1
$\begingroup$

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 $\lambda$.

Thanks in advance!

$\endgroup$
  • $\begingroup$ What do you call a generalized Eigenvector ? $\endgroup$ – Yves Daoust Sep 5 '16 at 10:38
  • $\begingroup$ I know that eigenvectors are generalized eigenvectors but the converse is not necessarily true... $\endgroup$ – Kierra Sep 5 '16 at 10:44
  • 1
    $\begingroup$ 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. $\endgroup$ – DonAntonio Sep 5 '16 at 10:51
  • $\begingroup$ 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? $\endgroup$ – Kierra Sep 5 '16 at 11:01
  • 2
    $\begingroup$ 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 $\lambda$. $\endgroup$ – Dave Sep 5 '16 at 13:34
1
$\begingroup$

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 $$ \mathbf{T} v_{k} = \lambda_{k} v_{k} \qquad \Rightarrow \qquad \left( \mathbf{T} - \lambda_{k} \mathbf{I}_{3} \right) v_{k} = \mathbf{0} $$

$\lambda = 3$

$$ \begin{align} \left( \mathbf{T} - 3 \mathbf{I}_{3} \right) v_{1} &= \mathbf{0} \\ \left[ \begin{array}{rrr} -4 & 2 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -3 \\ \end{array} \right] % \left[ \begin{array}{c} v_{1_{1}} \\ v_{1_{2}} \\ v_{1_{3}} \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] % \qquad \Rightarrow \qquad v_{1} = \left[ \begin{array}{c} 1 \\ 2 \\ 0 \end{array} \right] % \end{align} $$

$\lambda = -1$

$$ \begin{align} \left( \mathbf{T} + \mathbf{I}_{3} \right) v_{2} &= \mathbf{0} \\ \left[ \begin{array}{rrr} 0 & 2 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array} \right] % \left[ \begin{array}{c} v_{2_{1}} \\ v_{2_{2}} \\ v_{2_{3}} \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] % \qquad \Rightarrow \qquad v_{2} = \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] % \end{align} $$

$\lambda = 0$

$$ \begin{align} \left( \mathbf{T} + 0 \mathbf{I}_{3} \right) v_{3} &= \mathbf{0} \\ \left[ \begin{array}{rrr} -1 & 2 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \\ \end{array} \right] % \left[ \begin{array}{c} v_{3_{1}} \\ v_{3_{2}} \\ v_{3_{3}} \end{array} \right] &= \left[ \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right] % \qquad \Rightarrow \qquad v_{3} = \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right] % \end{align} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.