# Let f1(t)=e^t, f2(t)=te^t, f3(t)=t^2e^t, and let V=Span(f1,f2,f3) in

1. Apr 11, 2012

### renolovexoxo

Let f1(t)=e^t, f2(t)=te^t, f3(t)=t^2e^t, and let V=Span(f1,f2,f3) in the infinite continuous functions. Let T:V-->V be give by T(f)=f''-2f'+f. Decide whether T is diagonalizable.

We learned a theorem that this will be diagonalizable if and only if the geometric multiplicity of each eigenvalue equals its algebraic multiplicity.

What I am having trouble with is translating this into a way to find geometric and algebraic multiplicity. I'm not entirely sure what to do when I'm not given a matrix, since that's how we did it in class.

2. Apr 11, 2012

### micromass

Staff Emeritus
Re: Diagonalization

Try to find the matrix associated to this linear map.

Choose a basis of V, and try to construct the matrix.

3. Apr 11, 2012

### renolovexoxo

Re: Diagonalization

So would I choose f1, f2, f3 as a basis for V and then construct a matrix
[f1 f2 f3]? Something like that?

4. Apr 11, 2012

### micromass

Staff Emeritus
Re: Diagonalization

You need to construct a matrix with respect to the basis $(f_1,f_2,f_3)$. Do you know how to do that?

5. Apr 12, 2012

### renolovexoxo

Re: Diagonalization

I know I should know how to do that, but I get tripped up in functions, especially involving e. After looking at it, I can't figure it out.

6. Apr 12, 2012

### HallsofIvy

Staff Emeritus
Re: Diagonalization

You are given the basis vectors for a space, V, and a linear transformation from V to itself. To find the matrix representation of the linear transformation, apply the transformation to each basis vector, in turn, and write the result as a linear combination of the basis vectors. The coefficients give a column of the matrix.

Here, the given basis is $\{e^t, te^t, t^2e^t\}$ and the linear transformation is T(f)= f''- 2f'+ f. Applying that to, say, $t^2e^t$, gives $2e^t= 2(e^t)+ 0(te^t)+ 0(t^2e^t)$ so that the third column of the matrix is
$$\begin{bmatrix}2 \\ 0 \\ 0\end{bmatrix}$$