# How to determine the eigenvectors for this matrix

I have the matrix $$\left( \begin{array}{ccc} -\alpha & \beta \\ \beta/K & -\alpha/K \end{array} \right)$$ for which the eigenvalues are $$\lambda_{1,2}=-\dfrac{\alpha}{2}-\dfrac{\beta}{2K}\pm\dfrac{\sqrt{(K\alpha-\beta)^2+4K\alpha\beta }}{2K}$$ I have a problem with finding the eigenvectors. It seems that the eigenvectors are zero vectors but then eigenvectors cannot be zero. What am I missing?

• If $\lambda$ is an eigenvalue, then $A-\lambda I$ is a singular matrix, and the corresponding eigenvector is a non-zero vector in its nullspace. How are you calculating eigenvectors? – G Tony Jacobs Jul 4 '14 at 14:51
• I believe you said eigenvectors when you meant so say eigenvalues so I have changed it for you. – Brad Jul 4 '14 at 15:09
• I would say that after adjustment be: $\lambda_1=0,\,\,\lambda_2=-\frac{K\alpha+\beta}{K}$ – georg Jul 4 '14 at 15:12
• When I calculate these eigenvalues, I get $\lambda_{1,2}=-\frac{(K+1)\alpha}{2K} \pm \frac{\sqrt{(K-1)^2\alpha^2 + 4K\beta^2}}{2K}$. Are you sure of your eigenvalues? – G Tony Jacobs Jul 4 '14 at 15:24
• Are you sure you didn't exchange $\alpha$ and $\beta$ in the second row? – egreg Jul 4 '14 at 15:30

## 2 Answers

Your matrix is of the form $$M=\begin{pmatrix} a & b \\ b/k & a/k \end{pmatrix},$$ now let $x=(x_1,x_2) \neq (0,0)$ be such that $Mx = \lambda x$, then $$\begin{pmatrix} 0 \\ 0 \end{pmatrix} =Mx-\lambda x =\begin{pmatrix} (a-\lambda)x_1+bx_2 \\ \frac{1}{k}(bx_1 +(a-k\lambda)x_2) \end{pmatrix},$$ This a two unknown linear system of two equations. Solving it shows directly that your eigenvectors must have the form $$\alpha\begin{pmatrix} 1 \\ b/(\lambda-a)\end{pmatrix}, \quad \alpha \in \mathbb{R}.$$ if $\lambda \neq a$ (use the first equation) and $$\alpha\begin{pmatrix} 1 \\ 0\end{pmatrix}, \quad \alpha \in \mathbb{R}.$$ if $\lambda = a$ (from the first equation you get $x_2=0$). Since you aleready computed the eigenvalues, it shouldn't be hard to recover the eigenvectors for your special case.

• Your matrix is not quite the same as posters, regards the positions of $a$ and $b$. But I think the original question should be checked for mistakes, I am not confident in the calculation of the eigenvalues given. – Rene Schipperus Jul 4 '14 at 15:11
• @ReneSchipperus It has been edited (thanks I mixed the components). Seems that georg checked the eigenvalues (see comments of the OP). – Surb Jul 4 '14 at 15:21
• Yeah I thought something was off, I am deleting my answer which was based on the false computation. Thanks. – Rene Schipperus Jul 4 '14 at 15:46

To find the eigenvalues of a matrix $A$, we solve $\mathrm{det}(A-\lambda I)=0$ for $\lambda$. In this case:

\begin{align} \left|\begin{array}{cc} -\alpha-\lambda & \beta \\ \frac{\beta}{K} & \frac{-\alpha}{K}-\lambda \end{array}\right| &= (-\alpha-\lambda)\left(\frac{-\alpha}{K}-\lambda\right)-\frac{\beta^2}{K} \\ &=\frac{\alpha^2}{K}+\left(\alpha+\frac{\alpha}{K}\right)\lambda + \lambda^2 - \frac{\beta^2}{K} \\ &=\lambda^2 + \left(\alpha+\frac{\alpha}{K}\right)\lambda + \frac{\alpha^2-\beta^2}{K} = 0, \end{align}

implies that our eigenvalues are:

\begin{align} \lambda_{1,2} &= -\frac{\alpha+\frac{\alpha}{K}}{2} \pm \frac{\sqrt{\left(\alpha+\frac{\alpha}{K}\right)^2 - 4\frac{\alpha^2-\beta^2}{K}}}{2} \\ &=-\frac{K\alpha+\alpha}{2K} \pm \frac{\sqrt{(K\alpha+\alpha)^2 - 4K(\alpha^2-\beta^2)}}{2K} \\ &=-\frac{(K+1)\alpha}{2K} \pm \frac{\sqrt{\left((K+1)^2-4K\right)\alpha^2 + 4K\beta^2}}{2K} \\ &=-\frac{(K+1)\alpha}{2K} \pm \frac{\sqrt{(K-1)^2\alpha^2 + 4K\beta^2}}{2K}. \end{align}

Now, an eigenvector corresponding to the eigenvalue $\lambda_i$ is in the nullspace of $\left(\begin{array}{cc} -\alpha-\lambda_i & \beta \\ \frac{\beta}{K} & \frac{-\alpha}{K}-\lambda_i \end{array}\right).$

It appears that the vector $\bf{v}_i=\left(\begin{array}{c} \beta \\ \alpha + \lambda_i \end{array}\right)$ would work.

• How are you getting v? Row operations give $\left( \begin{array}{ccc} 1 & -\frac{\beta}{\alpha+\lambda_i}\\ 0 & \frac{\alpha-K\lambda_i}{\beta} \end{array} \right)$ from which $v_2$ is zero – Vaolter Jul 4 '14 at 16:33
• Since $A-\lambda I$ is singular, I'm not bothering with row operations, and just writing down a vector that kills row 1. If you multiply my $v_i$ by my matrix $A-\lambda I$, you get a $0$ on top. – G Tony Jacobs Jul 4 '14 at 22:40