Proof of ##M^n## (matrix multiplication problem)

[member 731016]

Homework Statement

Please see below

Relevant Equations

Please see below

For,

Does anybody please know why they did not change the order in the second line of the proof? For example, why did they not rearrange the order to be $M^n = (DP^{-1}P)(DP^{-1}P)(DP^{-1}P)(DP^{-1}P)---(DP^{-1}P)$ for to get $M^n = (DI)(DI)(DI)(DI)---(DI) = D^n$

Many thanks!

 

[fresh_42]

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

For,
View attachment 326261
Does anybody please know why they did not change the order in the second line of the proof? For example, why did they not rearrange the order to be $M^n = (DP^{-1}P)(DP^{-1}P)(DP^{-1}P)(DP^{-1}P)---(DP^{-1}P)$ for to get $M^n = (DI)(DI)(DI)(DI)---(DI) = D^n$

Many thanks!

They did exactly what you proposed only with one more step that shows how the associativity law is necessary here. We first need $(AB)B^{-1}=A(BB^{-1})=AI=A$ before we are allowed to write $A$. It is important to recognize which laws of matrix multiplication are actually used. Here it's the law of associativity $(AB)B^{-1}=A(BB^{-1})$, the definition of an inverse $BB^{-1}=I$ and the definition of the neutral element $AI=A.$ So every single property of matrix multiplication, except for the distributive law since there is no addition here, has actually been used.

 

[FactChecker]

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

For,
View attachment 326261
Does anybody please know why they did not change the order in the second line of the proof? For example, why did they not rearrange the order to be $M^n = (DP^{-1}P)(DP^{-1}P)(DP^{-1}P)(DP^{-1}P)---(DP^{-1}P)$ for to get $M^n = (DI)(DI)(DI)(DI)---(DI) = D^n$

Many thanks!

Because that would not be correct. Matrix multiplication is not commutative, and it is definitely not true that a matrix, $M^n$ that is probably non-diagonal, is equal to the diagonal matrix $D^n$.

 

[fresh_42]

@FactChecker has a good point here. I didn't see that you switched the order of $P$ and $D.$

The only matrices that can be swapped are the diagonal matrices with the same value on the entire diagonal.
$$
A\cdot B = B\cdot A \text{ for all matrices } A \Longleftrightarrow D=\operatorname{diag}(d,d,\ldots,d)
$$

So if we do not have any specific information about $A,$ we must treat it like an arbitrary matrix. And that leaves us with $\begin{pmatrix}d&0&\ldots&0\\0&d&\ldots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\ldots&d\\ \end{pmatrix}$ as the only matrix that commutes with $A.$ If we consider a specific matrix $A,$ then there are possibly more matrices that commute with $A$. (Commute means $A\cdot B=B\cdot A.$) However, if all these matrices are as before, then $PD\neq DP.$