Is the linear transformation matrix T invertible

[war485]

Homework Statement

T: M₂₂ --> M₂₂ defined by T(A) = AB where B =
[ 3 2 ]
[ 2 1 ]

Is the linear transformation matrix T invertible with respect to the standard bases? If so, find it.

Homework Equations

none

The Attempt at a Solution

This is going to sound stupid, but I need help in finding what the transformation matrix T actually is before I can say if it is invertible or not. So far I got:

T(A) = AB =
[a b ] [ 3 2 ]
[c d ] [ 2 1 ]

=
[ 3a+2b 2a+b ]
[ 3c+2d 2c+d ]

and I found T(e₁₁) =
[3 2 ]
[0 0 ]

T(e₁₂) =
[ 2 1 ]
[ 0 0 ]

T(e₂₁) =
[ 0 0 ]
[ 3 2 ]

T(e₂₂) =
[ 0 0 ]
[ 2 1 ]

and now I'm stuck. Thinking ahead a little, I know that if the kernal of T = 0, it will be invertible, or if the determinant of the matrix is not 0, it is invertible.

P.S. is there an easier way to write matrices here?

 

[yyat]

What is the inverse of T? Hint: T^-1 will look a lot like T, but with something replaced by its inverse.

 

[Mark44]

Homework Statement

T: M₂₂ --> M₂₂ defined by T(A) = AB where B =
[ 3 2 ]
[ 2 1 ]

Is the linear transformation matrix T invertible with respect to the standard bases? If so, find it.

Homework Equations

none

The Attempt at a Solution

This is going to sound stupid, but I need help in finding what the transformation matrix T actually is before I can say if it is invertible or not.

No, it doesn't sound stupid to me. This transformation takes a 2x2 matrix A as input, and produces another 2x2 matrix AB as its output, where B is as you showed.

So far I got:

T(A) = AB =
[a b ] [ 3 2 ]
[c d ] [ 2 1 ]

=
[ 3a+2b 2a+b ]
[ 3c+2d 2c+d ]

and I found T(e₁₁) =
[3 2 ]
[0 0 ]

T(e₁₂) =
[ 2 1 ]
[ 0 0 ]

T(e₂₁) =
[ 0 0 ]
[ 3 2 ]

T(e₂₂) =
[ 0 0 ]
[ 2 1 ]

and now I'm stuck.

I think what is meant by asking if T is invertible with respect to the standard basis is this: Your basis vectors (the matrices e_ij) are linearly independent. Are the new vectors (the matrices T(e_ij)) also linearly independent?

Thinking ahead a little, I know that if the kernal of T = 0, it will be invertible, or if the determinant of the matrix is not 0, it is invertible.

The determinant of AB is zero some of the time, and this is completely determined by the matrix A. This means that whether T is invertible depends completely on the matrix A that is used. Calculate the determinant of AB to see this.

P.S. is there an easier way to write matrices here?

There's a nicer way using tex script, but I wouldn't call it easier.

 

[war485]

I think what is meant by asking if T is invertible with respect to the standard basis is this: Your basis vectors (the matrices e_ij) are linearly independent. Are the new vectors (the matrices T(e_ij)) also linearly independent?

Yes they're also linearly independent.

Is the inverse =
[ -1 2 ] [ a b ]
[ 2 -3 ] [ c d ]

=
[ 2c-a 2d-b ]
[ 2a-3c 2b-3d ]
?

 

[Mark44]

It's easy enough to check. Multiply what you got by AB, and it should come out to I.

I think I remember a theorem that says, if A and B are invertible, then (AB)^(-1) = B^(-1)A^(-1). The matrix you show on the left in your product looks like it might be B^(-1), but the one on the right definitely isn't A^(-1).

 

[war485]

It's starting to look really aweful. Is there another way to find the inverse?

 

[Dick]

You don't actually need to find the matrix corresponding to T. You just have to define it's inverse. It looks like you've already found the inverse of B. Think about S(X)=XB^(-1). Could S be T^(-1)?

 

[war485]

what's that X in S(X)=XB^(-1) ?

 

[Dick]

X is any matrix, just like the 'A' in T(A)=AB. I could have written S(A)=AB^(-1) as well, they both mean the same thing.

 

[war485]

This is what I got after what you said:
[ a b ] [ -1 2 ]
[ c d ] [ 2 -3 ]
=
-a+2b 2a-3b
-c+2d 2c-3d
You hinted towards this may be being the inverse and I multiplied it out by

[ 3a+2b 2a+b ]
[ 3c+2d 2c+d ]

And it doesn't look like the inverse in the end.

 

[Dick]

S is the inverse of T if S(T(A))=A=T(S(A)). S(T(A))=S(AB)=ABB^(-1)=A. You seem to be multiplying T(A)*S(A) and trying to get A or something. That's not what the inverse means.