# Homework Help: Finding a diagonalizable matrix

1. Oct 23, 2009

### DanielJackins

1. The problem statement, all variables and given/known data

Let M = [{-1, -6}{3, 8}] ( meaning {row1} {row2}

Find formulas for the entries of M^n, where n is a positive integer.

3. The attempt at a solution

So I found the eigenvalues and eigenvectors; e.value1 = 2, e.value2 = 5, e.vector1 = [{-2}{1}], e.vector2 = [{-1}{1}]. I'm 99% sure these are all correct.

So I proceed to write it as A = PDP^-1, with P as [{-2, -1}{1, 1}], D as [{2, 0}{0, 5}], and P^-1 as 1/10[{8,6}{-3,-1}].

So next I believe you're supposed to multiply them all together, with the 2 and 5 in D raised to the power of n. I did this and came up with a huge ugly jumble of numbers, which I entered and were incorrect. Am I doing this right? Or did I make a mistake somewhere?

Thanks

2. Oct 23, 2009

### Staff: Mentor

M = PDP-1
Then
Mn =(PDP-1)n
= (PDP-1)(PDP-1)...(PDP-1)
= (PDP-1PDP-1...PDP-1)
= PDnP-1

All the interior PP-1 products simplify to I, and you're left with what's shown above.

The idea is that, instead of raising M to the power n, you can raise D to the power n (easy to do, since it is a diagonal matrix), and then multiply that on the left by P and on the right by P-1.

I haven't checked your work, so once you get a result, verify it by comparing Mn for a small value of n (like 2), to see if it agrees with what you have on the right.

3. Oct 23, 2009

### DanielJackins

I did raise D to the power n, and then multiplied on the left by P and on the right by P^-1, and it gave me a big mess

4. Oct 23, 2009

### Staff: Mentor

Did you try it to the power 2? Compare M2 and PD2P-1. If they're not equal, here are some things to check.
1. Your eigenvectors need to be in the matrix P in the same order that the eigenvalues are in the matrix D. IOW, in the same columns.
2. Are your eigenvalues and eigenvectors correct? It should be that (A - (eigenvalue_1)I)(eigenvector_1) = 0, and the same for the other eigenvalue/eigenvector pair.
3. You might have a mistake in your calculation for P-1.

5. Oct 23, 2009

### HallsofIvy

Your first problem is that 2 and 5 are NOT eigenvalues of that matrix.
If 2 were an eigenvalue then we must have
$$\begin{bmatrix}-1 & -6 \\ 3 & -8\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}=2\begin{bmatrix}x \\ y\end{bmatrix}$$
Which gives the two equations -x- 6y= 2x and 3x- 8y= 2y. Those are the same as -3x- 6y= 0 and 3x- 10y= 0. Adding the two equations, -16y= 0 which is satisfied only if y= 0 and then we get x= 0. There is no non-trivial vector for which that is true. The same happens if you try 5 rather than 2.