91Ó°ÊÓ

Prove Theorem 10.12: Let \(Z(v, T)\) be a \(T\) -cyclic subspace, \(T_{v}\) the restriction of \(T\) to \(Z(v, T),\) and \(m_{v}(t)=t^{k}+a_{k-1} t^{k-1}+\cdots+a_{0}\) the \(T\) -annihilator of \(v .\) Then (i) The set \(\left\\{v, T(v), \ldots, T^{k-1}(v)\right\\}\) is a basis of \(Z(v, T) ;\) hence, \(\operatorname{dim} Z(v, T)=k\) (ii) The minimal polynomial of \(T_{v}\) is \(m_{v}(t)\) (iii) The matrix of \(T_{v}\) in the above basis is the companion matrix \(C=C\left(m_{v}\right)\) of \(m_{v}(t)\) [which has 1 's below the diagonal, the negative of the coefficients \(a_{0}, a_{1}, \ldots, a_{k-1}\) of \(m_{v}(t)\) in the last column, and \(0 \text { 's elsewhere }]\). (i) \(\quad\) By definition of \(m_{v}(t), T^{k}(v)\) is the first vector in the sequence \(v, T(v), T^{2}(v), \ldots\) that, is a linear combination of those vectors that precede it in the sequence; hence, the set \(B=\left\\{v, T(v), \ldots, T^{k-1}(v)\right\\}\) is linearly independent. We now only have to show that \(Z(v, T)=L(B)\), the linear span of \(B\). By the above, \(T^{k}(v) \in L(B) .\) We prove by induction that \(T^{n}(v) \in L(B)\) for every \(n .\) Suppose \(n>k\) and \(T^{n-1}(v) \in L(B)-\) that \(\quad\) is, \(\quad T^{n-1}(v) \quad\) is \(\quad\) a \(\quad\) linear \(\quad\) combination \(\quad\) of \(\quad v, \ldots, T^{k-1}(v) . \quad\) Then \(T^{n}(v)=T\left(T^{n-1}(v)\right)\) is a linear combination of \(T(v), \ldots, T^{k}(v) .\) But \(T^{k}(v) \in L(B) ;\) hence, \(T^{n}(v) \in L(B)\) for every \(n .\) Consequently, \(f(T)(v) \in L(B)\) for any polynomial \(f(t) .\) Thus, \(Z(v, T)=L(B),\) and \(\operatorname{so~} B\) is a basis, as claimed. (ii) Suppose \(m(t)=t^{s}+b_{s-1} t^{s-1}+\cdots+b_{0}\) is the minimal polynomial of \(T_{v} .\) Then, because \(v \in Z(v, T)\) \\[ 0=m\left(T_{v}\right)(v)=m(T)(v)=T^{s}(v)+b_{s-1} T^{s-1}(v)+\cdots+b_{0} v \\] Thus, \(T^{s}(v)\) is a linear combination of \(v, T(v), \ldots, T^{s-1}(v),\) and therefore \(k \leq s\). However, \(m_{v}(T)=\mathbf{0}\) and so \(m_{v}\left(T_{v}\right)=\mathbf{0} .\) Then \(m(t)\) divides \(m_{v}(t),\) and so \(s \leq k .\) Accordingly, \(k=s\) and hence \(m_{v}(t)=m(t)\) (iii) \\[ \begin{array}{ll} T_{v}(v) & =\quad T(v) \\ T_{v}(T(v)) & = \\ & = \\ T_{v}\left(T^{k-2}(v)\right) & = \\ T_{v}\left(T^{k-1}(v)\right) & =T^{k}(v)=-a_{0} v-a_{1} T(v)-a_{2} T^{2}(v)-\cdots-a_{k-1} T^{k-1}(v) \end{array} \\] By definition, the matrix of \(T_{v}\) in this basis is the tranpose of the matrix of coefficients of the above system of equations; hence, it is \(C,\) as required.

Step by step solution

01

(i) Proving basis of Z(v, T)

First, we need to show the set \(B=\{v, T(v), \ldots, T^{k-1}(v)\}\) is linearly independent. By definition of \(m_v(t)\), \(T^k(v)\) is the first vector in the sequence \(v, T(v), T^2(v), \ldots\) that is a linear combination of those vectors that precede it in the sequence. Therefore, the set \(B\) must be linearly independent. Next, we need to show that \(Z(v, T) = L(B)\) (the linear span of \(B\)). We know that \(T^k(v) \in L(B)\). We can prove by induction that \(T^n(v) \in L(B)\) for every \(n\). Suppose for some \(n>k\), \(T^{n-1}(v) \in L(B)\). Then, \(T^n(v) = T(T^{n-1}(v))\) is a linear combination of \(T(v), \ldots, T^k(v)\). Since \(T^k(v) \in L(B)\), it follows that \(T^n(v) \in L(B)\) for every \(n\). Therefore, \(f(T)(v) \in L(B)\) for any polynomial \(f(t)\). Thus, \(Z(v, T) = L(B)\), and so \(B\) is a basis with dimension \(k\).

