# Eigenvector of simple matrix

I want to find the eigenvector of the matrix $$A \equiv \left(\begin{array}{rr} -\,{1 \over k} & 0 \\ 1 & -\,{2 \over k} \end{array}\right)$$ I found the eigenvalue $\lambda_1=-1/k$, $\lambda_2=-2/k$

Then, I have to found the eigenvector, $V_1=1/(1+k^2)^{1/2}(1, k)$ and $V_2=(0,1)$. I can't get them properly.

Thank.

• So, you already know what they are, but you are having trouble deriving them. Did I understand that right? Feb 1 '14 at 14:44
• Yes exactly, especillay the pre-factor, $1/(1+k^2)^{1/2}$. Feb 1 '14 at 15:30
• The pre-factor is useless: any nonzero multiple of an eigenvector is an eigenvector, so with or without the factor makes no difference for this question. Feb 1 '14 at 15:44

For the first eigenvalue, $\lambda_1=-\dfrac{1}{k}$, we have a RREF for $[A - \lambda I]v_1 = 0$ of:

$$\begin{bmatrix}1 & -\dfrac{1}{k}\\0 & 0\end{bmatrix}v_1 = 0$$

We choose $v_1 = \left(\dfrac{1}{k},1\right)$

Note: they normalized both eigenvectors and the easiest way to do that is to divide by its length, $|v_1| = \sqrt{a^2 + b^2}$, which yields:

$$\dfrac{1}{\sqrt{1 + \dfrac{1}{k^2}}} = \dfrac{k}{\sqrt{1+k^2}}$$

Multiply that by the eigenvector and you have your result.

For the second eigenvalue, $\lambda_2=-\dfrac{2}{k}$, we have a RREF for $[A - \lambda I]v_2=0$ of:

$$\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}v_2 = 0$$

We choose $v_2 = (0,1)$

Note: the second eigenvector is already normalized, else we would have done the same process as we did for the first.

• Ok ! So the pre-factor, $1/(1+k^2)^{1/2}$ could be anything else ? Thank a lot. Feb 1 '14 at 14:58
• @MarcRaquil That is a normalization factor. It is there to make the vector have length 1. Note that any vector that points in the same direction is also an eigenvector with the same eigenvalue. Feb 1 '14 at 15:07
• @TimSeguine, but how can I found it ? Feb 1 '14 at 15:30

By definition of characteristic polynomial

$$\det(tI-A)=\begin{vmatrix}t+\frac1k&0\\-1&t+\frac2k\end{vmatrix}=\left(t+\frac1k\right)\left(t-\frac2k\right)$$

To find say the general form of an eigenvector with regard to the first eigenvalue, form the corresponding homogeneous linear system substituting $\;\lambda=-\frac1k\;$ in the above matrix:

$$\begin{cases}0\cdot x+0\cdot y=0\\{}\\-x+\frac1ky=0\end{cases}\iff x=\frac1ky$$

so we have that a possible eigenvector in this case is

$$\binom 1k$$

• Thank you too for your help. Feb 1 '14 at 14:59