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...such that $\mathbf{A}$, $\mathbf{X}$ and $\mathbf{A} - \mathbf{A}\mathbf{X}$ are invertible. Now, suppose $$(\textbf{A} - \textbf{A}\textbf{X})^{-1} = \textbf{X}^{-1}\textbf{B}$$

I've proven that B is invertible as follows:

  1. $\because \mathbf{A} - \mathbf{A}\mathbf{X}$ is invertible, and given that $(\textbf{A} - \textbf{A}\textbf{X})^{-1} = \textbf{X}^{-1}\textbf{B}$,
  2. $\implies \textbf{X}^{-1}\textbf{B}$ is also invertible, and
  3. $((\textbf{A} - \textbf{A}\textbf{X})^{-1})^{-1} = (\textbf{X}^{-1}\textbf{B})^{-1}$
  4. $\mathbf{A} - \mathbf{A}\mathbf{X} = (\textbf{X}^{-1})^{-1}\textbf{B}^{-1}$
  5. $\mathbf{A} - \mathbf{A}\mathbf{X} = \textbf{X}\textbf{B}^{-1}$.
  6. Therefore there must exist a matrix $\textbf{B}^{-1}$, and $\textbf{B}$ is invertible.

Is this 'proof' of sorts sufficient, and logically sound? If not, how could it be improved?

Furthermore, how might I go about solving for $\textbf{X}$ in terms of $\textbf{A}$ and $\textbf{B}$? From the last line in the proof above, I've gotten to: $$\textbf{A}\textbf{X} + \textbf{X}\textbf{B}^{-1} = \textbf{A}$$ $$\textbf{A}^{-1}\textbf{A}\textbf{X} + \textbf{A}^{-1}\textbf{X}\textbf{B}^{-1} = \textbf{A}^{-1}\textbf{A}$$ $$\textbf{I}\textbf{X} + \textbf{A}^{-1}\textbf{X}\textbf{B}^{-1} = \textbf{I}$$ $$\textbf{X} + \textbf{A}^{-1}\textbf{X}\textbf{B}^{-1} = \textbf{I}$$ but I'm stuck. I believe I have to somehow 'pull out' the $\textbf{X}$, but I was told by my lecturer that the order of matrix multiplication is important, and is not commutative, unlike normal algebraic/arithmetic multiplication, so I can't just 'factorise out' the $\textbf{X}$ in the middle term.

Thanks very much.

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  • $\begingroup$ $B=X(A-AX)^{-1}$ if you multiply first equation with$ X$ from the right. Now show $B^{-1}=(A-AX)X^{-1}$ $\endgroup$ Sep 12, 2018 at 18:16
  • $\begingroup$ Also Matrix multiplikation IS distributive. $\endgroup$ Sep 12, 2018 at 18:23

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Your proof is flawed because you are assuming $\mathbf{B}^{-1}$ exists, which is what you must prove.

The proof that $\mathbf{B}$ is invertible is easier: $$ \mathbf{B}=\mathbf{X}(\mathbf{A}-\mathbf{A}\mathbf{X})^{-1} $$ is the product of two invertible matrices, so it is invertible. Indeed, if $\mathbf{C}$ and $\mathbf{D}$ are invertible, it is readily seen that $\mathbf{D}^{-1}\mathbf{C}^{-1}$ is the inverse of $\mathbf{C}\mathbf{D}$.

Now, how to express $\mathbf{X}$ in terms of $\mathbf{A}$ and $\mathbf{B}$ (and inverses thereof)?

You make another error: from $(\mathbf{A}-\mathbf{A}\mathbf{X})^{-1}=\mathbf{X}^{-1}\mathbf{B}$ you can deduce that $$ \mathbf{A}-\mathbf{A}\mathbf{X}=(\mathbf{X}^{-1}\mathbf{B})^{-1}= \mathbf{B}^{-1}\mathbf{X} $$ (not the other way around). Therefore $$ (\mathbf{A}+\mathbf{B}^{-1})\mathbf{X}=\mathbf{A} $$ Since the right hand side is invertible and $\mathbf{X}$ is invertible, also $\mathbf{A}+\mathbf{B}^{-1}$ is and so $$ \mathbf{X}=(\mathbf{A}+\mathbf{B}^{-1})^{-1}\mathbf{A} $$

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