need help finding eigenvectors

**mancillaj3** · September 12th 2009, 07:29 PM

$M=\left| \begin{array}{cc} 2 & -3 \\ 1 & 5 \end{array} \right|$
the eigenvalues are
$\frac{7+\sqrt{3}i}{2} and \frac{7-\sqrt{3}i}{2}$ , i need help finding the eigenvectors

**garymarkhov** · September 13th 2009, 03:03 AM

Originally Posted by mancillaj3

$M=\left| \begin{array}{cc} 2 & -3 \\ 1 & 5 \end{array} \right|$
the eigenvalues are
$\frac{7+\sqrt{3}i}{2} and \frac{7-\sqrt{3}i}{2}$ , i need help finding the eigenvectors

I'm not sure why you've used the notation $M=\left| \begin{array}{cc} 2 & -3 \\ 1 & 5 \end{array} \right|$ , but I'll assume that you just mean you have some matrix $A=\left( \begin{array}{cc} 2 & -3 \\ 1 & 5 \end{array} \right)$ and you want to find the eigenvectors.

Anyway, you do so in the same way you always find the eigenvectors. Nothing changes. First, find the eigenvalues as you've done. Choose one of the eigenvalues, say, $\frac{7+\sqrt{3}i}{2}$ , and subtract it from the diagonal entries:

$ A-\lambda I=\left( \begin{array}{cc} 2-\frac{7}{2}-\frac{i\sqrt{3}}{2} & -3 \\ 1 & 5-\frac{7}{2}-\frac{i\sqrt{3}}{2} \end{array} \right) $

$ =\left( \begin{array}{cc} \frac{4}{2}-\frac{7}{2}-\frac{i\sqrt{3}}{2} & -3 \\ 1 & \frac{10}{2}-\frac{7}{2}-\frac{i\sqrt{3}}{2} \end{array} \right) $

$ =\left( \begin{array}{cc} \frac{-3}{2}-\frac{i\sqrt{3}}{2} & -3 \\ 1 & \frac{3}{2}-\frac{i\sqrt{3}}{2} \end{array} \right) $

Solving for $x$ in $\left( A-\lambda I \right)x=0$ will give us the eigenvectors just like it does when we have real eigenvalues. [Added: This is from the equation $Ax=\lambda x$ , which can be rewritten as $\left( A-\lambda I \right)x=0$ ].

$ \left( \begin{array}{cc} \frac{-3}{2}-\frac{i\sqrt{3}}{2} & -3 \\ 1 & \frac{3}{2}-\frac{i\sqrt{3}}{2} \end{array} \right) \left( \begin{array}{c} x_{1} \\ x_{2} \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right) $

We know that this matrix is singular because we made it singular. We did that by subtracting just the right eigenvalues that would make the determinant zero! So Just set $x_{1}=1$ and solve for $x_{2}$ . In other words, we want to make sure this is true:

$ \left( \begin{array}{cc} \frac{-3}{2}-\frac{i\sqrt{3}}{2} & -3 \\ 1 & \frac{3}{2}-\frac{i\sqrt{3}}{2} \end{array} \right) \left( \begin{array}{c} 1 \\ x_{2} \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right) $

Super. That means we just need to solve for $x_{2}$ in $\left( \frac{-3}{2}-\frac{i\sqrt{3}}{2} \right) \left(1 \right) + (-3)x_{2}=0$

It turns out that $x_{2}=-\frac{1}{2}-\frac{i\sqrt{3}}{6}$ so we can write $ \left( \begin{array}{cc} \frac{-3}{2}-\frac{i\sqrt{3}}{2} & -3 \\ 1 & \frac{3}{2}-\frac{i\sqrt{3}}{2} \end{array} \right) \left( \begin{array}{c} 1 \\ -\frac{1}{2}-\frac{i\sqrt{3}}{6} \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \end{array} \right) $ and know it's true. Do the matrix multiplication to convince yourself.

So you've found one basis eigenvector, which is $\left( \begin{array}{c} 1 \\ -\frac{1}{2}-\frac{i\sqrt{3}}{6} \end{array} \right)$ . There's one more basis for the other eigenvalue, but I think you can work that out yourself now

**HallsofIvy** · September 13th 2009, 03:56 AM

I prefer a more basic method. The definition of "eigenvalue" is that " $\lambda$ is an eigenvalue of matrix A if and only if there exist a non-zero vector v, an eigenvector corresponding to eigenvalue $\lambda$ , such that $Av= \lambda v$ ."

Since $\frac{7+ i\sqrt{3}}{2}$ is an eigenvalue, for any eigenvector we must have
$\begin{bmatrix}2 & -3 \\ 1 & 5\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}\frac{7+ i\sqrt{3}}{2}x \\ \frac{7+ i\sqrt{3}}{2}y\end{bmatrix}$
$\begin{bmatrix}2x- 3y \\ x+ 5y\end{bmatrix}= \begin{bmatrix}\frac{7+ i\sqrt{3}}{2}x \\ \frac{7+ i\sqrt{3}}{2}y\end{bmatrix}$

That gives us two equations, $2x- 3y= \frac{7+ i\sqrt{3}}{2}x$ and $x+ 5y= \frac{7+ i\sqrt{3}}{2}y$ . The first equation is the same as $y= 2x- \frac{7+ i\sqrt{3}}{6}x= \frac{-3- i\sqrt{3}}{6}x$ . The second equation is the same as $x= \frac{-3+ i\sqrt{3}}{2}y$ . Dividing both sides by $\frac{-3+ i\sqrt{3}}{2}$ (which is the same as multiplying by $\frac{-3- i\sqrt{3}}{6}$ , we get $y= \frac{-3- i\sqrt{3}}{6}$ exactly the same as the first equation!

That had to happen. There exist an infinite number of eigenvectors corresponding to a given eigenvalue. The problem must reduce to one equation in one unknown. Take x to be anything you like and solve for y. If we take x= 6 to get rid of the fraction, $y= 3- i\sqrt{3}$ so an eigevector corresponding to eigenvalue $\frac{7+ i\sqrt{3}}{2}$ is

$\begin{bmatrix}6 \\ 3- i\sqrt{3}\end{bmatrix}$

and any eigevector corresponding to that eigenvalue is of the form

$\begin{bmatrix}6 \\ 3- i\sqrt{3}\end{bmatrix}x$
for some number, x.

Do the same thing to find an eigenvector corresponding to eigenvalue $\frac{7- i\sqrt{3}}{2}$ .

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