If possible, compute the Jordan normal form of

$\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & a & b \end{pmatrix}\in\mathbb{R}^{3\times 3}$ with $a,b\in\mathbb{R}$.

In the case that $a,b=0$ the matrix already has Jordan normal form. However, the case that one or both, $a,b\neq0$ seems more complicated. How do I continue?

Edit: The eigenvalues are $b/2\pm\sqrt{b^2/4+a}$. Using this in order to find the kernel of $(A-\lambda_i)^j$ using Gaussian elimination doesn't seem like the intended approach. That is what I meant.

  • $\begingroup$ Simpler than what? $\endgroup$ Jul 21 '20 at 9:15
  • $\begingroup$ Simpler way than computing manually is having it calculated by a calculator... $\endgroup$
    – VIVID
    Jul 21 '20 at 9:20
  • $\begingroup$ If you want to find Jordon canonical form, then first try to find the eigenvalue and then the eigenspace! $\endgroup$
    – A learner
    Jul 21 '20 at 9:20
  • $\begingroup$ Jordon block depends on G.M and A.M of an eigenvalue! I think without these, it is not possible to creat a block! $\endgroup$
    – A learner
    Jul 21 '20 at 9:23
  • $\begingroup$ Clearly, Jordon canonical form varries with the values of a and b, first see, if you take a=0=b, then eigenvalue 0 has only one 3×3 block here, and which represents the Jordon canonical form, but if you take a=0, b some non zero real, then b have 1×1 block and 0 have 2×2 block! $\endgroup$
    – A learner
    Jul 21 '20 at 9:34

We can avoid computing generalized eigenvectors. Separate this into $3$ cases.

Case 1: $a = b = 0$. As you said, the matrix is already in Jordan form

Case 2: $a=0$, $b \neq 0$. The matrix is upper triangular, so we quickly see that its only eigenvalues are $0$ and $b$. The block associated with $b$ has size $1$, and the because the matrix $M$ has rank $2$, the block associated with $0$ has size $2$.

Now, if $a \neq 0$, then $M$ is block upper triangular with the form $$ M = \left[\begin{array}{c|cc} 0&1&0\\ \hline 0&0&1\\0&a&b\end{array}\right] $$

Case 3: If $a = -(b/2)^2 \neq 0$, then the lower-right block has a repeated eigenvalue, and its Jordan form consists of a single block. Thus, the overall Jordan form has a $0$ followed by a size-$2$ block.

Case 4: In the remaining cases, the lower-right block has eigenvalues that are distinct and non-zero. Thus, $M$ has distinct eigenvalues, which means that its Jordan form is diagonal.

As an alternative, it would suffice to note that $M$ is the transpose of the companion matrix associated with the polynomial $p(t) = t^3 - bt^2 - at$. It follows that its Jordan form consists of a single block of maximal size for each of the roots of $p$.

  • $\begingroup$ I just noticed that the determinant of the lower right matrix only is negative if $a>0$. If $a<0$ we can still invert the lower right block but I'm not sure what that tells us about the eigenvalues. $\endgroup$
    – user731634
    Jul 21 '20 at 10:04
  • $\begingroup$ That's a good point. We know that $0$ won't be a repeated eigenvalue, but we run into trouble if $a = -(b/2)^2$. $\endgroup$ Jul 21 '20 at 10:07
  • $\begingroup$ @user See my latest edit $\endgroup$ Jul 21 '20 at 10:11
  • $\begingroup$ Thanks for the edit, the approach with the companion matrix is really good. $\endgroup$
    – user731634
    Jul 21 '20 at 10:16

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