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.
 A: 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
$$\binom 1k$$
A: 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.
