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

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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.
Can someone please assist?

Thanks in advance.
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.
Can someone please assist?

Thanks in advance.

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