Diagonalization

[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.

 

[renegade05]

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.

Hmm... I have said it before and i will say it again, there should be a linear algebra section.

Anyway, You can get the geometric multiplicity very easily. What is the dimension of V ? f1, f2, and f3 form a basis for V and there is three of them.

As for algebraic multiplicity. I would try to find a the standard matrix for T. Unless I am missing something here.

 

[HallsofIvy]

The "algebraic multiplicity" of a linear transformation is the number of eigenvalues, counting the multiplicity which is the same as the degree of its characteristic equation. For transformation on an n-dimensional vector space over the comlex numbers, that is always n. The "geometric multiplicity" of a linear transformation is the number of indendent eigenvectors. If all n eigenvalues are distinct, then, because eigenvectors corresponding to distinct eigenvalues are indendent, the "algebraic multiplicity" and "geometric multiplicity" are the same. If the eigenvalues are not all distinct, then eigenvectors corresponding to the same eigenvalue are not necessarily independent so the geometric multiplicity might be less than the algebraic multiplicity.

The simplest way to determine all that is to, as renegade05 suggests, write the transformation as a matrix. And the simplest way to do that is to apply the transformation to each basis function in turn, writing the result as a linear combination of the basis functions. The coefficients will be a column of the matrix.

For example, \(\displaystyle T(e^t)= e^t= 1(e^t)+ 0(te^t)+ 0(t^2e^t)\) so the first column is \(\displaystyle \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\).

[/tex]T(te^t)= e^t+ te^t[/tex] so the second column is \(\displaystyle \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}\).

Can you finish?