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

In summary, the problem is asking to show that the degree of the minimal polynomial u_L is equal to the dimension of the vector space V, where V is spanned by the linear operator L applied to powers of x. One approach to proving this is by showing that if n is the degree of u_L, then L^n(x) can be expressed as a linear combination of x, L(x), and so on up to L^(n-1)(x).

## 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.

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

## 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.

Dick said:
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?

## 1. What is linear algebra?

Linear algebra is a branch of mathematics that focuses on the study of linear equations and their representations in vector spaces. It deals with the manipulation and analysis of vectors, matrices, and linear transformations.

## 2. What is the deg of a minimal polynomial?

The degree of a minimal polynomial is the highest power of the variable in the polynomial that is used to define a linear transformation. It is the smallest degree polynomial that can be used to express a linear transformation over a given field.

## 3. Why is it important to show that the deg of the minimal polynomial is equal to the dimension of V?

It is important to show that the deg of the minimal polynomial is equal to the dimension of V because it provides a way to uniquely determine the structure of a linear transformation. This is crucial in understanding and solving problems in various fields, such as physics, engineering, and computer science.

## 4. How do you show that the deg of the minimal polynomial is equal to the dimension of V?

To show that the deg of the minimal polynomial is equal to the dimension of V, we use the concept of characteristic polynomial. By finding the characteristic polynomial of a linear transformation, we can determine its eigenvalues and eigenvectors, which in turn can help us find the minimal polynomial and its degree.

## 5. Can the deg of the minimal polynomial be greater than the dimension of V?

No, the deg of the minimal polynomial cannot be greater than the dimension of V. This is because the minimal polynomial represents the smallest degree polynomial that can define a linear transformation, and the dimension of V represents the number of independent vectors needed to span the vector space. If the deg of the minimal polynomial is greater than the dimension of V, it would mean that there are more independent vectors needed to span the vector space than the minimal polynomial suggests, which is not possible.

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