Eigenvalue and vector quick question

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The matrix A has eigenvalues λ = 8 and λ = 0. The calculated eigenvector for λ = 8 is -1, 3, and for λ = 0 is 3, -1. These results are correct, as eigenvectors can be multiplied by any constant, including -1, and still be valid. To verify the correctness, one can calculate Ax and check if it equals λx for the respective eigenvalues. The signs of the eigenvectors do not affect their validity.
Tzabcan
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So, I have the matrix:

A = -1 -3
3 9

Eigenvalues i calculated to be λ = 8 and 0

Now when i calculate the Eigenvector for λ = 8, i get the answer -1
3

Then when solve for the eigenvector for λ= 0 I get the eigenvector 3
-1

Both incorrect, they're supposed to have the signs the opposite way round. I don't understand how I am getting it wrong.

This is how i am attemption to solve it.

IMG_1940.jpg


Any pointers would be appreciated.
 
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Tzabcan said:
So, I have the matrix:

A = -1 -3
3 9

Eigenvalues i calculated to be λ = 8 and 0

Now when i calculate the Eigenvector for λ = 8, i get the answer -1
3

Then when solve for the eigenvector for λ= 0 I get the eigenvector 3
-1

Both incorrect, they're supposed to have the signs the opposite way round. I don't understand how I am getting it wrong.

This is how i am attemption to solve it.

IMG_1940.jpg


Any pointers would be appreciated.
I didn't check your work, but from what you said, your eigenvectors are fine. Your answers are the -1 multiples of the ones shown in your answers. If you have an eigenvector, any constant multiple of it will also be an eigenvector.

It's easy to check whether you work is correct. Calculate Ax, where A is your matrix and x is an eigenvector. If you get ##\lambda x##, where ##\lambda## is the eigenvalue associated with that eigenvector, your work is find. Note that if x is an eigenvector, -x will also be one as well.
 
Mark44 said:
I didn't check your work, but from what you said, your eigenvectors are fine. Your answers are the -1 multiples of the ones shown in your answers. If you have an eigenvector, any constant multiple of it will also be an eigenvector.

It's easy to check whether you work is correct. Calculate Ax, where A is your matrix and x is an eigenvector. If you get ##\lambda x##, where ##\lambda## is the eigenvalue associated with that eigenvector, your work is find. Note that if x is an eigenvector, -x will also be one as well.
Oh ok, thank you :)
 

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