# Showing that a linear transformation from P3 to R4 is an isomorphism?

1. May 4, 2014

### elo

I have a linear transformation, T, from P3 (polynomials of degree ≤ 3) to R4 (4-dimensional real number space). I have a second linear transformation, U, from R4 back to P3.

In the first step of this four-step problem, I have shown that the composition TU from R4 to R4 is the identity linear transformation by sending a vector x through U, then sending the product of that through T, and ending up with x again.

What I don't understand is the next step: show that T is an isomorphism. The problem gives the hint that I should first show that the image of T spans R4: okay, fair enough. After some misguided attempts to prove linear independence of the entries of the column vector of T(f), I skipped ahead to the solutions, and was given this:

"Given a vector x in R4, we have T(U(x)) = x, by the previous part, and so x is a value of T, and im(T) = R4."

I do not understand this answer. I did T(U(x)) = x, so I get that. But what does it mean that "x is a value of T?" And how does that mean that the image of T spans R4?

2. May 4, 2014

### jbunniii

If you have already shown that $TU=I$, then $T$ is surjective: if not, then the range of $T$ would not be all of $\mathbb{R}^4$, and so the range of $TU$ could not be either, contradicting $TU=I$. Can you use the rank-nullity theorem to conclude that $T$ is also injective?

3. May 4, 2014

### jbunniii

I think it's a typo or misstatement. It would be correct if it said "and so x is an element of im(T)".

4. May 4, 2014

### elo

Okay: since I can get a 4-dimensional x out of T, the image of T must span R4. The starting space P3 has four dimensions. By rank-nullity, the dimension of the kernel of the transformation plus the dimension of its image must equal the dimension of P3; since the dimension of the image is four, the dimension of the kernel must be zero. Thus, T is both surjective and injective, and is thus an isomorphism.

Do I have the right idea?

5. May 5, 2014

### jbunniii

Yes, it sounds right to me!