# Math Help - Eigenvalues and eigen vectors

1. ## Eigenvalues and eigen vectors

The question is, for which x is the equation $A_x= lambda_x$ valid? And then I am a given a square matrix of say (9 5)(first row) (6 7) (second row.)Then you are given a lambda value of say 4 which apparently is the eigenvalue and $x_1 and x_2$. I don't understand the question properly. I can calculate for the eigenvectors since I have the eigenvalue by subtracting them from the diagonal matrix so I will have 9-4 and 7-4 and so on and then I can find the value of x to get the eigenvector. But I don't understand the question properly. I can't multiply the vector by the matrix since it is not a square matrix so, how do you go about it? Thanks all in advance.

2. Hi

It seems not very clear to you ...

http://www.mathhelpforum.com/math-he...envectors.html

3. Thank you gag. I understand the equation and can get the eigenvectors and we are already given the eigenvalue. What I don't understand is just the question which is: for which equation is the $A_x= lambda_x$ valid? We already have lambda which is the eigenvalue, we can get x which is the eigenvector and we have A which is the matrix. This equation is also the same as the det (A - lambdaI)=0. I can calculate for the eigenvectors of both, we already have the eigenvalue so what does for which x is $A_x= lambda_x$ valid mean then?

4. For each eigenvalue $\lambda$, you can find an eigenvector $X$

Thus you have for instance $AX_1 = \lambda_1\:X_1$

All the vectors $Y_1 = k\:X_1$ are such that $AY_1 = \lambda_1\:Y_1$ because $AY_1 = A\: k\:X_1 = k\:AX_1 = k\:\lambda_1\:X_1 = \lambda_1\:k\:X_1 = \lambda_1\:Y_1$

In other words all the vectors parallel to X1 are solutions

5. Originally Posted by running-gag
For each eigenvalue $\lambda$, you can find an eigenvector $X$

Thus you have for instance $AX_1 = \lambda_1\:X_1$

All the vectors $Y_1 = k\:X_1$ are such that $AY_1 = \lambda_1\:Y_1$ because $AY_1 = A\: k\:X_1 = k\:AX_1 = k\:\lambda_1\:X_1 = \lambda_1\:k\:X_1 = \lambda_1\:Y_1$

In other words all the vectors parallel to X1 are solutions
Thank you. It is just the question and not solving the equation. I know how to get the eigenvalues and eigenvectors but I just didn't get the question: for which x does the equation $A_X=lambda_X$valid? What does that mean? The matrices I have all have eigen vectors and we are already given their eigen values so, what does for which x (eigenvektor) is the equation valid mean? I think maybe I will try to seek clarification from the teacher but I appreciate your help. Thanks.

6. Originally Posted by Keep
Thank you. It is just the question and not solving the equation. I know how to get the eigenvalues and eigenvectors but I just didn't get the question: for which x does the equation $A_X=lambda_X$valid? What does that mean? The matrices I have all have eigen vectors and we are already given their eigen values so, what does for which x (eigenvektor) is the equation valid mean? I think maybe I will try to seek clarification from the teacher but I appreciate your help. Thanks.
Assuming that by $A_X= lambda_X$, you mean $AX= \lambda X$, then asking "for what X is this equation valid" is just asking "what values of x satisfy this equation". If $\lambda$ is NOT an eigenvalue, only x= 0 satisfies that equation. If $\lambda$ is an eigenvalue the all eigenvectors, as well as 0, satisfy the equation.

7. Originally Posted by HallsofIvy
Assuming that by $A_X= lambda_X$, you mean $AX= \lambda X$, then asking "for what X is this equation valid" is just asking "what values of x satisfy this equation". If $\lambda$ is NOT an eigenvalue, only x= 0 satisfies that equation. If $\lambda$ is an eigenvalue the all eigenvectors, as well as 0, satisfy the equation.
Oh OK, I think I get it now. Looks easier than I thought. Thanks!