How to find eigenvectors of a 3 by 3 matrix

[gboff21]

Question: find the eigenvectors of {{2,-1,-1},{-1,2,-1},{-1,-1,2}}.

Equations: none

Attempted answer:

I have the characteristic equation as x^3-6x^2+9 which gives eigenvalues as 0 3 and 3. This is correct says wolfram alpha

for x=0 {{2,-1,-1},{-1,2,-1},{-1,-1,2}}•{x,y,z}=0
I get 3 different equations! How does this give an answer?

For x=3 I get -{{1,1,1},{1,1,1},{1,1,1}•{x,y,z}=0
which gives v=anything
So what the hell??

Please help!

 

[Sethric]

For x=0 you get three equations for three unknowns. That's exactly what you need. Use whatever method you prefer to solve (substitute, gaussian elimination, etc.) for x,y, and z.

For x=3 you have repeated roots. Since this multiplicity is two, all you need to do is find two eigenvectors that work that are linearly independent. As you demonstrated, anything works, so just pick something simple for the first and then use that one to determine a second.

 

[gboff21]

For x=0. How do you get the right answer cos I constantly get {3,1,1} which is wrong.
And for x=3 how do you do do it. I don't understand how you use one randomly picked one, say {1,1,1} (which is one of the answers), to work out another!

 

[gboff21]

Never mind I've done it. Thanks

 

[rwisz]

To find the eigenvectors of a 3 by 3 matrix, we first need to find the eigenvalues. This can be done by solving the characteristic equation, which in this case is x^3-6x^2+9=0. As you correctly calculated, the eigenvalues for this matrix are 0, 3, and 3.

To find the eigenvectors for each eigenvalue, we need to solve the equation (A-λI)v=0, where A is the original matrix, λ is the eigenvalue, and v is the eigenvector. This will give us a system of equations, and the solutions to these equations will be the components of our eigenvector.

For the eigenvalue of 0, we have the equation (A-0I)v=0, which simplifies to Av=0. Solving this system of equations will give us the eigenvector associated with this eigenvalue.

For the eigenvalue of 3, we have the equation (A-3I)v=0, which simplifies to (A-3I)v=0. Solving this system of equations will give us the eigenvector associated with this eigenvalue.

In this specific example, the eigenvectors for the eigenvalue of 0 are any vector in the null space of the matrix, since multiplying the matrix by this vector will result in 0. For the eigenvalue of 3, we can solve the system of equations to find the eigenvector associated with it.

In summary, to find the eigenvectors of a 3 by 3 matrix, we first find the eigenvalues by solving the characteristic equation. Then, for each eigenvalue, we solve the equation (A-λI)v=0 to find the eigenvector associated with it. I hope this helps clarify the process for finding eigenvectors.