# Thread: Basic probelm on 2x2 Eigenvector

1. ## Basic probelm on 2x2 Eigenvector

Hi

can someone explain to be how they get the eigenvectors for the following matrix

$\begin{matrix}2 & -1 \\ -4 & 2 \end{matrix}$

$\begin{matrix}1 & \frac{-1}{2} \\ 0 & 0 \end{matrix}$

hence

$v1 = \frac{-1}{2}v2$

$v2 = -2v1$

eigenvalues are $\lambda = 4 \lambda = 0$

i get $\begin{matrix} \frac{-1}{2} \\ -2 \end{matrix}$

however answers says when $\lamba =4$

matrix is $\begin{matrix} 1 \\ -2 \end{matrix}$

i don't understand how they got this result, someone please explain this to me.

P.S

2. ## Re: Basic probelm on 2x2 Eigenvector

(2 - L)^2 - 4 = 0 is the characteristic polynomial.

3. ## Re: Basic probelm on 2x2 Eigenvector

You are making two errors. The first is in thinking that the row reduced matrix $\begin{bmatrix}0 & -\frac{1}{2} \\ 0 & 0 \end{bmatrix}$ leads to $v2= -\frac{1}{2}v2$. This really the reduced form of the "augmented matrix" $\begin{bmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0\end{bmatrix}$
for solving the matrix equation $\begin{bmatrix}2 & -1 \\ -4 & 2\end{bmatrix}\begin{bmatrix}v1 \\ v2\end{bmatrix}= \begin{bmatrix}4v1 \\ 4v2\end{bmatrix}$.

That is, it corresponds to the equation $v1- \frac{1}{2}v2= 0$ so that you get $v1= \frac{1}{2}v2$, not $v1= -\frac{1}{2}v2$.

Your second error is in thinking you can set v2= 1 to get $v_1= \frac{1}{2}$ (using the correct $v1= \frac{1}{2}v2$) and then set $v1= 1$to get $v_2= 2$. You can do one or other, not both! If you set v2= 1, you get v1= 1/2 and so have the eigenvector $\begin{bmatrix}\frac{1}{2} \\ 1\end{bmatrix}$. If you take v1= 1, you get v2= 2 and so have the eigenvector $\begin{bmatrix}1 \\ 2\end{bmatrix}$.

Since one is a multiple of the other, both are eigenvectors corresponding to eigenvalue 4.

4. ## Re: Basic probelm on 2x2 Eigenvector

ok i understand now thanks for the explanation