# Linear Algebra: 2x2 matrix raised to the power of n

1. Jun 24, 2016

### Miguel Guerrero

1. The problem statement, all variables and given/known data
If n is a positive integer, then 2x2 matrix [-32,252] [-4,32] raised to the power of n is....

2. Relevant equations
I know that first I should diagonalize the given matrix, something I also seem to have a hard time with.

3. The attempt at a solution
I determined the eigenvalues to be 32 and -32 and determined eigenvectors to be 16,1 and 252,1 , respectively. I am not sure if this is right or where to even go from there....

2. Jun 24, 2016

### blue_leaf77

I think you are supposed to calculate
$$\left( \begin{array}{cc} -32 & 252 \\ -4 & 32 \\ \end{array} \right)^n$$
with the help of the diagonal form of the original matrix. By the way, the eigenvalues you got are not right.

3. Jun 25, 2016

### ehild

You need not diagonalize the matrix. What is the square of it? What is its cube? Try, it is surprising.
To make it easier, pull out the factor 4 from the matrix.

Last edited: Jun 25, 2016
4. Jun 25, 2016

### Ray Vickson

If you have a 2x2 matrix $A$ with two distinct eigenvalues $r_1, r_2$, then there are two constant 2x2 matrices $E_1, E_2$ such that $p(A) = E_1 p(r_1) + E_2 P(r_2)$ for any polynomial $p$, so $A^n = E_1 r_1^n + E_2 r_2^n$. You can figure out $E_1$ and $E_2$ by applying that to two simple examples of $p$, such as $p(x) = x^0 =1 \Rightarrow p(A) = E_1 r_1^0 + E_2 r_2^0 = I$ (the identity matrix) and $p(x) = x \Rightarrow p(A) = A = E_1 r_1 + E_2 r_2$. That gives you two equations in the two "unknowns" $E_1$ and $E_2$.

Your eigenvalues are incorrect; start again.

5. Jun 25, 2016

### Miguel Guerrero

So, I redid my calculations for the eigenvalues and after taking out the factor of 4 i found -1 and 1, do i need to multiply theses values by the factor of 4 that I took out? There is also a hint that says my answer will be a formula that involves n

Last edited: Jun 25, 2016
6. Jun 25, 2016

### ehild

Yes, the eigenvalues of the original matrix are 4 and -4.
BUT: try my previous hint. The original matrix is A=4B. A2=42B2. Determine B2. What matrix do you get?

7. Jun 25, 2016

### Miguel Guerrero

[1,0],[0,1]

8. Jun 25, 2016

### Miguel Guerrero

I was able to solve the problem guys, thanks for your help

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9. Jun 25, 2016

### ehild

If B=(1/4)A

B2=E

and B3=B.

So the resulting matrix is 4n E if n is even and 4n B if n is odd.

10. Jun 26, 2016

### Ray Vickson

Alternatively, using the method in #4 we have
$$A^n = 4^n E_1 + (-4)^n E_2, \; n= 0,1,2, \ldots ,$$
giving
$$I = E_1+E_2\\ A = 4E_1 - 4E_2.$$
Thus
$$E_1 = \frac{1}{8}A + \frac{1}{2}I = \pmatrix{-7/2 & 63/2 \\ -1/2 & 9/2}$$
and
$$E_2 = -\frac{1}{8} A + \frac{1}{2} I = \pmatrix{9/2 & -63/2 \\ 1/2 & -7/2}$$

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