i have a 3x3 matrix
i got the eigenvalues of 2, 1, and 0.
im having a big problem with how to get the corresponding eigenvectors
if anyone can help me that would be great!
How we do this is place lambda on the diagonal in the matrix! In other words,
Now we need to row reduce!
At this point I don't think we can reduce any more.
Now we know there exists a set of vectors such that
Carry out the multiplication and find your values. There might be trivial solutions but that's okay!
Find more at: Eigenvalues and Eigenvectors
Or, use the definition of "eigenvalue" and "eigenvector".
If you are right that 2 is an eigenvalue, then there must exist a vector, , x, y, z not all 0, such that
That gives the three equations -2x-8 y- 12z= 2x, x+ 4y+ 4z= 2y, and z= 2z.
Of course, that last equations tells us that z= 0 so the first two become -2x- 8y= 2x, or 8y= -4x so that x= -2y, and x+ 4y= 2y and, again, x= -2y. The fact that there exist an infinite number of solutions to those equations confirms that 2 is an eigenvalue. Any eigenvector corresponding to eigenvalue 2 is of the form .
Now, to find the eigenvectors corresponding eigenvalues 1 and 0, just solve -2x- 8y- 12z= x, x+ 4y= y, z= z, and -2x- 8y- 12z= 0, x+ 4y= 0, z= 0.
On the contrary, eigenvectors are great fun! (Not to mention extremely useful.)