Finding eigenvectors for a given matrix

[cooltowns]

Homework Statement

Finding Eigenvectors of the given matrix

Homework Equations

The matrix is
$A=\begin{pmatrix} 5.956 & -1.284\\ -1.284&0.435 \end{pmatrix}$

The Attempt at a Solution

I have found the eigenvalues to be

$\lambda _{1}=0.624001$ and $\lambda _{2}=0.150994$

and

$A-\lambda_{1} I =\begin{pmatrix} -0.2840 &-1.284 \\ -1.284 &-5.8050 \end{pmatrix}$

$A-\lambda_{2} I =\begin{pmatrix} 5.8050 &-1.284 \\ -1.284 &0.2840 \end{pmatrix}$

where A is the given matrix, I have rounded up the values slightly.

Using wolfram online I know the eigenvectors I should get are
$E_{1}(-0.9764,0.2161)$ and $E_{2}(-0.2161, 0.9764)$

However I can't seem to get them.

Could somone please show me how to get those eigenvectors

Thanks

 

[jncarter]

To get the eigenvectors of a matrix A use this equation and solve the system of equations for each eigenvector:

$(A - \lambda_{\alpha}I)E_{\alpha} = 0$​

 

[HallsofIvy]

Assuming that 0.624001 really is an eigenvalue, then there should be an infinite number of values of x and y such that
$$\begin{bmatrix}5.956 & -1.284 \\ -1.284 & 0.435\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 0.624001x \\ 0.624001 y\end{bmatrix}$$
which is the same as the pair of equations
5.956x- 1.284y= 0.624001x and -1.284x+ 0.435y= 0.624001y.
Solve either of those for y= ax and take, say, x=1 to get an eigenvector.

 

[lurflurf]

^It is not
The eigenvalues are approximately 6.24058 and 0.151.
Probably a decimal shift to blame.

 

[SeannyBoi71]

To get the eigenvectors of a matrix A use this equation and solve the system of equations for each eigenvector:

$(A - \lambda_{\alpha}I)E_{\alpha} = 0$​

Actually shouldn't that be det(A-λI)? Then he should find the roots of that characteristic polynomial and then proceed to find the bases for the Nullspace of those eigenvalues. Each basis should be the eigenvector for whatever eigenvalue you used.

 

[Mark44]

To get the eigenvectors of a matrix A use this equation and solve the system of equations for each eigenvector:

$(A - \lambda_{\alpha}I)E_{\alpha} = 0$​

Actually shouldn't that be det(A-λI)?

No, what jncarter wrote was correct. The equation above is a matrix equation, and 0 on the right side is a vector.

The equation above implies the equation you show with the determinant. The idea is that if Mx = 0, where M is a square matrix and x is an n-vector, then |M| = 0.

Then he should find the roots of that characteristic polynomial and then proceed to find the bases for the Nullspace of those eigenvalues. Each basis should be the eigenvector for whatever eigenvalue you used.

 

[SeannyBoi71]

My bad, just never seen that notation before. Learn something new every day, hope I didn't confuse OP. Thanks for clarification.