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I'm having trouble simplifying the following expression in matrix form:

$$\mathbf{X}(\mathbf{X}+a\mathbf{I})^{-1}$$

Where $\mathbf{X}$ is an invertible $n \times n$ matrix, $a$ is a scalar value, and $\mathbf{I}$ is the identity matrix.

I reasoned that since the product of a matrix times its inverse is the identity matrix, the product of a matrix times the inverse of a "shifted" matrix is simply the shift value. In other words, the above would simplify to $a\mathbf{I}$. However, I don't know if that is correct, and if it is I don't know the linear algebra steps to prove it.

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    $\begingroup$ $\mathbf{X}(\mathbf{X}+a\mathbf{I})^{-1} = ((\mathbf{X}+a\mathbf{I})(\mathbf{X}^{-1}))^{-1} = (\mathbf{I}+a\mathbf{X}^{-1})^{-1}$. $\endgroup$ – njguliyev Oct 1 '13 at 19:51
  • $\begingroup$ I think that your are missing the hypothesis $\det (\mathbf X + a \mathbf I )\neq 0.$ For example, if $\mathbf X = -a\mathbf I $ the above expression doesn't make sense. $\endgroup$ – pppqqq Oct 1 '13 at 20:17
  • $\begingroup$ When somebody asks "please do something with this," one normally assumes the existence of "this" implicitly. $\endgroup$ – Algebraic Pavel Oct 2 '13 at 1:47
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$X(X+aI)^{-1} = ((X+aI) X^{-1})^{-1} = (I+ a X^{-1})^{-1}$.

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