# linear transformation eigenvectors and eigenvalues

Consider the operator defined by $T(x, y, z) = (-x+2y, 3y, 0)$.

1. Find the eigenvalues of T and all corresponding eigenvectors.

2. 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 $\lambda$.

• What do you call a generalized Eigenvector ? – Yves Daoust Sep 5 '16 at 10:38
• I know that eigenvectors are generalized eigenvectors but the converse is not necessarily true... – Kierra Sep 5 '16 at 10:44
• Since the matrix is $\;3\times 3\;$ and it has three different eigenvalues, it is diagonalizable and it thus has no generalized eigenvectors that are not standard eigenvectors. – DonAntonio Sep 5 '16 at 10:51
• But how do I compute these? like the second column of my matrix w.r.t the standard basis is a linear combination of 2(1, 0, 0) + 3(0, 1, 0) so does that mean these are my eigenvectors? – Kierra Sep 5 '16 at 11:01
• Solving the equation $(A-\lambda I)x=0$ where $A$ is the matrix of $T$ relative to the standard basis, and $\lambda$ is the eigenvalue, will give you the corresponding eigenvector for $\lambda$. Solve this equation with each eigenvalue in place of $\lambda$. – Dave Sep 5 '16 at 13:34

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}