I-\frac{\|\vec{v}\|^{2}}{\|\vec{v}\|^{2}+\langle\vec{v}, A \vec{v}\rangle} A
\]