Diagonalization, which eigenvector is found?

[FredMadison]

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?

 

[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

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

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

 

[FredMadison]

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?

 

[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.

 

[FredMadison]

Ofcourse! It WAS a silly question. This is what happens when you stop doing things by hand and start relying too much on Mathematica to do the thinking for you.

Thanks guys!