Given an $n\times n$ matrix $A$ (for $n$ small), I can find the Jordan Canonical Form (JCF) quickly by working with the characteristic polynomial to find the elementary divisors and then create the Jordan blocks which are the direct summands of the JCF for $A$. However, I also want to find a matrix $P$ so that $P^{-1}AP$ is in JCF. In general, finding the generalized eigen vectors for $A$ and using these vectors as the columns of $P$ is not to difficult, although I'm not quite sure how to do this in the case of having two eigenvectors and one generalized eigen vector.

Here's the example I'm working on: $$A= \begin{bmatrix} 0&0&-1&2\\ 2&-2&-1&2\\ 0&0&-2&0\\ 0&0&1&-2 \end{bmatrix} $$ which has characteristic polynomial $\chi(x)=x(x+2)^3$. The eigen values of $A$ are $\lambda_1=0$ and $\lambda_2=-2$. The eigen space for $\lambda_1$ is spanned by $v_1=\begin{bmatrix}1\\ 1\\ 0\\ 0 \end{bmatrix}$ and the eigen space for $\lambda_2$ is spanned by $v_2=\begin{bmatrix}-1\\ 0\\ 0\\ 1 \end{bmatrix}$ and $v_3=\begin{bmatrix}0\\ 1\\ 0\\ 0 \end{bmatrix}$. The generalized eigen space for $\lambda_2$ has dimension $3$, so there is a generalized eigen vector. The idea is to find an element $v_4$ so that $$(A+2I)v_4=w $$ with $w\in \text{Span }\{v_2,v_3\}$, but $v_4\not\in \text{Span }\{v_2,v_3\}$. I think $v_4$ should give the desired vector so that $P=\left[ v_2 \ v_3 \ v_4 \ v_1 \right]$ conjugates $A$ into its JCF. But this doesn't work. For the system $(A+2I)v_4=w$ to have solutions, we need $w=\begin{bmatrix}1\\ 1\\ 0\\ -1 \end{bmatrix}$ (or some multiple of this). When I solve for $v_4$ using this particular $w$, I obtain $v_4=\begin{bmatrix}0\\ 0\\ -1\\ 0 \end{bmatrix}$. The matrix $P$ does not conjugate $A$ into its JCF, which is $$JCF(A)=\begin{bmatrix} -2&0&0&0\\ 0&-2&1&0\\ 0&0&-2&0\\ 0&0&0&0 \end{bmatrix} $$ (or some permutation of the three Jordan blocks).

I believe I am picking the generalized eigen vector $v_4$ incorrectly. Can someone explain how to choose $v_4$ in general and/or point out what I am doing incorrectly? Thanks


1 Answer 1


Actually it's $v_2,v_3$ that need to change, not $v_4$. You want $(A+2I)v_4$ to be one of the elements of your basis for $\ker(A+2I)$. Let's change $v_3$ to $$ v_3'=(A+2I)v_4=\begin{bmatrix}1\\1\\0\\-1\end{bmatrix}. $$ Note that $\{v_2,v_3'\}$ is still a basis for $\ker(A+2I)$. With respect to the basis $\{v_1,v_2,v_3',v_4\}$, the matrix for $A$ is $$ \begin{bmatrix}0&0&0&0\\0&-2&0&0\\0&0&-2&1\\0&0&0&-2\end{bmatrix}. $$

  • $\begingroup$ Thanks, this worked. I'm guessing the basis vector that you change $v_3$ to doesn't change the solution, that is we can pick any basis element $v_3'$ of $\text{Ker}(A+2I)$ so that the system has a solution, and the matrix $P$ which we form from the $v_i$ will still conjugate $A$ into $JCF(A)$. $\endgroup$
    – MAM
    Aug 5, 2016 at 1:43
  • 1
    $\begingroup$ Yes, but note that there is a constraint between $v_3$ and $v_4$; eg if you doubled $v_3$, you would need to also double $v_4$ to still get $JCF(A)$. $\endgroup$
    – stewbasic
    Aug 5, 2016 at 1:53

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