# Thread: Eigenvalues, eigenvectors, finding an orthogonal matrix

1. ## 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.

2. 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.

3. Can you fix you LaTex errors? 'Cause this looks really helpful

4. Originally Posted by chella182
Can you fix you LaTex errors? 'Cause this looks really helpful
I think it's fixed now.

5. Cheers I'll have a look at this when I've tackled the other problems and see if I understand.