I have the matrix $A=\begin{bmatrix} -2 & -3 & 6 \\ 1 & 2 & -2\\ -1 & -1 &3 \end{bmatrix}$ and I have to find the transformation matrix and its Jordan normal form. This is what I did so far:

Char. polynomial: $p_A=(\lambda-1)^3$ so I have eigenvalue $\lambda=1$

Then I calculated the kernel: $\ker(A-1.I_3)=\ker\begin{pmatrix} -3 & -3 & 6 \\ 1 & 1 & -2 \\ -1 & -1 &2 \end{pmatrix} = \operatorname{span}\left\{\begin{pmatrix} -1\\1\\0\end{pmatrix};\begin{pmatrix} 2\\0\\1\end{pmatrix}\right\}$

Then I have to calculate a third vector $v_3$, such that: $(A-I_3)v_3=v_2$ but the system doesn't give me a solution for this vector, am I missing something?


The eigenspace corresponding to $\ker(A-I)$ is two dimensional and we have one eigenvalue therefore there must be a generalized eigenvector in the kernel of $(A-I)^2$.
As if $(A-I)^2v=0$, then $(A-I)v$ must be one of your eigenvectors (as there are only 2 corresponding to the eigenvalue and we would have $(A-I)[(A-I)v]=0$.
Thus in fact you are looking for the basis of the kernel of $(A-I)^2$. This will give you the $v_3$ you are looking for.

For a great reference on computing and understanding Jordan-Normal form at a basic level please see https://www.maths.tcd.ie/~pete/ma1212/chapter2.pdf.

  • $\begingroup$ Thanks, but actually $(A-I)^2$ gives a zero matrix, does that mean that there is no third vector? $\endgroup$ – Dada Mar 5 '19 at 9:23
  • $\begingroup$ That means you can take any vector that is linearly independent to the vectors you have already found $\endgroup$ – Matthew Mar 5 '19 at 10:45
  • $\begingroup$ For example (1,0,0) $\endgroup$ – Matthew Mar 5 '19 at 10:46

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