# Diagonalization, which eigenvector is found?

1. Nov 27, 2009

Hi!

This might be a silly question, but I can't seem to figure it out and have not found any remarks on it in the literature.

When diagonalizing an NxN matrix A, we solve the characteristic equation:

Det(A - mI) = 0

which gives us the N eigenvalues m. Then, to find the eigenvectors v of A, we solve the eigenvalue problem

Av = mv

Now, any scalar multiple of an eigenvector of A is itself an eigenvector with the same eigenvalue. So, which eigenvector do we find when solving the eigenvalue problem? It can't be totally random, can it? Is there a way of determining which of the infinitude of eigenvectors (all with the same "direction") the algorithm chooses?

2. Nov 27, 2009

### trambolin

Usually 3 things are done.

Normalize the vector so that $\|v\| = 1$
Make the smallest entry other than zero equal to 1
Make the largest equal to one

\begin{align*} (1) \Longrightarrow v_0 &= \frac{1}{\sqrt{v^*v}}v\\ (2) \Longrightarrow v_0 &= \frac{1}{\min_{v_i\neq 0}(v_i)}v\\ (3) \Longrightarrow v_0 &= \frac{1}{\max(v_i)}v \end{elign*}

It is a choice... You can come up yourself with something else anyway

3. Nov 27, 2009

Thanks trambolin for your answer. However, that was not really my question. I'm aware this is what one usually does when the eigenvectors are found.

My problem is this: When we solve

Av = mv (1)

for the eigenvectors v, we find a set of N eigenvectors. But these are not unique. How does equation (1) "choose" which eigenvector in the 1-d subspace spanned by an eigenvector to "return" since all of them are equally valid?

4. Nov 27, 2009

### HallsofIvy

Equation (1) doesn't "choose" any specific eigenvector. What typically happens is that you get something like y= 2x, z= 3x- that is, all but one of the components (in the case that there is only one eigenvector corresponding to each eigenvalue) so that the eigenvector is of the form <x, 2x, 3x>= x<1, 2, 3>. You then choose a value of x.

As far as the diagonalization is concerned, it doesn't matter which you choose- using any of the eigenvectors corresponding to the eigenvalues as columns of "P" will still give you a matrix such that $P^{-1}AP$ is diagonal.

5. Nov 27, 2009