# Linear operator proof

1. Oct 12, 2009

### guroten

1. The problem statement, all variables and given/known data

Let V be a finite-dimensional vector space over the field F and let S and T
be linear operators on V. We ask: When do there exist ordered bases a and b
for V such that a = [T]b? Prove that such bases exist if and only if there is
an invertible linear operator U on V such that T = USU-1. (Outline of proof:
If a = [T]b, let U be the operator which carries a onto b and show that
S = UTU-1. Conversely, if T = USU-1 for some invertible U, let a be any
ordered basis for V and let b be its image under U. Then show that a = [T]b.
where [T]b means T with relative to b

3. The attempt at a solution

I'm not sure where to start. Is there a particular theorem to use here?

2. Oct 12, 2009

### jbunniii

[Note: I think the $U$ and $U^{-1}$ should be reversed in the problem statement.]

I don't think you need any particular theorem. The proof outline looks good. So let's suppose that there do exist two bases, $a$ and $b$, such that $a = [T]b$.

Let's give the basis elements some names:

$$a = \{a_1,\ldots,a_n\}$$
$$b = \{b_1,\ldots,b_n\}$$

As the outline suggests, let us define a new operator U as follows:

$$U(a_j) = b_j, \textrm{ for }j=1,\ldots,n$$

Since we have specified $U$ on a basis for $V$, that uniquely defines $U$ on all of $V$. By the way, note that $U$ is invertible.

Now consider the operator $U^{-1}TU$. In particular, can you find a useful way to express this operator in terms of the basis $a$?

Hint: you want to express $[U^{-1}TU]_a$ as the product of three matrices, where the middle one is $[T]_b$.