Eigenvalue and vector quick question

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SUMMARY

The discussion revolves around the calculation of eigenvalues and eigenvectors for the matrix A = [[-1, -3], [3, 9]]. The eigenvalues determined are λ = 8 and λ = 0. The user initially calculated the eigenvectors as [-1, 3] for λ = 8 and [3, -1] for λ = 0, but expressed confusion over the signs being opposite. The response clarified that the calculated eigenvectors are indeed correct, as any scalar multiple of an eigenvector is also valid, including the negative of the vector.

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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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