# Diagonalization-ordered basis

1. May 16, 2009

### jeff1evesque

I am trying to perpare for my final for linear algebra and am attempting to catch up in readings- in particular about diagonlization.

Theorem 5.1: A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis B for V consisting of eigenvectors of T. Furthermore, if T is diagonalizable, B = {$$v_1, v_2, ..., v_n$$} is an ordered basis of eigenvectors of T, and D = $$[T]_B$$, then D is a diagonal matrix and $$D_j_j$$ is the eigenvalue corresponding to $$v_j$$ for 1<= j <= n.

Question: For some reason in my textbook, they omitted the proof. I'm guessing it's fairly straightfoward, but I was wondering if someone could aid me with a proof.

Idea for a proof: If T is diagonalizable, then for some $$v_j$$, $$T(v_j)$$ = $$\lambda_j$$$$v_j$$ (by definition- from last post).
So it follows, for each eigenvalue $$\lambda_j$$ ( 1 <= j <= n, where n is the number of elements in our particular basis B ), there exists corresponding eigenvectors $$v_j$$ by definition. Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors, our basis B = { lambda_1, lambda_2, ..., lambda_n } is composed of eigenvectors.

Questions:
1. Can someone tell me if my proof above is correct? I feel as if I'm assuming to much when I said "Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors... "
2. Also can someone justify the remaining portion of the theorem "... Furthermore, if T is diagonalizable... is the eigenvalue corresponding to $$v_j$$ for 1 <= j <= n."

THanks again,

JL

Last edited: May 16, 2009
2. May 17, 2009

### jbunniii

"Since a basis for a diagonalizable matrix by definition is composed of the eigenvectors..."

This isn't "by definition" - in fact, it's one of the things that you are supposed to prove!

First suppose that T is diagonalizable. That means that there exists a basis

$$B = \{v_1,v_2,\ldots,v_n\}$$

with respect to which the matrix $$D = [T]_B$$ is diagonal.

Call the entries of the matrix $$D_{ij}$$, for $$i,j=1,2,\ldots,n$$. Then for each j,

$$Tv_j = \sum_{i=1}^n D_{ij} v_i$$ (this is the definition of what the matrix elements mean)

Because D is diagonal, $$D_{ij} = 0$$ for $$i \neq j$$, so this reduces to

$$Tv_j = D_{jj} v_j$$

Since each $$v_j$$ is a member of the basis B, in particular, $$v_j \neq 0$$, so $$D_{jj}$$ is an eigenvalue of T and $$v_j$$ is a corresponding eigenvector.

The converse is proved the same way. (If there exists an ordered basis B for V consisting of eigenvectors of T, then T is diagonalizable.) All you have to do is show that T has a diagonal matrix with respect to that basis B.

3. May 17, 2009

### jeff1evesque

Wow, that was very straightfoward. Thanks for the help.