What Does Commutation Mean for Matrices A and B?

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this,

Dose someone pleas know where they get $C = CI$ from?

Also,

What dose it mean when A and B commute?

Many thanks!

 

[fresh_42]

Homework Statement: Please see below
Relevant Equations: Please see below

For this,
View attachment 325350
Dose someone pleas know where they get $C = CI$ from?

Also,
View attachment 325351
What dose it mean when A and B commute?

Many thanks!

Commuting matrices means that $A\cdot B = B\cdot A.$ Most matrices do not commute. That means
$$
(A\cdot B)^{-1} =B^{-1} \cdot A^{-1} \neq A^{-1}\cdot B^{-1} = (B\cdot A)^{-1}
$$

Inversion and transposition, too, change the order. You can see this by checking $(A\cdot B)\cdot (A\cdot B)^{-1} =I.$

 

[ChiralSuperfields]

Commuting matrices means that $A\cdot B = B\cdot A.$ Most matrices do not commute. That means
$$
(A\cdot B)^{-1} =B^{-1} \cdot A^{-1} \neq A^{-1}\cdot B^{-1} = (B\cdot A)^{-1}
$$

Inversion and transposition, too, change the order. You can see this by checking $(A\cdot B)\cdot (A\cdot B)^{-1} =I.$

Thank you for your reply @fresh_42 !

Is my checking correct $A \cdot A^{-1} \cdot B \cdot B^{-1} = I_A \cdot I_B = I$?

Many thanks!

 

[fresh_42]

Thank you for your reply @fresh_42 !

Is my checking correct $A \cdot A^{-1} \cdot B \cdot B^{-1} = I_A \cdot I_B = I$?

Many thanks!

Yes, but a bit short. The long version is:

\begin{align*}
(A\cdot B) \cdot (A\cdot B)^{-1}&=(A\cdot B) \cdot (B^{-1}\cdot A^{-1}) \\
&= A \cdot (B \cdot (B^{-1}\cdot A^{-1}))\\
&= A\cdot (( B\cdot B^{-1})\cdot A^{-1})\\
&= A\cdot (I\cdot A^{-1})\\
&= A \cdot A^{-1} \\
&= I
\end{align*}
This proves by using the associative law of multiplication that $B^{-1}\cdot A^{-1}$ is a inverse of $(AB)^{-1}.$

I leave it to you to show that there cannot be more than one inverse, making $B^{-1}\cdot A^{-1}$ the inverse of $(AB)^{-1}.$ Same with the identity matrix. There can only be one so we do not need to distinguish between $I_A$ and $I_B$ or between left-identity $I_L\cdot A=A$ and right-identity $A\cdot I_R=A.$ Both are the same. This can also be proven.

These proofs are a bit like a puzzle playing around with the associative, possibly distributive law. A nice Sunday afternoon exercise. The trick is to proceed step by step and only use these laws plus the definitions, e.g. that $I_L\cdot A=A$ and $A\cdot I_R=A.$ Show that $I_L=I_R\,!$