# Eigenvalues, eigenvectors, finding an orthogonal matrix

• May 5th 2009, 09:42 AM
chella182
Eigenvalues, eigenvectors, finding an orthogonal matrix
Sorry if this is the wrong section - there seems to be a few matrices questions in here though. The question goes:

Let A be the $2\times2$ matrix given by
$A=\left( \begin{array}{cc} 0.8 & 0.2 \\ 0.2 & 0.8 \end{array} \right).$
i) Find the eigenvalues and eigenvectors of the matrix A.
ii) Find an orthogonal matrix O, which diagonalizes A.
iii) Compute $A^{100}$.

I'm rubbish with stuff like this.
• May 5th 2009, 10:28 AM
HallsofIvy
Quote:

Originally Posted by chella182
Sorry if this is the wrong section - there seems to be a few matrices questions in here though. The question goes:

Let A be the $2\times2$ matrix given by
$A=\left( \begin{array}{cc} 0.8 & 0.2 \\ 0.2 & 0.8 \end{array} \right).$
i) Find the eigenvalues and eigenvectors of the matrix A.
ii) Find an orthogonal matrix O, which diagonalizes A.
iii) Compute $A^{100}$.

I'm rubbish with stuff like this.

The characteristic equation is $\left|\begin{array}{cc}.8- \lambda & 0.2 \\ 0.2 0.8- \lambda\end{array}\right|= (\lambda- .8)^2- .04= 0$. Then [tex](\lambda-.8)^2= .04[tex] so $\lambda- 0.8= \pm .2$ and $\lambda= .8+ .2= 1$ or $\lambda= .8- .2= .6$.

The find the eigenvectors corresponding to those, use the basic definition of "eigenvector" and "eigenvalue". Since 1 is an eigenvector there exist non-zero vectors $\begin{bmatrix}x \\ y\end{bmatrix}$ such that $\begin{bmatrix}.8 & .2 \\ .2 & .8\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}$ which gives the two equations .8x+ .2y= x and .2x+ .8y= y. The first equation can be written .2y= .2x so y= x. The second equation can be written .2y= .2x also which is also satisfied by y= -x. We can write any eigenvector corresponding to $\lambda= 1$ as x<1, 1>. Since .6 is an eigenvector there exist non-zero vectors $\begin{bmatrix}x \\ y\end{bmatrix}$ such that $\begin{bmatrix}.8 & .2 \\ .2 & .8\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}.6x \\ .6y\end{bmatrix}$ which gives the two equations .8x+ .2y= .6x and .2x+ .8y= .6y. Both equations now reduce to .2y= -.2x or y= -x. Any eigenvector corresponding to eigenvalue x<1, -1>.

Taking eigenvectors as columns gives a matrix P such that $P^-1AP$ is the diagonal matrix with the eigenvalues 1 and .6. Since you want P and $P^{-1}$ to be orthogonal, the columns must be "orthonormal". It is easy to see that <1, 1> and <1, -1> are orthogonal. To normalize them, find the length of each and divide each vector by its length.

If $P^{-1}AP= D$, then $A= PDP^{-1}$. Now, $A^2= (pDP^{-1})(PDP^{-1})= PD^2P^{-1}$ because the $P^{-1}P$ in the middle cancel each other. Similarly, $A^3= (PDP^{-1})A^2= PDP^{-1}(PD^2P^{-1})= PD^3P^{-1}$ and it is easy to see that continues: for any positive integer, n, $A^n= PD^nP^{-1}$ so, in particular $A^100= PD^{100}P^{-1}$ and $D^{100}$ is easy: it is just the diagonal matrix with the 100th power of the eigenvalues on the diagonal.
• May 5th 2009, 10:35 AM
chella182
Can you fix you LaTex errors? 'Cause this looks really helpful :)
• May 5th 2009, 10:59 AM
Jameson
Quote:

Originally Posted by chella182
Can you fix you LaTex errors? 'Cause this looks really helpful :)

I think it's fixed now.
• May 5th 2009, 11:06 AM
chella182
Cheers :) I'll have a look at this when I've tackled the other problems and see if I understand.