Lagrange to find eigen values and vectors?

[seto6]

Homework Statement

im given a matrix A= 1 -2
///////////////////////-2 4
im told to find the eigen values and the vectors... but the thing is i have never came across this, i learned lagrange multipliers but never used it to find eigen values and vector..

Homework Equations

The Attempt at a Solution

im really lost..

can someone hint me pls
thanks in advanve

 

[Winzer]

To find the eigenvalues use:
$det|A - I \lambda| = 0$
Where I is the identity matrix.
Solve for lambda.
http://www.miislita.com/information-retrieval-tutorial/matrix-tutorial-3-eigenvalues-eigenvectors.html" is more information for finding Eigenvalues and their Eigenvectors:

 

[seto6]

To find the eigenvalues use:
$det|A - I \lambda| = 0$
Where I is the identity matrix.
Solve for lambda.:

the mothod you states above, i know how to do it that way,but i need to do it using lagrange multiplier

 

[HallsofIvy]

Please state the problem exactly as it is given. Normally, "Lagrange multipliers" are used to find maximum and minimum values, not for finding eigenvalues or eigenvectors.

However, you can think of a matrix as "stretching" or "compressing" directions in the xy-plane. Specifically, if $\lambda$ is the largest eigenvalue of matrix A, and v is a corresponding eigenvector, then $Av= \lambda v$ maximizes the length of Av. Given
$$v= \begin{bmatrix}x \\ y \end{bmatrix}$$,

$$\begin{bmatrix}1 & -2 \\ -2 & 4\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x- 2y \\ -2x+ 4y\end{bmatrix}$$
will have maximum length (among all vectors with the same length) if and only if v is an eigenvector corresponding to the largest eigenvector.

That is, the problem of finding the larger and smaller eigenvalues can be construed as "Maximize (minimize) $(x- 2y)^2+ (-2x+ 4y)^2$ subject to the constraint $x^2+ y^2= 1$". That is a problem that can be handled with Lagrange multipliers.

 

[seto6]

i thought i stated...but i see now that i stated in the title only sorry,
thanks for your help!