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The following question was asked in my linear algebra quiz and I was unable to think which result I should use to solve it.

Question: Let A be a $6 \times 6$ matrix over $\mathbb{R}$ with chacterstic polynomial =$(x-3)^2 (x-2)^4$ and minimal polynomial = $(x-3)(x-2)^2$ .Then Jordan canonical form of A can be? I have to find all the matrix which are possible.

Algebraic multiplicity of 2 and 3 are 4,2 respectively and Geometric multiplicity of 2 and 3 are 2 and 1 respectively.

But i don't know which result should be used to move foreward. I calculated these in case they are useful.

Can you please shed some light on this?

  • So how big is the matrix? What's the sum of the sizes of the Jordan blocks for $3$? for $2$? Tell us these and we can progress .... – ancient mathematician Nov 19 '20 at 11:55
  • Think of Jordan chains. If $(A-2)^2$ kills all Jordan blocks with diagonal $2$, then how long can a Jordan chain be allowed to be? – Joppy Nov 19 '20 at 13:53

2 Answers2

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If a factor $(x-\lambda)^k$ appears in the characteristic polynomial, then $k$ is total size of all Jordan $\lambda$-blocks.

If a factor $(x-\lambda)^k$ appears in the minimal polynomial, then $k$ is the size of the largest Jordan $\lambda$-block.

In our case we get:

  • size of the largest $3$-block is $1$ and the total size is $2$ so there are two $3$-blocks of size $1$
  • size of the largest $2$-block is $2$ and the total size is $4$ we can have two $2$-blocks of size $2$ or one $2$-block of size $2$ and two $2$-blocks of size $1$

$$\begin{bmatrix} 3 & & & & & \\ & 3 & & & & \\ & & 2 & 1 & & \\ & & & 2 & & \\ & & & & 2 & 1 \\ & & & & & 2 \end{bmatrix}\quad\text{ or } \quad \begin{bmatrix} 3 & & & & & \\ & 3 & & & & \\ & & 2 & 1 & & \\ & & & 2 & & \\ & & & & 2 & \\ & & & & & 2 \end{bmatrix}$$

mechanodroid
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  • From which book have you studied these results:If a factor $(x−λ)^k$ appears in the characteristic polynomial, then k is total size of all Jordan λ-blocks.

    If a factor $(x−λ)^k$ appears in the minimal polynomial, then k is the size of the largest Jordan λ-block.

     –  Nov 23 '20 at 15:41
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    @User The result for the characteristic polynomial is evident from calculating $\det(J-xI)$ where $J$ is a matrix in Jordan form. Namely, every Jordan $\lambda$-block in $J$ of size $j$ contributes with $(x-\lambda)^j$ in the characteristic polynomial. For the second result, see https://math.stackexchange.com/q/82607/144766 – mechanodroid Nov 23 '20 at 18:51
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By assumption, $A$'s group of elementary divisors can be one of the following:

  1. $\lambda - 3, \lambda - 3, (\lambda - 2)^2, (\lambda - 2)^2$;
  2. $\lambda - 3, \lambda - 3, (\lambda - 2)^2, \lambda - 2, \lambda - 2$.

Therefore, up to row/column permutations, there are two possibilities of $A$'s Jordan form $\newcommand{\diag}{\mathrm{diag}}$ \begin{align*} & \diag(3, 3, J_{(2)}(2), J_{(2)}(2)), \\ & \diag(3, 3, J_{(2)}(2), 2, 2), \end{align*} where $J_{(m)}(\lambda)$ denotes the Jordan block of size $m$ associated with eigenvalue $2$.

Zhanxiong
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