Show deg of minimal poly = dimension of V

[catsarebad]

if V = C_x for some x belongs to V then show

deg(u_L) = dim(V)

here,

L: V -> V linear operator on finite dimensional vector space

C_x = span {x, L(x), L^2(x),....}

u_L = minimal polynomial

my thought:

since C_x = V, it spans V. if it spans v, then degree of minimal poly that we somehow obtain from L should be dim(V). i dunno.

i can't figure out how to connect C_x and L with minimal poly.

 

[Deveno]

If $C_x = V$ for some $x$ (such an $x$ is clearly non-zero), then consider the smallest $m$ for which:

$\{x,Lx,\dots,L^mx\}$ is linearly dependent.

By the definition of linearly dependent, this means we have $c_0,c_1\dots,c_m$ not all 0 with:

$c_0x + c_1Lx +\cdots + c_mL^mx = 0$.

By the minimality of $m$, $c_m \neq 0$.

Thus $L$ satisfies the polynomial (since $x$ is non-zero):

$p(x) = c_0 + c_1x + \cdots + c_mx^m$

Therefore (since $p(L) = 0$) we conclude that $\mu_L|p$.

But if $L^mx$ is a linear combination of $\{x,Lx,...,L^{m-1}x\}$ then any higher power of $L$ is likewise a linear combination of these (you may want to supply an inductive proof of this..up to you).

Thus $V$ = span($C_x$) = span($\{x,Lx,...,L^{m-1}x\}$).

and we have $m = $ dim$(V)$.

The above shows that deg($\mu_L$) $\leq$ dim($V$).

Now argue that $\{x,Lx,...,L^{m-1}x\}$ is linearly independent, to show that:

deg($\mu_L$) $> m - 1$.