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I'm curious to know if this assumption is correct,

Consider $A_{n \times n}$ is invertible, so $AA^{-1} = A^{-1}A = I$,

Now, what I wonder is if we have lets say $(I + AB)_{n\times n}$ (an arbitrary matrix),
So is it possible to diminish $A$ and $A^{-1}$ in this situation:

$A(I+AB)A^{-1} \rightarrow (I+AB)$

I think it's possible due to the fact that $AA^{-1} = I $,
Because it's the same as saying $4 \times X \times {1 \over 4}$ is the same as $X$

Is this the same for $_{n \times n}$ matrices?
if so how can it be proven?

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2 Answers 2

up vote 5 down vote accepted

Matrix multiplication is not commutative in general. Hence, in general $$A(I + AB)A^{-1} \neq I + AB$$ If $A$ and $B$ commute i.e. if we have $AB = BA$, then $$A(I + AB)A^{-1} = A(I + BA)A^{-1} = \underbrace{A \cdot A^{-1}}_I + A \cdot B \underbrace{A \cdot A^{-1}}_I = I + AB$$ Hence, if $A$ and $B$ commute then it is true that $$A(I + AB)A^{-1} = I + AB$$

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Thanks for the explanation, if (I + AB) is invertible also, does it affect it somehow? –  res Nov 21 '12 at 1:40
1  
res: no. If $(I + AB)$ and $A$ commute, then you can obtain your result. –  amWhy Nov 21 '12 at 1:48
    
@amWhy: I see, thanks. –  res Nov 21 '12 at 1:53

Consider $A_{n \times n}$ is invertible, so $AA^{-1} = A^{-1}A = I$.
Now, what I wonder is if we have lets say $(I + AB)_{n\times n}$ (an arbitrary matrix),
So is it possible to diminish $A$ and $A^{-1}$ in this situation:

$$A(I+AB)A^{-1} \rightarrow (I+AB)\;?$$

  • Answer: Not in general, and not in most cases.

Matrix multiplication is not commutative, as is scalar multiplication. So trying to simplify $A(I+AB)A^{-1}$ to obtain $I + AB$ is not "the same as saying" $$4 \times X \times {1 \over 4} = 4 \times (\frac{1}{4} \times X) = (4\times \frac{1}{4}) \times X = X$$ which is true simply because scalar multiplication and scalar multiplication of a matrix are commutative.

That said, your result holds whenever $AB = BA$, for matrices $A, B$.


One added thought: You are correct that a matrix $A$ is invertible if and only if there is a matrix $A^{\prime}$ such that $AA^{\prime} = A^{\prime} A = I.$ Then we call $A^{\prime}$ the inverse of $A$ and denote it $A^{-1}$. By the definition of an inverse of a matrix $A$, if invertible, it is true that $A$ commutes with its inverse. But it does not follow that $A$ then commutes with every matrix.

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