Strange question regarding eigenvectors / eigenvalues

In summary, the homework equation is Ax = lambda X, where Ax is the eigenvalue corresponding to the eigenvector [2,-1]^T and P is the corresponding entry in A^n. To find A and A^-1, you would use the PDP^-1 equation.
  • #1
lubricarret
34
0

Homework Statement



Suppose that the 2x2 matrix A has eigenvalues lambda = 1,3 with corresponding eigenvectors [2,-1]^T and [3,2]^T. Find a formula for the entries of A^n for any integer n. And then, find A and A^-1 from your formula.

Homework Equations



Ax = lambda X
(P^-1)AP = D
A = PDP^-1

The Attempt at a Solution



I've been trying to figure this out for a long time. I'm really not sure where to start... If anyone could provide some sort of guidance on how to begin this problem... that would be really helpful. I've tried using the above formulas, but I'm just not sure how to get a generalized formula. I could find A and A^-1 individually without the generalized formula, but the problem asks for it; thanks in advance for even the slightest hint.
 
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  • #2
D=P^(-1)AP is [[1,0],[0,3]], right? And you have P and P^(-1), I hope. Then isn't A^n=P(D^n)P^(-1)? Since A^n=PDP^(-1)*PDP^(-1)*... n times. D^n is pretty easy to figure out. It's not a strange question at all. It's easy.
 
  • #3
Hi Dick.

Thanks for your explanation! So, my formula would be:
A^(n) = P(D^n)P^(-1)

P is obviously just
[2 3
-1 2]

and P^(-1) would then just be its inverse, or
[(2/7) (-3/7)
(-1/7) (2/7)]

Then you just use this P, P^(-1) and the given D^n to get the answer of A^n.

So, then with A^(-1), I would just get the inverse of D, and plug it into the formula...?



Thanks!
 
  • #4
D^n is just [[1,0],[0,3^n]], right? Sure, then use P and P^(-1) to convert it back to A^n. Why do you need to invert D? Ohh. I see what you mean. Sure you would.
 
  • #5
Hi

Thanks. Just for clarification though, why would I not say

A^(n) = (PDP^(-1))^n

Why is it just D that I raise to the n, and not the whole right of the equation?
 
  • #6
Because they are both the same thing and it's much easier to raise D^n than A^n. Take A^3. A^3=(PDP^(-1))*(PDP^(-1))*(PDP^(-1)). Don't you see how the P^(-1)*P parts cancel in the middle?
 
  • #7
Ahhh, ok. Thanks for clearing that up for me; understand it now.

Thanks!
 

FAQ: Strange question regarding eigenvectors / eigenvalues

What are eigenvectors and eigenvalues?

Eigenvectors and eigenvalues are mathematical concepts that are used in linear algebra. An eigenvector is a vector that, when multiplied by a square matrix, remains in the same direction but is scaled by a constant factor. The corresponding constant factor is known as the eigenvalue.

How are eigenvectors and eigenvalues used in science?

Eigenvectors and eigenvalues are used in a variety of scientific fields, such as physics, engineering, and computer science. They are particularly useful in analyzing and solving systems of linear equations, as well as in understanding the behavior of complex systems.

What is the significance of finding eigenvectors and eigenvalues?

Finding eigenvectors and eigenvalues can reveal important information about a system, such as its stability, equilibrium points, and directional behavior. They are also used in data analysis and pattern recognition, where they can help identify underlying patterns and trends in data.

Are there any real-world applications of eigenvectors and eigenvalues?

Yes, there are many real-world applications of eigenvectors and eigenvalues. Some examples include analyzing the stability of chemical reactions, predicting the behavior of electrical circuits, and understanding the dynamics of population growth in biology.

Is it important to understand eigenvectors and eigenvalues in order to be a successful scientist?

While it may not be necessary to have a deep understanding of eigenvectors and eigenvalues in all scientific fields, having a basic understanding can be helpful in problem-solving and data analysis. In certain fields, such as physics and engineering, a thorough understanding of these concepts is crucial for success.

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