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

[Miguel Guerrero]

Homework Statement

If n is a positive integer, then 2x2 matrix [-32,252] [-4,32] raised to the power of n is...

Homework Equations

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

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...

 

[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.

 

[ehild]

Homework Statement

If n is a positive integer, then 2x2 matrix [-32,252] [-4,32] raised to the power of n is...

Homework Equations

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

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...

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.

 

[Ray Vickson]

Homework Statement

If n is a positive integer, then 2x2 matrix [-32,252] [-4,32] raised to the power of n is...

Homework Equations

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

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...

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.

 

[Miguel Guerrero]

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.

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

 

[ehild]

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

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?

 

[Miguel Guerrero]

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?

[1,0],[0,1]

 

[Miguel Guerrero]

[1,0],[0,1]

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

 

[ehild]

If B=(1/4)A

B²=E

[1,0],[0,1]

and B³=B.

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

 

[Ray Vickson]

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.

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\\<br /> A = 4E_1 - 4E_2.
Thus
E_1 = \frac{1}{8}A + \frac{1}{2}I = \pmatrix{-7/2 &amp; 63/2 \\ -1/2 &amp; 9/2}
and
E_2 = -\frac{1}{8} A + \frac{1}{2} I = \pmatrix{9/2 &amp; -63/2 \\ 1/2 &amp; -7/2}