- #1

- 84

- 0

For example, I want to show that given a linear operator T such that [tex]T(x_1,x_2,x_3) = (3x_3, 2x_2, x_1)[/tex] then T can be represented by a diagonal matrix with respect to some basis of [tex]V = \mathbb{R}^3[/tex].

So one approach is to use a theorem that says: If T has dim(V) distinct eigenvalues, then T has a diagonal matrix with respect to some basis V.

So we simply look for the solutions of: [tex]\lambda x_1 = 3 x_3[/tex], [tex]\lambda x_2 = 2 x_2[/tex], [tex]\lambda x_3 = x_1[/tex]

So we find that [tex]\lambda = 2, \pm \sqrt 3[/tex] with some corresponding eigenvectors.So we use the theorem and finish off the question.

So finally, my problem with all this. The matrix of T with respect to the standard basis {(0,0,1), (0,1,0),(1,0,0)} (purposely out of order) would be:

[tex]\begin{bmatrix} 3 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}[/tex]

which is diagonal right? But I have a theorem that says that the eigenvalues of T with respect to a uppertriangular matrix "consist precisely of the entries on the diagonal of that upper-triangular matrix".

But I didn't find 3 and 1 as eigenvalues. Is it because I switched the standard basis elements out of order? Why would that matter? And doesn't this last theorem imply that every linear operator precisely has dim V eigenvalues :S?

It was a long one but any comments are appreciated!