# Math Help - Eigenvectors

1. ## Eigenvectors

I have the matrix

$A=\left[ \begin{matrix} i & 1 \\ 0 & (-1+i) \end{matrix} \right]$

With the eigenvalues

$\lambda_1 = i, \lambda_2 = (-1+i)$

then the eigenvectors

$\lambda_1 : \left[ \begin{matrix} i - i & 1 \\ 0 & (-1+i) - i \end{matrix}\right] \sim \left[\begin{matrix} 0 & 1 \\ 0 & -1 \end{matrix}\right] \sim \left[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right]$

$\lambda_2 : \left[ \begin{matrix} i - (-1+i) & 1 \\ 0 & (-1+i) - (-1+i) \end{matrix}\right] \sim \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]$

Now I understand that for $\lambda_2$ the eigenvector is $\left[ 1 , -1\right]$ since $x_1 + x_2 = 0 \rightarrow x_1 = -x_2$.

However when it comes to the eigenvector for $\lambda_1$ all I have (in reduced form) $x_2 = 0$.
Why is the eigenvector for $\lambda_1 \rightarrow \left[ 1 , 0 \right]$?

Is it because the first column $(x_1)$ is a free variable?

2. ## Re: Eigenvectors

because any vector of the form (a,0) = a(1,0) satisfies x2 = 0, so this is an eigenvector iff a ≠ 0.

yes this is the same as being a "free variable", any choice of x1 (freely made) will do (but of course, the fact that an eigenvector must be non-zero means we can't choose x1 = 0, although (0,0) *is* in the eigenspace Eλ1).

3. ## Re: Eigenvectors

First, do you understand that there is no such thing as "the eigenvector" for given eigenvalue? Given any eigenvector, any (non-zero) multiple of it is also an eigenvector. In fact, the set of all eigenvectors for a given eigenvalue form a subspace. You say that "the eigenvector" for eigenvalue $\lambda_2$ is [1, -1]. In fact, any vector of the form [a, -a], for a any (non-zero) number is an eigenvector.

I had to include "(non-zero)" because Deveno says "but of course, the fact that an eigenvector must be non-zero". Some text (but, admittedly a minority) define an eigenvalue by saying " $\lambda$ is an eigenvalue for linear operator A if and only if there exist a non-zero vector $Av= \lambda v$" but then define an "eigenvector" as any vector satisfying $Av= \lambda v$. That has the advantage, in my opinion, of including the 0 vector as an "eigenvector" so that the set of all eigenvector corresponding to a given eigenvalue is a subspace without having to add the 0 vector separately.