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

  • Thread starter catsarebad
  • Start date
  • #1
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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



Homework Equations





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.
 

Answers and Replies

  • #2
52
7
If you were to prove that dim(Cx) = deg(uL) then you would be done. Does this seem like an easier problem?
 
  • #3
Dick
Science Advisor
Homework Helper
26,258
618

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



Homework Equations





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.
 
  • #4
52
7
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?
 

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