The characteristic equation is . Then [tex](\lambda-.8)^2= .04[tex] so and or .

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 such that 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 as x<1, 1>. Since .6 is an eigenvector there exist non-zero vectors such that 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 is the diagonal matrix with the eigenvalues 1 and .6. Since you want P and 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 , then . Now, because the in the middle cancel each other. Similarly, and it is easy to see that continues: for any positive integer, n, so, in particular and is easy: it is just the diagonal matrix with the 100th power of the eigenvalues on the diagonal.