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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.
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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
$(I+A)^{-1}=I-\frac{\|\vec{v}\|^{2}}{\|\vec{v}\|^{2}+\langle\vec{v}, A \vec{v}\rangle} A$
by Platinum (164,290 points)

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