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

  • #1

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....
 

Answers and Replies

  • #2
blue_leaf77
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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
ehild
Homework Helper
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1,827

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:smile:.
To make it easier, pull out the factor 4 from the matrix.
 
Last edited:
  • #4
Ray Vickson
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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.
 
  • #5
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
 
Last edited:
  • #6
ehild
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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?
 
  • #7
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]
 
  • #9
ehild
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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.
 
  • #10
Ray Vickson
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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
[tex] A^n = 4^n E_1 + (-4)^n E_2, \; n= 0,1,2, \ldots , [/tex]
giving
[tex] I = E_1+E_2\\
A = 4E_1 - 4E_2. [/tex]
Thus
[tex] E_1 = \frac{1}{8}A + \frac{1}{2}I = \pmatrix{-7/2 & 63/2 \\ -1/2 & 9/2} [/tex]
and
[tex] E_2 = -\frac{1}{8} A + \frac{1}{2} I = \pmatrix{9/2 & -63/2 \\ 1/2 & -7/2} [/tex]
 

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