Eigenvalues and Eigenvectors

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April 1st 2010, 08:47 PM
mybrohshi5

Eigenvalues and Eigenvectors

Given that $V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix}$ are Eigenvectors of the matrix

$A = \begin{bmatrix}-21&48\\-6&13\end{bmatrix}$

Find the Eigenvalues

I know how to find eigenvalues and using those i know how to find eigenvectors, but i am not sure how to find eigenvalues if eigenvectors and a matrix are given.

Thank you for any help :)
April 1st 2010, 09:19 PM
Prove It

Quote:

Originally Posted by mybrohshi5

Given that $V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix}$ are Eigenvectors of the matrix

$A = \begin{bmatrix}-21&48\\-6&13\end{bmatrix}$

Find the Eigenvalues

I know how to find eigenvalues and using those i know how to find eigenvectors, but i am not sure how to find eigenvalues if eigenvectors and a matrix are given.

Thank you for any help :)

You need to solve

$|A - \lambda I| = 0$

$\left|\left[\begin{matrix} - 21 & 48 \\ -6 & 13 \end{matrix}\right] - \left[\begin{matrix}\lambda & 0 \\ 0 & \lambda\end{matrix}\right]\right| = 0$

$\left|\begin{matrix}-21 - \lambda & 48\\ -6 & 13 - \lambda\end{matrix}\right| = 0$

$(-21 - \lambda)(13 - \lambda) - (48)(-6) = 0$

$-273 + 21\lambda - 13\lambda + \lambda^2 + 288 = 0$

$\lambda^2 + 8\lambda + 15 = 0$

$(\lambda + 3)(\lambda + 5) = 0$

$\lambda = -3$ or $\lambda = -5$ .
April 1st 2010, 09:22 PM
harish21

Quote:

Originally Posted by mybrohshi5

Given that $V_1 = \begin{bmatrix}3\\1\end{bmatrix}\: and\: \: V_2 = \begin{bmatrix}8\\3\end{bmatrix}$ are Eigenvectors of the matrix

$A = \begin{bmatrix}-21&48\\-6&13\end{bmatrix}$

Find the Eigenvalues

I know how to find eigenvalues and using those i know how to find eigenvectors, but i am not sure how to find eigenvalues if eigenvectors and a matrix are given.

Thank you for any help :)

If $I$ is your identity matrix, and $\lambda$ is your eigenvalue, then you have to find the matrix $A - {\lambda}I$ and find out $\lambda$ by equating the determinant of the matrix $|A - {\lambda}I| = 0$
April 1st 2010, 09:34 PM
Prove It

Alternatively, you use the property that if

$A\mathbf{x} = \lambda \mathbf{x}$

then $\mathbf{x}$ is the eigenvector and $\lambda$ is the corresponding eigenvalue.

You have been given a matrix $A$ and two eigenvectors.

So $\left[\begin{matrix}-21 & 48 \\ -6 & 13\end{matrix}\right] \left[\begin{matrix}3 \\ 1\end{matrix}\right] = \lambda \left[\begin{matrix}3 \\ 1\end{matrix}\right]$

$\left[\begin{matrix}-21 \cdot 3 + 48 \cdot 1 \\ -6 \cdot 3 + 13 \cdot 1\end{matrix}\right] = \left[\begin{matrix}3\lambda \\ \lambda\end{matrix}\right]$

$\left[\begin{matrix}-63 + 48 \\ -18 + 13\end{matrix}\right] = \left[\begin{matrix}3\lambda \\ \lambda\end{matrix}\right]$

$\left[\begin{matrix}-15 \\ -5 \end{matrix}\right] = \left[\begin{matrix}3\lambda \\ \lambda\end{matrix}\right]$

So that means $\lambda = -5$ .

Do the same for the other eigenvector.

Note that I've edited and fixed my other post too.
April 1st 2010, 09:41 PM
mybrohshi5

Thank you Prove It and harish21 for such fast responses :) i really appreciate it