Matrix Multiplication -- Commutivity versus Associativity

[Crystal037]

Homework Statement

If A is a square matrix, then A^2.A^3=A^3.A^2.
Is the above statement true or false????

Relevant Equations

A^2.A^3=A^3.A^2.

According to me matrix multiplication is not commutative. Therefore A^2.A^3=A^3.A^2 should be false. But at the same time matrix multiplication is associative so we can take whatever no. of A's we want to multiply i.e A^5=A.A^4 OR A^5=A^2.A^3

 

[RPinPA]

According to me matrix multiplication is not commutative. Therefore A^2.A^3=A^3.A^2 should be false.

Matrix multiplication is not commutative which means AB is NOT NECESSARILY BA. That doesn't mean that they have to be unequal.

There are many cases where the matrices commute, such as this one. Another one is a matrix and its inverse. The definition of the inverse of $A$ is a matrix $A^{-1}$ such that $A^{-1}A = A A^{-1} = I$. The product most definitely commutes, and in either order gives you the identity matrix.

Two additional trivial examples: the identity matrix, $IA = AI = A$ and the zero matrix $0 A = A 0 = 0$.

 

[Crystal037]

I have tried for some matrices an even for general matric having elements a11,a12...
The result shows that the above statement is true

 

[Orodruin]

I have tried for some matrices an even for general matric having elements a11,a12...
The result shows that the above statement is true

You cannot show a statement based on a couple of examples. The correct argumentation is using associativity as you outlined in the first post.

 

[RPinPA]

I have tried for some matrices an even for general matric having elements a11,a12...
The result shows that the above statement is true

Sure, for the simple reason that both expressions are equal to $A * A * A * A * A$, regardless of how $A$ is defined. And as you said, there's an associative property for matrices. So $A^2 A^3 = A^3 A^2 = A A^4 = A A^3 A = \cdots$

By the way, remember that a scalar can be interpreted as a 1 x 1 matrix. And scalars definitely commute under multiplication. So $ab = ba$ is another example of "matrices" that commute under multiplication.

 

[Crystal037]

So to conclude any matrix A , A^n=A*A^(n-1)=A^2*A^(n-2)...=A^n-1*A.

 

[Orodruin]

So to conclude any matrix A , A^n=A*A^(n-1)=A^2*A^(n-2)...=A^n-1*A.

Yes, and even further if the matrix is invertible:
$$
A^n = A^{n-k} A^k,
$$
where $k$ is any number in $\mathbb Z$ (including negative ones or ones larger than $n$).

 

[Crystal037]

But what about matrices which are non-invertible. Please explain through an example

 

[haruspex]

But what about matrices which are non-invertible. Please explain through an example

Then all the exponents must be nonnegative, i.e. your list in post #6.

 

[Crystal037]

Yes correct Also Is dividing a matrix A with B is same as multiplying A with additive inverse of B?

 

[haruspex]

Yes correct Also Is dividing a matrix A with B is same as multiplying A with additive inverse of B?

Multiplicative inverse.

 

[Crystal037]

yeah Yeah sorry multiplicative inverse
So is the above statement true for multiplicative inverse or simply inverse

 

[haruspex]

yeah Yeah sorry multiplicative inverse
So is the above statement true for multiplicative inverse or simply inverse

Yes. In fact, multiplying by the multiplicative inverse is the definition of division in groups generally.