# Linear algebra - need to show deg of minimal poly = dimension of V

## Homework Statement

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

## The Attempt at a Solution

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.

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If you were to prove that dim(Cx) = deg(uL) then you would be done. Does this seem like an easier problem?

Dick
Homework Helper

## Homework Statement

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

## The Attempt at a Solution

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.
I'll add one more clue to kduna's suggestion. If $n=deg(U_L)$ then $L^n(x)=0$ for all x in V.

I'll add one more clue to kduna's suggestion. If $n=deg(U_L)$ then $L^n(x)=0$ for all x in V.
If $n=deg(U_L)$ and $L^n(x)=0$ for all $x \in V$, this would imply that $U_L(x) = x^n$. Which is certainly not always true.

However it is true that $L^n(x) \in \text{span}\{x, L(x), ... , L^{n-1}(x) \}$. This is a big hint. Can you see why this is true?