# A proof

1. Jun 12, 2009

### evilpostingmong

1. The problem statement, all variables and given/known data
Let T: V---->V be a linear operator where dim V=n. Show that V
has a basis of eigenvectors if and only if V has a basis B such that
TB is diagonal.

2. Relevant equations

3. The attempt at a solution
Let T=[a1,1......an,1] ai,j=/=0
[a1,n......an,n]
Let TB=[a1,1v1......0n,1]
[01,n.......an,nvn]
Since this is diagonal, and ai,j=/=0,
then we have a basis of eigenvectors (these are meant to be vertical)<[v1,...0]......[0.....vn]>
that, after being mulitplied by T, formed the matrix TB.
To show that they are eigenvectors, a possible linear combination is
[v1,...0] and when multiplied by T gives a1,1[v1,...0]+...+a1,n[v1,....0]
=(a1,1+...+a1,n)[v1,....0]. Since [v1,....0] is an eigenvector, and
the others follow the same logic, mulitplying the basis of eigenvectors by T
should produce a diagonal matrix, as shown.

Last edited: Jun 12, 2009
2. Jun 12, 2009

### jbunniii

Suppose V has a basis $$B=\{b_1,\ldots,b_n\}$$ consisting of eigenvectors of T. Then for each j, $$T b_j = \lambda_j b_j$$. If I express this equation in terms of the basis B, then
$$[T]_B [b_j]_B = \lambda_j [b_j]_B$$
But $$[b_j]_B$$ is simply a column vector containing all zeros except for a 1 in the j'th position. Carrying out the matrix multiplication then shows that
$$[T]_B = \Lambda$$
where $$\Lambda$$ is the diagonal matrix whose diagonal entries are $$\{\lambda_1,\ldots,\lambda_n\}$$. This gives you one direction of the proof. For the other direction, you can essentially reverse the steps above to verify that the $$\lambda_j$$'s must be eigenvalues and the $$b_j$$'s must be eigenvectors.

3. Jun 12, 2009

### evilpostingmong

Yeah, I did the other direction by assuming that TB is diagonal then showing that
the basis consists of eigenvectors and that TB consists of eigenvalues, though I
think I messed up when I mentioned a1,1+...+a1,n is an eigenvalue since I multiplied wrong.
It should be a1,1 for the eigenvalue of [v1.....0]. Made a mistake with multiplication.
Oh and when I said "then we have a basis of eigenvectors" I shouldn't've called
the basis of eigenvectors a basis of eigenvectors, since I was proving it in the first place,
and I should've called the scalar an eigenvalue. It's these small mistakes that make a big
difference.

Last edited: Jun 12, 2009