Proof: Relationship between a linear map and the associated matrix

[Fochina]

Hi!
I don't understand how to demonstrate the following exercise.

Let $F: R^{n} \rightarrow R^{n}$ be a linear map which is invertible. Show that if $A$ is the matrix associated with $F$, then $A^{-1}$ is the matrix associated with the inverse of $F$.

 

[PeroK]

Hi!
I don't understand how to demonstrate the following exercise.

Let $F: R^{n} \rightarrow R^{n}$ be a linear map which is invertible. Show that if $A$ is the matrix associated with $F$, then $A^{-1}$ is the matrix associated with the inverse of $F$.

I've aksed that this be moved to Homework. In the meantime, we'd like to see your best effort at proving this.

 

[Fochina]

I am very confused and I don't know if my attempt makes sense:

Composing linear maps corresponds to multiplying the associated matrix each other.
So we know that $A$ is the associated matrix with $F$ and $F^{-1}F=I$.
Now we have that $XA$ must be the matrix associated with the identity map (the identity matrix)
and becouse $X$ must be equal to $A^{-1}$ and each matrix has a unique linear map associated, from this
it follow that $A^{-1}$ is the matrix associated to $f^{-1}$

 

[PeroK]

Composing linear maps corresponds to multiplying the associated matrix each other.

This is the key. If you know this, then the rest should follow.

So we know that $A$ is the associated matrix with $F$ and $F^{-1}F=I$.
Now we have that $XA$ must be the matrix associated with the identity map (the identity matrix)
and becouse $X$ must be equal to $A^{-1}$ and each matrix has a unique linear map associated, from this
it follow that $A^{-1}$ is the matrix associated to $f^{-1}$

This is the right idea. What you have done is:

Note that we are talking about the matrices representing the linear transformations in some fixed basis:

Let $A$ be the matrix representing $F$ and $X$ be the matrix representing $F^{-1}$.

$AX$ is the matrix representing $FF^{-1}$, which is the identity transformation, hence $AX = I$ (the identity matrix) and we see that $X = A^{-1}$.

That just seems a bit neater.

 

[Fochina]

okok the answer now seems clear to me. I've made a lot of confusion for something that now seems very simple. thank you very much

 

[HallsofIvy]

I think it should be pointed out that there is not ONE "matrix associated with a linear transformation" (that is, a matrix, A, associated with linear transformation, T, such that if T(x)= y then Ax= y). If T is a linear transformation from vector space V to vector space U then the matrix representation depends upon the bases selected for both V and U.

Of course, it is still true that if A is the matrix representing invertible linear transformation T in given bases then A is an invertible matrix and $$A^{-1}$$ is the matrix representing $$T^{-1}$$ using those same bases.