Basic probelm on 2x2 Eigenvector

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October 5th 2011, 03:51 AM
Paymemoney

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
October 5th 2011, 04:07 AM
TheChaz

Re: Basic probelm on 2x2 Eigenvector

(2 - L)^2 - 4 = 0 is the characteristic polynomial.
October 5th 2011, 09:21 AM
HallsofIvy

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.
October 6th 2011, 01:17 AM
Paymemoney

Re: Basic probelm on 2x2 Eigenvector

ok i understand now thanks for the explanation