02

(ii) Proving the minimal polynomial of Tv

Let \(m(t)=t^s+b_{s-1}t^{s-1}+\cdots+b_0\) be the minimal polynomial of \(T_v\). Since \(v \in Z(v, T)\), we have \[0 = m(T_v)(v) = m(T)(v) = T^s(v) + b_{s-1} T^{s-1}(v) + \cdots + b_0 v.\] This means \(T^s(v)\) is a linear combination of \(v, T(v), \ldots, T^{s-1}(v)\), thus \(k \le s\). However, \(m_v(T)=\mathbf{0}\), so \(m_v(T_v) = \mathbf{0}\). Therefore, \(m(t)\) divides \(m_v(t)\), and so \(s \le k\). Consequently, \(k=s\), and hence \(m_v(t)=m(t)\).

03

(iii) Proving the matrix of Tv is the companion matrix of mv(t)

Let's compute the action of the restricted transformation \(T_v\) on the basis vectors: \[\begin{array}{ll} T_v(v) & = T(v) \\ T_v(T(v)) & = T^2(v) \\ & = \vdots \\ T_v(T^{k-2}(v)) & = T^{k-1}(v) \\ T_v(T^{k-1}(v)) & = T^k(v) = -a_0v - a_1T(v) - a_2T^2(v) - \cdots - a_{k-1} T^{k-1}(v) \end{array}\] The matrix of \(T_v\) in this basis is the transpose of the matrix of coefficients of the above system of equations, which is the companion matrix \(C\) of \(m_v(t)\). This shows that the matrix of \(T_v\) in the basis \(B\) is the companion matrix of the minimal polynomial \(m_v(t)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Minimal Polynomial

The concept of a minimal polynomial plays a crucial role in understanding the structure of linear operators in linear algebra. It is the monic polynomial of least degree such that when the linear operator is applied to it, it returns the zero vector. In the context of the problem, the minimal polynomial of the restriction of transformation T to the cyclic subspace generated by v, denoted by m_v(t), is essential in determining the dynamic of the linear operator within the subspace.

When we say that m_v(t) is the T-annihilator of v, we mean it is the smallest polynomial such that m_v(T)(v) = 0. Finding the minimal polynomial is instrumental in establishing several key properties of T_v, including its matrix representation which turns out to be the companion matrix of m_v(t). Notably, the minimal polynomial also gives us insights into the characteristic polynomial and the eigenspaces associated with T.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations, linear functions, and their representations through matrices and vector spaces. It is foundational for understanding systems of linear equations, which are prevalent in various fields such as engineering, physics, computer science, and more.

In terms of cyclic subspaces and minimal polynomials, linear algebra provides us with the tools to examine the behavior of linear transformations and the structure of the spaces they act upon. The notions of independence, span, and basis, which are fundamental to linear algebra, enable us to characterize the cyclic subspace Z(v, T) and to demonstrate that the set {v, T(v), ..., T^k-1(v)} is indeed a basis for that subspace.

Companion Matrix

The companion matrix of a polynomial is a special type of square matrix that encapsulates the coefficients of the polynomial. The construction of a companion matrix is standard: ones on the subdiagonal, coefficients of the polynomial (with flipped signs) in the last column, and zeros elsewhere.

For instance, the companion matrix of the minimal polynomial m_v(t) = t^k + a_k-1t^k-1 + ... + a₀ is critical in defining the matrix representation of the linear transformation T_v restricted to the cyclic subspace. This matrix becomes a powerful computational tool to understand the effect of T_v and to solve for eigenvectors and eigenvalues that may arise from applications in differential equations, control theory, and more.

Basis of Subspace

In linear algebra, a basis of a subspace is a set of vectors that are linearly independent and span the entire subspace. Every element of the subspace can be written uniquely as a linear combination of basis vectors.

In the provided exercise, the basis of the cyclic subspace Z(v, T) consists of the vectors {v, T(v), ..., T^k-1(v)}, where k is the dimension of the subspace. The importance of identifying a basis lies in its ability to simplify problems by providing a framework for which every vector in the space can be expressed and operations such as linear transformations can be efficiently performed. Recognizing and proving that a set of vectors forms a basis is fundamental in transitioning between different vector spaces and in understanding the underlying structure of a given subspace.