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I have the following question :

Proof if $I+AB$ invertible then $I+BA$ invertible and $(I+BA)^{-1}=I-B(I+AB)^{-1}A$

I managed to proof that $I+BA$ invertible

My proof :

We know that $AB$ and $BA$ has the same eigenvalues, and Since $I+AB$ invertible $-1$ is not an eigenvalue for $I+AB$ since if $-1$ is an eigenvalue then $I+AB$ is singular which is a contradiction. and since $AB$ and $BA$ has the same eigenvalues then $-1$ is also not an eigenvalue for $I+BA$ therefore $I+BA$ is also invertible.

But how do I show that $(I+BA)^{-1}=I-B(I+AB)^{-1}A$

I tried to "play" with the equations to reach one end to other end meaning that $(I+BA)^{-1}=...=...=...=I-B(I+AB)^{-1}A$

Or to show that $ I=(I+BA)^{-1}(I-B(I+AB)^{-1}A)$

But wasn't successful.

Any ideas?

Thank you!


marked as duplicate by Brahadeesh, KReiser, Adrian Keister, max_zorn, metamorphy Jan 4 at 2:49

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    $\begingroup$ You don't need the first part at all, you can show that it's invertible by showing it has an inverse. Simply compute: $ (I+BA)(I-B(I+AB)^{-1}A) $ $\endgroup$ – user223391 Aug 15 '15 at 15:52
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    $\begingroup$ See this math.stackexchange.com/questions/1078781/… on the "Related" bar. $\endgroup$ – darij grinberg Aug 15 '15 at 15:53
  • $\begingroup$ @avid19 Can you explain why I don't need the first path, If I don't know that $I+BA$ is invertible then I can't multi by $I+BA$ to get $(I+BA)(I−B(I+AB)−1A)$ $\endgroup$ – JaVaPG Aug 15 '15 at 16:40
  • $\begingroup$ @darij grinberg I've seen that and actually I asked that, but it different question. $\endgroup$ – JaVaPG Aug 15 '15 at 16:41
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    $\begingroup$ @JaVaPG You can multiply $(I + BA)(1 - B (1 + AB)^{-1}A)$ out, as you know everything is defined, then when you get that this equals $I$, you showed that $(1 - B (1 + AB)^{-1}A)=(1+BA)^{-1}$. $\endgroup$ – user223391 Aug 15 '15 at 18:07

I take it that $A$ and $B$ are $n \times n$ matrices here. In that context, $X$ is invertible with inverse $Y$ iff $XY = I$. Here we have: $$ \begin{array}{rcl} (I + BA)(1 - B (1 + AB)^{-1}A) &=& I + BA - B(I + AB)^{-1}A - BAB(I + AB)^{-1}A \\ &=& I + BA - B((I+AB)(I+AB)^{-1})A \\ &=& I + BA - BIA \\ &=& I + BA - BA \\ &=& I \end{array} $$

  • $\begingroup$ $XY=I$ is not enough in a ring; you also need $YX=I$. $\endgroup$ – darij grinberg Aug 15 '15 at 15:51
  • $\begingroup$ This is true even if $A$ and $B$ are non-square matrices such that $A$ and $B^T$ have the same dimensions. If $A$ is a row vector and $B$ is a column vector, this is is known as the Sherman-Morrison-Woodbury fotmula. $\endgroup$ – Hans Engler Aug 15 '15 at 15:53
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    $\begingroup$ @darijgrinberg: you typed that comment in less time than it took to decide between deleting my overstrong claim or making an accurate claim. If an $n\times n$ matrix has a right-inverse then it has a left-inverse and the two are the same. $\endgroup$ – Rob Arthan Aug 15 '15 at 16:01
  • $\begingroup$ @RobArthan Thank you for your answer, can you explain in the second line where did the $B$ disappeared? the $B$ on the right of $BAB$ $$I + BA - B(I + AB)^{-1}A - BAB(I + AB)^{-1}A = I + BA - B((I+AB)(I+AB)^{-1})A$$ I think that this expression should be the result $$I+BA-B((I+AB)^{-1}-B(I+AB^{-1})A$$ in the second line. $\endgroup$ – JaVaPG Aug 15 '15 at 17:19
  • $\begingroup$ The $B$ you are talking about hasn't disappeared, The expression on the right of my first line has the form $P - BQA - BABQA$, which is equal to $P - B(I + A \underline{B})QA$, where $P = I + BA$, $Q = (I + AB)^{-1}$ and I've underlined the $B$ that you think has gone missing. $\endgroup$ – Rob Arthan Aug 15 '15 at 20:50

$\large{\text{Neumann}}:$ (supposing $\rho (AB) <1$) $$(I+AB)^{-1}=I-AB+(AB)(AB)-(AB)(AB)(AB)+\cdots$$ $$B(I+AB)^{-1}A=BA-(BA)(BA)+(BA)(BA)(BA)+(BA)(BA)(BA)(BA)+\dots$$ $$B(I+AB)^{-1}A=-(I+BA)^{-1}+I$$ $$(I+BA)^{-1}=I-B(I+AB)^{-1}A$$

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    $\begingroup$ Though, the series won't converge if $\sigma(AB) > 1$ $\endgroup$ – uranix Aug 15 '15 at 15:56
  • $\begingroup$ @uranix Thanks for pointing that out, I'll add the condition, though I've read somewhere that there are weaker conditions that allow the series to converge. $\endgroup$ – GeorgSaliba Aug 15 '15 at 16:19
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    $\begingroup$ Not in classic sense, since for $A = xI$ it is the usual Taylor series for $\frac{1}{1-x}$ which diverge for $|x| > 1$. $\endgroup$ – uranix Aug 15 '15 at 16:21

Just make the product: \begin{align*} (I+BA)(I-B(I+AB)^{-1}A) ={}& I-B(I+AB)^{-1}A + BA - BAB(I+AB)^{-1}A={} \\ {}={}& I-B(I+AB)^{-1}A + BA - B(I + AB - I)(I+AB)^{-1}A ={} \\ {}={}& I-B(I+AB)^{-1}A + BA - B((I + AB)(I+AB)^{-1} - (I+AB)^{-1})A ={} \\ {}={}& I-B(I+AB)^{-1}A + BA - B(I - (I+AB)^{-1})A {} \\ {}={}& I-B(I+AB)^{-1}A + BA - BA + B(I+AB)^{-1}A = I \end{align*}

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    $\begingroup$ Sorry, there is a similar answer $\endgroup$ – user261263 Aug 15 '15 at 16:00
  • $\begingroup$ This is hard to read. You could try using the align environment $\endgroup$ – Vim Aug 15 '15 at 16:27

Instead of multiplying the two matrices to get the identity matrix, we can also 'derive' the inverse by simple matrix multiplication. This way it would be clear why the inverse looks like the way it does.

It has been already shown by OP that $I+BA$ is invertible $\Leftrightarrow I+AB$ is invertible.

Let $ D={I+BA}$ $\qquad($$I,B$ and $A$ have orders such that $D$ is defined$)$

$\Rightarrow {DB}= B+{BAB}= B (I+{AB})$

$\Rightarrow (DB)(I+{AB})^{-1}= B$

$\Rightarrow ({DB})(I+{AB})^{-1} A={BA}$

$\Rightarrow I+({DB})(I+{AB})^{-1} A=I+{BA}= D$

$\Rightarrow I= D (I-{B}(I+{AB})^{-1} A)$

$\Rightarrow {D^{-1}}=I-{B}(I+{AB})^{-1} A$

$\text{i.e.,}\quad \boxed{( {I+BA})^{-1}=I-{B}(I+{AB})^{-1} A}$


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