Is this considered a valid proof that, if $A$ and $I+AB$ are invertible, then so is $I+BA$, and $(I+AB)^{-1}A=A(I+BA)^{-1}\,?$

Question

(Newbie Question Warning :D)

Prove that if $A$ and $I + AB$ are invertible, so is $I+BA$, and that $$(I+AB)^{-1}A=A(I+BA)^{-1}$$

I managed to work this out with the following calculation:

$$\begin{align} (I+AB)^{-1}A &= A(I+BA)^{-1} \\ \rightarrow \quad A &= (I+AB)A(I+BA)^{-1} \\ \rightarrow \quad A &= (A+ABA)(I+BA)^{-1} \\ \rightarrow \quad A &= A(I+BA)(I+BA)^{-1} \\ \rightarrow \quad A &= A \end{align}$$

But I have no idea how to specify my proof with words... the only thing I did is deconstruct the formula I was given, is that considered a legit proof by itself?

Answer 1

No, your proof is not valid.

You are asked to prove that (1) $I + BA$ is invertible and (2) $(I + AB)^{-1} A = A (I + BA)^{-1}$. But your proof assumes that both of these statements are true to begin with, a common beginner error. A valid proof starts with statements that are already known to be true, and works towards the statements that need to be proved.

It so happens that (assuming (1) has already been proved) your proof of (2) works if you reverse all the steps:

\begin{align*} A = A &\implies A = A (I + BA) (I + BA)^{-1} \\ &\implies A = (A + ABA) (I + BA)^{-1} \\ &\implies A = (I + AB) A (I + BA)^{-1} \\ &\implies (I + AB)^{-1} A = A (I + BA)^{-1} \end{align*}

(Warning: This is not always the case. Sometimes, statement A implies statement B but statement B does not imply statement A.)

You still need to prove (1). Use the hint provided by Kavi Rama Murthy:

$A^{-1} (I + AB)^{-1} A$ is a candidate for $(I + BA)^{-1}$.

More specifically: Multiply $(I + BA)$ by $A^{-1} (I + AB)^{-1} A$ and see if you can reduce it to $I$. (Since you are aiming for $I$, it does not matter which way around the multiplication is done.) The required manipulations are similar to the ones in your proof of (2).

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Your question says:

But I have no idea how to specify my proof with words

In general, the presence or absence of words does not determine whether a proof is valid. Words are very useful in mathematics, but they are not required here.

Answer 2

$x$ is an eigenvalue of $AB$ iff $x$ is an eigenvalue of $BA$. Since $AB+I=AB-(-I)$ is invertible, $-1$ is not an eigenvalue of $AB$, hence $-1$ is not an eigenvalue of $BA$ and $|BA-(-I)|\ne0$. Hence $I+BA$ is invertible.

Note that you have started working by assuming $(I+AB)^{-1}A=A(I+BA)^{-1}$, which is needed to be proven and is not known. To prove the matrix equality, just use your chain of equations backwards, i.e. start from $A=A$, multiply RHS by $I=(I+BA)(I+BA)^{-1}$ to get $A=A(I+BA)(I+BA)^{-1}...$ and so on.

Answer 3

(Weird Answer Warning :o)

Let $A$ be what she or he wants, invertible or something else$-$it doesn't matter.

$$\quad 1-ab\,\text{ is invertible} \;\iff\; 1-ba\,\text{ is invertible}\tag{abba}$$

holds true with $\,a,b\,$ being elements of a unital ring.
In case of invertibility you have $$(1-ba)^{-1}\:=\: 1+b(1-ab)^{-1}a\tag{-1}$$ and $$\begin{align}a\,(1-ba)^{-1} & \:=\:(1-ab)^{-1}a\tag{1} \\[1ex] (1-ba)^{-1}b & \:=\:b\,(1-ab)^{-1}\,.\tag{2} \end{align}$$

Proving $($-$1)$:
If $\,1-ab\,$ is invertible then let $\,u=(1-ab)^{-1}.\,$ Notice $\,a(1-ba)=(1-ab)a\,$ and compute $$\begin{align}(1+bua)(1-ba) \:=\:1 -ba +b\,\underbrace{u(1-ab)}_{\quad =1}\,a & \:=\:1 \quad\implies 1-ba\,\text{ is left-invertible} \\[1ex] (1-ba)(1+bua) \:=\:1 -ba +b\,(1-ab)u\,a & \:=\:1 \quad\implies 1-ba\,\text{ is right-invertible}\,, \end{align}$$ hence $\,1-ba\,$ is invertible with inverse $1+b\,(1-ab)^{-1}a\,.\quad\blacksquare$

This shows "$\implies$" in $(\text{abba})$, and this is sufficient to validate the opposite direction as well since $(\text{abba})$ is clearly a symmetrical statement. $\quad\blacksquare$

There is a constructive approach to derive $($-$1)$ via formal power series, see the worthwhile
How-would-you-solve-this-tantalizing-Halmos-problem on MathOverflow.

Proving $(1,2)$: Use $($-$1)$ to get $$a\,(1-ba)^{-1} \:=\: a+ab(1-ab)^{-1}a\:=\:\big[(1-ab) +ab\big](1-ab)^{-1}a \:=\:(1-ab)^{-1}a\,,$$ which is $(1)$. Swapping $a$ and $b$ yields $(2).\quad\blacksquare$

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Writing this down happened under some sort of unification spirit because I managed to find out the following duplicates & their parents:

• $I_m - AB$ is invertible if and only if $I_n - BA$ is invertible.
• Finding inverse in non-commutative ring
  with bounty-decorated answer

Ordered by decreasing age the dupes are:

• Proof If $AB-I$ Invertible then $BA-I$ invertible.
  Its most upvoted answer contains the worthwhile link Matrix inverse identity

• $1+ab$ is a unit if and only if $1+ba$ is a unit.

• Inverse of $I+BA$ is $(I+BA)^{-1} = I-B(I+AB)^{-1}A$

• Inverse of $1-ba$ when $a,b$ are the elements of unital algebra $A$

• In a ring, $1-ab$ is invertible $\implies$ $1-ba$ is invertible (from scratch)

• if $a,b$ are elements of a unital algebra $A,$ then $1-ab$ is invertible if and only if $1-ba$ is invertible.

• $(1-ba)$ left-invertible $\implies (1-ab)$ left-invertible

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Another popular math.SE question also fits under this umbrella: it asks for the spectra of $\,ab\,$ and $\,ba\,$ where $a,b$ are elements of a unital algebra over $\mathbb C$.
In this context, $($-$1)$ generalises to a scaled version$-$you only have to stay away from zero$-$and allows to Prove that $\sigma(AB) \backslash \{0\} = \sigma(BA)\backslash \{0\} $ in a straightforward way.