# 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.

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So, you already know what they are, but you are having trouble deriving them. Did I understand that right? – Tim Seguine Feb 1 '14 at 14:44
Yes exactly, especillay the pre-factor, $1/(1+k^2)^{1/2}$. – Marc Raquil 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. – Marc van Leeuwen 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.

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Ok ! So the pre-factor, $1/(1+k^2)^{1/2}$ could be anything else ? Thank a lot. – Marc Raquil 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. – Tim Seguine Feb 1 '14 at 15:07
@TimSeguine, but how can I found it ? – Marc Raquil Feb 1 '14 at 15:30
Perfect. Thank you – Marc Raquil Feb 1 '14 at 15:40

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

$$\binom 1k$$

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Thank you too for your help. – Marc Raquil Feb 1 '14 at 14:59