I have a matrix:


solving $\det|A-\lambda{I}|$ I got characteristic polynom that equals to $(2-\lambda)^3 = 0$ for eigenvalue found two eigenvectors and one generalized eigenvector: $v_1=(1,2,0)\quad v_2=(0,0,1) \quad v_3=(1,0,0)$

What do I have to do to find Jordan basis here? (and what do I need to find Jordan basis in general, I mean is there appropriate alghoritm?, What I read did not make things more clear).

  • $\begingroup$ The eigenvectors and generalized eigenvectors, together, form the Jordan basis. But I think your first eigenvector is wrong. $\endgroup$ – Ian Sep 17 '17 at 21:49
  • $\begingroup$ The characteristic polynomial of this matrix is not $(2-\lambda)^3$. $\endgroup$ – Bernard Sep 17 '17 at 22:01
  • $\begingroup$ @Bernard, I confused one sign in the matrix, should be fixed now $\endgroup$ – M.Mass Sep 17 '17 at 22:04

Here the way to go: consider the sequence of kernels: $$\{\,0\,\}\varsubsetneq\ker(A-2I)\varsubsetneq\ker(A-2I)^2\subset\dots$$ The sequence stops after step $2$ since $$A-2I=\begin{bmatrix}-2&1&0\\-4&2&0\\-2&1&0\end{bmatrix}\qquad (A-2I)^2=\begin{bmatrix} 0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ $A-2I$ has rank $1$, hence its kernel (the eigenspace) has codimension $1$, i.e. has dimension $2$.

$(A-2I)^2$ is the null matrix, hence its kernel has dimension $3$. Take any vector in $\ker(A-2I)^2\smallsetminus\ker(A-2I)$, i.e. any vector of $\mathbf R^3$ which is not an eigenvector. As the eigenspace is defined by the equation $\; y=2x$, we'll take, say $$e_3=(0,1,0). $$ Note $e'_2=(A-2I)e'_3=(1,2,1),\;$ is an eigenvector by construction1,2,0. We complete this set of two vectors to a basis, by choosing another eigenvector, linearly independent from $e'_2$, say $$e'_1=(1,2,0).$$ The definition of $e'_2$ from $e'_3$ can be written as $\; Ae'_3=2e'_3+e'_2$, so the matrix of the lineap map in basis $(e'_1,e'_2,e'_3)$ is the Jordan form: $$J=\begin{bmatrix}2&0&0\\0&2&1\\0&0&2\end{bmatrix}.$$


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.