[SOLVED] Invertible and diagonal matrix

• Apr 3rd 2010, 03:00 PM
mybrohshi5
[SOLVED] Invertible and diagonal matrix
Let A = $\displaystyle \begin{bmatrix}3&-\frac{1}{2}\\-6&5 \end{bmatrix}$

Find an invertible $\displaystyle S$ and a diagonal $\displaystyle D$ such that $\displaystyle S^{-1}AS = D$

S = $\displaystyle \begin{bmatrix}.&. \\.&. \end{bmatrix}$

D = $\displaystyle \begin{bmatrix}.&.\\.&. \end{bmatrix}$

I have never seen a problem like this and i am just a little confused on where to begin..

Thank you for any help :)
• Apr 3rd 2010, 07:39 PM
NonCommAlg
Quote:

Originally Posted by mybrohshi5
Let A = $\displaystyle \begin{bmatrix}3&-\frac{1}{2}\\-6&5 \end{bmatrix}$

Find an invertible $\displaystyle S$ and a diagonal $\displaystyle D$ such that $\displaystyle S^{-1}AS = D$

S = $\displaystyle \begin{bmatrix}.&. \\.&. \end{bmatrix}$

D = $\displaystyle \begin{bmatrix}.&.\\.&. \end{bmatrix}$

I have never seen a problem like this and i am just a little confused on where to begin..

Thank you for any help :)

on the main diagonal of D put the eigenvalues of A and the columns of S are the eigenvectors corresponding to the eigenvalues of A.
• Apr 3rd 2010, 08:28 PM
mybrohshi5
This is what i got but it says it is wrong.

for my eigenvalues i got

$\displaystyle \lambda_1 = 2$ and $\displaystyle \lambda_2 = 6$

from my eigenvalue of $\displaystyle \lambda_1 = 2$ i got the eigenvector of
$\displaystyle \begin{bmatrix}\frac{1}{2}\\1\end{bmatrix}$

from my eigenvalue of $\displaystyle \lambda_2 = 6$ i got the eigenvector of
$\displaystyle \begin{bmatrix}\frac{1}{6}\\1\end{bmatrix}$

so for my S matrix i got

S = $\displaystyle \begin{bmatrix}\frac{1}{2}&\frac{1}{6} \\ 1&1\end{bmatrix}$

D = $\displaystyle \begin{bmatrix}2&0\\0&6\end{bmatrix}$

Can you see what i did wrong?

Thanks for the help
• Apr 3rd 2010, 08:35 PM
mybrohshi5
Never mind i got it. the 1/6 is supposed to be a -1/6 (Rofl)

Thanks again for the help