# Isomorphism to subspaces of different dimensions

1. Jan 25, 2016

### says

1. The problem statement, all variables and given/known data
Given the linear transformations
f : R 3 → R 2 , f(x, y, z) = (2x − y, 2y + z), g : R 2 → R 3 , g(u, v) = (u, u + v, u − v), find the matrix associated to f◦g and g◦f with respect to the standard basis. Find rank(f ◦g) and rank(g ◦ f), is one of the two compositions an isomorphism? If yes find its inverse.

2. Relevant equations
f=
[2 -1 0]
[0 2 1]

g=
[1 0]
[1 1]
[1 -1]

3. The attempt at a solution
f◦g =
[1 -1]
[3 1]

g◦f=
[ 2 -1 0 ]
[ 2 1 1 ]
[ 2 -3 -1 ]

I row reduced both matrices, I don't want to write them out here, but the rank (g ◦ f) = 2, it has one free variable. Rank(f ◦g) = 2. I think the real question I have here is with the last part of the question.

'Is one of the two compositions an isomorphism?' For a linear transformation to be an isomorphism is has to be injective and surjective. Is the very nature that this L.T maps from R3 to R2 and vice versa reason enough to say it is not an isomorphism?

I found an inverse of
f◦g =
[1/4 1/4]
[-3/4 1/4]

2. Jan 25, 2016

### Dick

Yes, you can show g◦f cannot be an isomorphism without doing any row reduction. Can you show that? Think rank-nullity theorem.

3. Jan 25, 2016

### Samy_A

To answer the question: "Is the very nature that this L.T maps from R3 to R2 and vice versa reason enough to say it is not an isomorphism?"
It is correct that a linear transformation from $\mathbb R²$ to $\mathbb R³$ or from $\mathbb R³$ to $\mathbb R²$ cannot be an isomorphism.
But: g ◦ f maps $\mathbb R³$ to $\mathbb R³$,and f ◦ g maps $\mathbb R²$ to $\mathbb R²$, so these could conceivably be isomorphisms. There you have (a little) more work to do. As Dick suggested, for g ◦ f it is relatively easy to show it is not an isomorphism just by reasoning.

4. Jan 25, 2016

### says

Ahhh yes! f ◦ g is an isomorph. I've got my working in a photo attached.

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