Consider the operator defined by $T(x, y, z) = (-x+2y, 3y, 0)$.
Find the eigenvalues of T and all corresponding eigenvectors.
Find each generalised eigenvector corresponding to each eigenvalue.
So for 1, I found the matrix with respect to the standard basis $(1,0,0), (0,1,0), (0,0,1)$. It was upper triangular so I read the eigenvalues from the diagonal entries, which gave me eigenvalues of $-1, 3$ and $0$. I'm just so lost with how to find the corresponding eigenvectors, I feel like this should be really simple.
I know once I have found these eigenvectors for 2, I can simply compute the generalized eigenvectors by solving the equation $(T- \lambda I)^j (v) = 0$ where $v$ is my particular eigenvector with corresponding eigenvalue $\lambda$.
Thanks in advance!
Best Answer
Operator
$$ \mathbf{T} = \left[ \begin{array}{rrr} -1 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] $$
Eigenvalues
To find the eigenvalues, compute $p(\lambda)$, the characteristic polynomial. $$ p (\lambda ) = \det \left( \mathbf{T} - \lambda \mathbf{I}_{3} \right) = \det \left[ \begin{array}{ccr} -\lambda -1 & 2 & 0 \\ 0 & 3-\lambda & 0 \\ 0 & 0 & -\lambda \\ \end{array} \right] = -\lambda \left( 3 - \lambda \right) \left( -1 - \lambda \right) $$ The roots $p(\lambda) = 0$ are the eigenvalues: $\lambda = \left\{ 3, -1, 0 \right\}$.
Eigenvectors
Solve $$ \mathbf{T} v_{k} = \lambda_{k} v_{k} \qquad \Rightarrow \qquad \left( \mathbf{T} - \lambda_{k} \mathbf{I}_{3} \right) v_{k} = \mathbf{0} $$
$\lambda = 3$
$$ \begin{align} \left( \mathbf{T} - 3 \mathbf{I}_{3} \right) v_{1} &= \mathbf{0} \\ \left[ \begin{array}{rrr} -4 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -3 \\ \end{array} \right] % \left[ \begin{array}{c} v_{1_{1}} \\ v_{1_{2}} \\ v_{1_{3}} \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] % \qquad \Rightarrow \qquad v_{1} = \left[ \begin{array}{c} 1 \\ 2 \\ 0 \end{array} \right] % \end{align} $$
$\lambda = -1$
$$ \begin{align} \left( \mathbf{T} + \mathbf{I}_{3} \right) v_{2} &= \mathbf{0} \\ \left[ \begin{array}{rrr} 0 & 2 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array} \right] % \left[ \begin{array}{c} v_{2_{1}} \\ v_{2_{2}} \\ v_{2_{3}} \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] % \qquad \Rightarrow \qquad v_{2} = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] % \end{align} $$
$\lambda = 0$
$$ \begin{align} \left( \mathbf{T} + 0 \mathbf{I}_{3} \right) v_{3} &= \mathbf{0} \\ \left[ \begin{array}{rrr} -1 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] % \left[ \begin{array}{c} v_{3_{1}} \\ v_{3_{2}} \\ v_{3_{3}} \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] % \qquad \Rightarrow \qquad v_{3} = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] % \end{align} $$