Eigenvector problem

**geft** · October 25th 2009, 10:12 PM

Never mind guys, I found the mistake. $0 - \lambda \neq 0$

Find the eigenvalues and eigenvectors of $\left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 1 & 1 & 0 \end{array} \right)$

Here's what I did.

$A - \lambda I = \left( \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 4 \\ 1 & 1 & 0 \end{array} \right)$

$det(A - \lambda I) = (2 - \lambda)(-4) + 1 = 0$

$-8 + 4 \lambda + 1 = 0$

$4\lambda = 7$

$\lambda = \frac{4}{7}$

$A - \lambda I = \left( \begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 0 & \frac{5}{7} & 4 \\ 1 & 1 & 0 \end{array} \right)$

$\frac{1}{4}x_1 = 0$

$\frac{5}{4}x_2 + 4x_3 = 0$

Eigenvector = $\left( \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array} \right)$

I'm pretty sure this is wrong.

**Defunkt** · October 26th 2009, 12:36 AM

Originally Posted by geft

Find the eigenvalues and eigenvectors of $\left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & 3 & 4 \\ 1 & 1 & 0 \end{array} \right)$

Here's what I did.

$A - \lambda I = \left( \begin{array}{ccc} 2-\lambda & 0 & 0 \\ 0 & 3-\lambda & 4 \\ 1 & 1 & 0 \end{array} \right)$

$det(A - \lambda I) = (2 - \lambda)(-4) + 1 = 0$

$-8 + 4 \lambda + 1 = 0$

$4\lambda = 7$

$\lambda = \frac{4}{7}$

$A - \lambda I = \left( \begin{array}{ccc} \frac{1}{4} & 0 & 0 \\ 0 & \frac{5}{7} & 4 \\ 1 & 1 & 0 \end{array} \right)$

$\frac{1}{4}x_1 = 0$

$\frac{5}{4}x_2 + 4x_3 = 0$

Eigenvector = $\left( \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array} \right)$

I'm pretty sure this is wrong.

First, the eigenvector can never be $\left( \begin{array}{ccc} 0 \\ 0 \\ 0 \end{array} \right)$ , so this is wrong (can you see why?).

Second, $det(A-\lambda I) = (2-\lambda)((3-\lambda)(-\lambda) -4) =$ $(2-\lambda)(\lambda(\lambda-3)-4) = (2-\lambda)(\lambda^2-3\lambda-4) = (2-\lambda)(\lambda+1)(\lambda-4)$

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