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I am self-studying the book Linear Algebra from Hoffman and Kunze. The authors make the following comment in the page 196. (Second edition)

If $f=(x-c_{1})^{d_{1}}\cdots(x-c_{k})^{d_{k}}$, $c_{1},...,c_{k}$ distinct, $d_{i}\geq 1$ and $p=(x-c_{1})^{r_{1}}\cdots(x-c_{k})^{r_{k}}$, $1\leq r_{j}\leq d_{j}$. We can find an $n\times n$ matrix which has $f$ as its characteristic polynomial and $p$ as its minimal polynomial. We shall not prove this now.

How do we prove this theorem?

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Dear spohreis: I would start by observing that we can assume $k=1,c_1=0$. –  Pierre-Yves Gaillard Feb 4 '12 at 18:00
2  
Hint: Jordan's canonical form. –  Yuval Filmus Feb 4 '12 at 18:28

1 Answer 1

up vote 4 down vote accepted

The simplest way to do it is with the Jordan canonical form, as Yuval notes. The important facts:

  1. The characteristic polynomial of an upper triangular matrix is just $$\prod_{i=1}^n (t-d_{ii})$$ where $d_{ii}$ are the diagonal entries.

  2. The minimal polynomial of a block diagonal matrix is the gcd of the minimal polynomials of the blocks of the matrix.

  3. The minimal polynomial of a Jordan block associated to $\lambda$ of size $k\times k$ is $(t-\lambda)^k$.

The three facts above are fairly easy to establish. That will tell you how to construct a matrix with the desired properties.

You can also use the rational canonical form by using companion matrices instead. Then you would replace 1 above with

The characteristic polynomial of a block diagonal matrix is the product of the characteristic polynomials of the blocks.

and 3 with

The characteristic and minimal polynomials of the companion matrix of $p(x)$ are both $p(x).

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