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$A$ is a positive definitive Matrix:

$$ A = \begin{bmatrix} 2 & 4 \\ 3 & 8 \\ \end{bmatrix} $$

So let try $$ A^{-1}A=I \\ A^{-1/2}A^{-1/2}A=I \\ $$

Now let's use the following substitutions: $$ X=A^{-1/2} \\ Y=A^{-1/2} A \\ $$

So it follows:

$$ XY=I $$

I was looking at a proof and it claims that

$$ YX=I $$

this didn't make any sense to me -- but when I tried a few numeric examples -- the property seemed to hold.

Would appreciate any insight (maybe I'm just seeing double)

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3 Answers 3

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$Y=YI=Y(XY)=(YX)Y$
Since $Y=YI=IY$,
$I=YX$

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If $XY = I$ where $I$ is the identity matrix, then $Y=X^{-1}$ so $YX=X^{-1}X=I$

In other words, multiplication of a matrix and its inverse is commutative.

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Suppose that $A_1, A_2 \in \mathbb{R}^{n \times n}$ are diagonalizable and have the same eigenspaces. Hence, they have eigendecompositions of the form $A_1 = Q \Lambda_1 Q^{-1}$ and $A_2 = Q \Lambda_2 Q^{-1}$. Multiplying $A_1$ and $A_2$,

$$A_1 A_2 = Q \Lambda_1 Q^{-1} Q \Lambda_2 Q^{-1} = Q \Lambda_1 \Lambda_2 Q^{-1} = Q \Lambda_2 \Lambda_1 Q^{-1} = Q \Lambda_2 Q^{-1} Q \Lambda_1 Q^{-1} = A_2 A_1$$

because diagonal matrices always commute. Thus, we conclude that if two given matrices have the same eigenspaces, then they do commute. One notable case is $A_2 = A_1^{-1}$.

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