# Eigenvalues and eigenvectors of a 3x3 matrix A? A unknown?

• March 16th 2010, 10:45 AM
chocaholic
Eigenvalues and eigenvectors of a 3x3 matrix A? A unknown?
I need some help with the following problem please?

Let A be a 3x3 matrix with eigenvalues -1,0,1 and corresponding eigenvectors
l1l . l0l . l0l
l0l ; l1l ; l1l respectively.
l0l . l1l . l2l

Find A.

I know that I need to work backwards on this problem so I set up the characteristic equation with th known eigenvalues ending up with x^3-x=0 but now I'm stuck I don't know where to go from here and how to use the eigenvectors. I can set up A with all unknowns and subtract it from kI where k is the eigenvalue and I is the identity matrix then find the determinant, but there are too many unknowns and too few equations. I'm totally confused.

• March 16th 2010, 11:22 AM
tonio
Quote:

Originally Posted by chocaholic
I need some help with the following problem please?

Let A be a 3x3 matrix with eigenvalues -1,0,1 and corresponding eigenvectors
l1l . l0l . l0l
l0l ; l1l ; l1l respectively.
l0l . l1l . l2l

Find A.

I know that I need to work backwards on this problem so I set up the characteristic equation with th known eigenvalues ending up with x^3-x=0 but now I'm stuck I don't know where to go from here and how to use the eigenvectors. I can set up A with all unknowns and subtract it from kI where k is the eigenvalue and I is the identity matrix then find the determinant, but there are too many unknowns and too few equations. I'm totally confused.

Put $P=\begin{pmatrix}1&0&0\\0&1&1\\0&1&2\end{pmatrix}\ Longrightarrow PAP^{-1}=\begin{pmatrix}\!\!-1&0&1\\\;0&0&0\\\;0&0&1\end{pmatrix}$ ...and now find A. (Cool)

Tonio
• March 17th 2010, 12:13 AM
chocaholic
Hi Tonio isn't that the formula for diagonalization? I don't understand how it relates here and how you got the answer for PAP^-1 ? Further insight would be appreciated.
Thanks
• March 17th 2010, 08:58 AM
tonio
Quote:

Originally Posted by chocaholic
Hi Tonio isn't that the formula for diagonalization? I don't understand how it relates here and how you got the answer for PAP^-1 ? Further insight would be appreciated.
Thanks

Of course that's the formula for diagonalization! We know A is diagonalizable because we're given three LINEAR INDEPENDENT eigenvectors (how we know they're lin. indep. without directly checking this?), which are then a basis for the vector space, and thus we can apply that formula...(Nerd)

Now, sometimes the formula is $P^{-1}AP=D$ and other times it is $PAP^{-1}=D$ , with $D=$ the diagonal matrix with the eigenvalues on its main diagonal (it all depends on how we define a matrix of a trnasformation), but since we know what is $D$ we can now find A . (Nod)

Tonio