# Linear Algebra; Isomorphism

1. May 19, 2015

### Myr73

Pre-knowledge

If V and W are finite-dimensional vector spaces, and dim(V) does not equal dim(W) then there is no bijective linear transformation from V to W.

An isomorphism between V and W is a bijective linear transformation from V to W. That is, it is both an onto transformation and a one to one.

1- Question
Let W be a proper subset of an vector space V, and let T be the projections onto W. Prove that T is not an isomorphism.

Since T is a projection onto W then, T(v)=w, therefore dim(V) > dim(W)

However since W is a proper subset of the vector space V, W is missing an element of V and therefore dim (w) is smaller then dim (V) and is not equal. Therefore it is not a one to one, and so is not an isomorphism.

This is my answer, however I am unsure if that is correct. It makes sense in my mind that a subset of a vector space is smaller then the vector space itself in dimension, however I am uncertain. Can you help me please?

Thank you,

2. May 20, 2015

### Ssnow

Hi,
only a few suggestions, in general a projection is from a vector space $V$ in a subspace (subset that is a vector space) $W$, so as you said we have that $dim(W) \leq dim(V)$, in the $=$ case your projection is the identity and is bijective but if $dim(W) < dim(V)$ the two dimensions are different so is not a bijection...
The second question is a conseguence of the first, in general projections are surjective and linerar operators so you can define $T:V\rightarrow W$ specifing where you send the basis of $V$ in $W$, for example $\{(x,y,0):x,y\in\mathbb{R}^{3}\}$ is a subspace of dimension $2$ of $\mathbb{R}^{3}$ you can project from $\mathbb{R}^{3}$ sending $e_{1}=(1,0,0)\mapsto (1,0,0),e_{2}=(0,1,0)\mapsto (0,1,0)\mapsto (0,1,0),e_{3}=(0,0,1)\mapsto (0,0,0)$.

I hope to clarify something but in general your answer was correct,
by
Simone

3. May 20, 2015

### WWGD

Sure, for one, an n-dim space does not have non-trivial n-dimensional subspaces.

4. May 20, 2015

### WWGD

Maybe Rank-nullity will work: Given two bases with n vectors , the nullity will be 0 , so the dimension of the image is the entire space.

5. May 22, 2015

### HallsofIvy

Since you are given that W is a proper subset of V, it follows that there is some vector in V that is NOT in W and therefore T is not surjective.

6. May 22, 2015

### Myr73

This is what I had assumed, I just waanted to verify that this is correct. Thanks all