Question

Find the eigenvalues and eigenvectors of
A = [ 1 1 1 ]
1 1 1
1 1 1

Answer 1

Your matrix is \[\left[\begin{matrix}1 & 1 & 1 \\ 1 & 1 & 1\\ 1 & 1 & 1\end{matrix}\right]\]?

In order to calculate eigenvectors, we first need the eigenvalues. We do this by forming the matrix \[(A-\lambda I)\] where A is your matrix and I is the identity matrix, and then taking the determinant of that new matrix.

So the new matrix would be \[\left[\begin{matrix}1-\lambda & 1 & 1 \\ 1 & 1-\lambda & 1\\ 1 & 1 & 1-\lambda\end{matrix}\right]\]Can you take the determinant of that?

Answer 2

Ok, so far I have the characteristic polynomial down to \[\lambda ^{2}(\lambda-3)\]
So the eigenvalues are 0 and 3?

And after that, I'm not sure how to find the eigenvectors...

Answer 3

Ok, we we have our eigenvalues. 3, 0, and 0 (yes, two of them are 0).

Now, in order to get our eigenvectors, we form this matrix:
\[\left[\begin{matrix}1-\lambda _{1} & 1 & 1 \\ 1 & 1-\lambda _{1} & 1\\ 1 & 1 & 1-\lambda _{1}\end{matrix}\right]\] If we let lambda 1 be 3, then your new matrix will look like this:\[\left[\begin{matrix}-2 & 1 & 1 \\ 1 & -2 & 1\\ 1 & 1 & -2\end{matrix}\right]\]Now you just need to row-reduce this down to it's reduced row-echelon form. This will give you the first set of eigenvectors. Then you do the same thing for 0. Note that since we have two eigenvalues that are 0, we need to get two eigenvectors out of it. When you get here I will help you (but for now row-reducing the above matrix is your goal).

Answer 4

Does what I've said make sense so far?

Answer 5

Or am I going too slowly?

Answer 6

yeah, but I got two leading variables so i figured i made a mistake somewhere so i'm trying to find it... one second

Answer 7

ok i did it twice, but i still get