Thread: finding eigenvectors of a diagno matrix

I have a three by three diagnol matrix
[1,2,-4;0,2,1;0,0,3]
1,2,3
i found the eigenvectors for 2 and 3 but i am having finding the eigenvector for when lambda equals 1.
When I try to solve I can only find the vector x=0 which it cannot be.

I realize that this would be a problem with any diagonal matrix and was wondering how to go about finding the eigenvector.

Originally Posted by mathleticman

I have a three by three diagnol matrix
[1,2,-4;0,2,1;0,0,3]
1,2,3
i found the eigenvectors for 2 and 3 but i am having finding the eigenvector for when lambda equals 1.
When I try to solve I can only find the vector x=0 which it cannot be.

I realize that this would be a problem with any diagonal matrix and was wondering how to go about finding the eigenvector.

That is NOT a "diagonal" matrix- it is an "upper triangular" matrix!

Personally, I have always prefered to find eigenvectors directly from the definition. An eigenvector of linear operator A, corresponding to eigenvalue must satisfy

So any eigenvector of this matrix corresponding to eigenvalue 2 must satisfy

which is the same as the three equations x+ 2y- 4z= 2x, 2y+ z= 2y, and 3z= 2z. (Since the matrix is upper triangular, those three equations are "partially uncoupled")- the last involves only z and it is easy to see that z= 0. then the second equation says 2y= 2y which is true for all y. The third is the same as x+ 2y- 4(0)= 2x or 2y= x. That is, such an eigenvector can be written <2y, y, 0>= y<2, 1, 0> so any multiple of <2, 1, 0> is an eigenvector corresponding to eigenvalue 2.