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I've tried my best with the MathJax thing but it's my first time using it!

V = $\mathbb{C}[x]_{90}$ is the $\mathbb{C}$-vector space consisting of all polynomials of degree $\leq$ 90. $D:V\to V$ is the linear map given by

$D(f) = \frac{d^3f}{dx^3}$,

find $ch_D(x)$ & $m_D(x)$ (characteristic and minimal polynomials)

B={ $\frac{x^n}{n!}$|0$\leq$ n $\leq$ 90} is a basis for V

Divide B into a series of sets $B_i$ such that $B_1 \cup ...\cup B_i$ is a basis for the generalized eigenspaces $V_i(0)$, and show that D($B_i) \subset B_{i-1}$, so that B is a Jordan basis for B. Hence find the Jordan normal form of D.

I think that $ch_D(x)$ & $m_D(x)$ both are $x^{91}$? I have no idea at all how to continue though! Any help would be appreciated.

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What are $ch_D(x)$, $m_D(x)$? Characteristic and minimal polynomial? These are not standard symbols as far as I know, so you should probably explain what you mean. –  tomasz Nov 14 '12 at 22:43
    
less than or equal to, sorry! i'll change –  user49447 Nov 14 '12 at 22:43
    
yes characteristic and minimal polynomial. I thought they were standard, sorry –  user49447 Nov 14 '12 at 22:44
    
How should I read "so that $B$ is a Jordan basis for $B$? For $D$? If so, what is the role of the $B_i$ exactly? –  Marc van Leeuwen Feb 10 at 15:18

1 Answer 1

up vote 1 down vote accepted

If you write your basis $B$ as $\{e_0,e_1,\ldots,e_{90}\}$, then $D$ is the operator $$ D(e_k)=\begin{cases}0,&\text{ if }k\leq 2\\ e_{k-3},&\text{ if }k\geq3\end{cases} $$ So, in this basis, the matrix of $D$ is an upper triangular matrix with main diagonal zero (the nonzero entries have value $1$ at $k-3,k$ for $k\geq3$). In any case, the determinant of $\lambda I-D$ is $\lambda^{91}$. So the characteristic polynomial is $p(x)=x^{91}$.

If we agree to write $e_j=0$ when $j<0$, $D$ is simply the operator $D:e_k\mapsto e_{k-3}$. It is easy to see then that $D^n:e_k\mapsto e_{k-3n}$. Then $D^{30}\ne0$, as it maps $e_{90}\mapsto e_0$. But $D^{31}=0$, as $3\times 31>91$. So the minimal polynomial is $x^{31}$.

If we reorder the basis as $$ e_0,e_1,e_2,e_3,e_6,e_9,\ldots,e_{90},e_4,e_7,\ldots,e_{88},e_5,e_8,\ldots,e_{89} $$ then $D$ is in Jordan form.

More formally, let $B_0=\text{span}\{e_0\}$, $B_1=\text{span}\{e_1\}$, $B_2=\text{span}\{e_2\}$, $B_3=\text{span}\{e_{3k}: k=1,\ldots,30\}$, $B_4=\text{span}\{e_{3k+1}: k=1,\ldots,29\}$, $B_5=\text{span}\{e_{3k+2}: k=1,\ldots,29\}$. Then, on each $B_j$ the operator $D$ acts as the backward shift and $B_0\cup\cdots B_5=B$, so $D$ is in Jordan form in this basis.

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