# Jordan normal form of $\;\begin{pmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & a & b \end{pmatrix},\; a,b\in\mathbb{R}$

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.

• Simpler than what? Jul 21 '20 at 9:15
• Simpler way than computing manually is having it calculated by a calculator... Jul 21 '20 at 9:20
• If you want to find Jordon canonical form, then first try to find the eigenvalue and then the eigenspace! Jul 21 '20 at 9:20
• Jordon block depends on G.M and A.M of an eigenvalue! I think without these, it is not possible to creat a block! Jul 21 '20 at 9:23
• 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! 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 $$p$$.

• 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.
– user731634
Jul 21 '20 at 10:04
• That's a good point. We know that $0$ won't be a repeated eigenvalue, but we run into trouble if $a = -(b/2)^2$. Jul 21 '20 at 10:07
• @user See my latest edit Jul 21 '20 at 10:11
• Thanks for the edit, the approach with the companion matrix is really good.
– user731634
Jul 21 '20 at 10:16