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Chapter 10: Problem 16

Prove Theorem 10.10: Let \(T: V \rightarrow V\) be a nilpotent operator of index \(k .\) Then \(T\) has a unique block diagonal matrix representation consisting of Jordan nilpotent blocks \(N\). There is at least one \(N\) of order \(k,\) and all other \(N\) are of orders \(\leq k\). The total number of \(N\) of all orders is equal to the nullity of \(T\).

Short Answer

Expert verified
To prove Theorem 10.10, first, we obtain a block diagonal matrix representation of T with Jordan nilpotent blocks N by considering a sequence of invariant subspaces and utilizing properties of cyclic subspaces. Next, we prove its uniqueness by comparing Jordan canonical forms. We then show that there exists at least one Jordan nilpotent block of order k by considering a vector v with specific properties and its subspace. We also prove that all other Jordan nilpotent blocks have orders 鈮� k based on the index of T. Finally, we demonstrate that the total number of blocks of all orders is equal to the nullity of T by connecting the blocks with the eigenspace dimensions.

Step by step solution

01

Show that T has a block diagonal matrix representation with Jordan nilpotent blocks N

Let's consider a sequence of subspaces of V, \(V_{k} \supset V_{k-1} \supset \cdots \supset V_{1} \supset V_{0} = \{0\}\). Each \(V_{i}\) is invariant under T, i.e., \(T(V_{i}) \subset V_{i}\), and \(V_{i} / V_{i-1}\) is cyclic, which means it is generated by a single vector. By the properties of cyclic subspaces, a basis for \(V_{i}\) can be chosen such that the matrix representation of T restricted to \(V_{i}\) is a Jordan block of size i. Thus, putting these blocks together in a diagonal arrangement, we obtain a block diagonal matrix representation of T with Jordan nilpotent blocks N.
02

Prove that this representation is unique

If T has another block diagonal matrix representation with Jordan nilpotent blocks N', then we can compare their Jordan canonical forms. Since the Jordan canonical form is unique, the block diagonal arrangement of Jordan blocks must be the same for both representations. Therefore, the matrix representation with Jordan nilpotent blocks N is unique.
03

Show that there is at least one N of order k

Since T has index k, there exists a vector v such that \(T^k(v) = 0\) and \(T^{k-1}(v) \neq 0\). Let's consider the subspace generated by v and its images under successive powers of T. The dimension of this subspace is k since the vectors form a linearly independent set. Therefore, the Jordan block corresponding to this subspace has order k, and there is at least one Jordan nilpotent block N of order k in the matrix representation of T.
04

Prove that all other N are of orders 鈮� k

Let N' be another Jordan nilpotent block in the matrix representation of T. If N' has order > k, then there exists a vector v' such that \(T^k(v') \neq 0\). But according to the definition of the index of T, this is not possible, as T has index k. Therefore, all other Jordan nilpotent blocks N' are of orders 鈮� k.
05

Show that the total number of N of all orders is equal to the nullity of T

The nullity of T is the dimension of the kernel of T, which is equal to the sum of the dimensions of the eigenspaces of T with eigenvalue 0. Each Jordan nilpotent block N corresponds to an invariant subspace, and the number of such blocks is equal to the dimension of the corresponding eigenspace. Since T is nilpotent, 0 is its only eigenvalue. Thus, the total number of Jordan nilpotent blocks in the matrix representation of T is equal to the sum of the dimensions of the eigenspaces of T with eigenvalue 0, which is the nullity of T.

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Key Concepts

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

Nilpotent Operator
A Nilpotent Operator is a linear transformation that eventually becomes the zero map when raised to some power. More specifically, if \( T: V \rightarrow V \) is such that \( T^k = 0 \) for some positive integer \( k \), then \( T \) is said to be nilpotent with index \( k \).

This means that after applying \( T \) multiple times, at least \( k \) times, the result will always be a zero vector, regardless of which vector in \( V \) you start with. This property makes nilpotent operators special because they simplify the structure of a vector space when a basis exists such that \( T \) can be represented with a matrix that contains only zeroes above a certain diagonal. Nilpotent operators play an essential role in the decomposition of linear transformations, particularly when discussing Jordan canonical forms in linear algebra.
Block Diagonal Matrix
A Block Diagonal Matrix is a type of matrix composed of smaller square matrices, called blocks, along its diagonal, with all other elements being zero. This format allows simplification of complex operations and better understanding of matrix structure. In the context of Jordan canonical form, each block would correspond to a Jordan block that represents certain invariant subspaces.

The importance of these blocks lies in their ability to isolate parts of a transformation, making it easier to describe or solve problems related to the linear operator. By breaking down a nilpotent operator into a block diagonal form with Jordan blocks, we can identify the structure of the operator and understand its behavior more deeply. This is particularly crucial when analyzing eigenvectors and eigenvalues of matrics.
Cyclic Subspace
A Cyclic Subspace is a subspace generated by applying a linear transformation to a vector repeatedly. For an operator \( T \) and a vector \( v \), the vectors \( \{ v, T(v), T^2(v), \ldots \} \) generate a cyclic subspace. This means that any vector in this subspace can be expressed as a linear combination of the vectors in this sequence.

Cyclic subspaces are significant when constructing Jordan forms because they allow us to handle invariant subspaces effectively. Each Jordan block corresponds to a cyclic subspace in the matrix representation of a nilpotent operator. Identifying cyclic subspaces helps build the block diagonal structure, which is fundamental to simplifying the operator into a form that is easier to analyze and to perform computations on.
Nullity of an Operator
The Nullity of an Operator refers to the dimension of its kernel, which is the set of vectors that are mapped to zero by the operator. For any linear map \( T: V \rightarrow V \), the nullity is a measure of how many independent solutions there are to the equation \( T(v) = 0 \).

In the context of the exercise with nilpotent operators, the nullity is significant because it is directly related to the number of Jordan blocks in its canonical form. Specifically, the total number of Jordan blocks equals the nullity, reflecting the structure of the operator under transformation. This connection allows us to infer properties about the operator's spectrum and helps in the practical decomposition of transformations into more manageable pieces.

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Most popular questions from this chapter

Let \(V\) be a seven-dimensional vector space over \(\mathbf{R}\), and let \(T: V \rightarrow V\) be a linear operator with minimal polynomial \(m(t)=\left(t^{2}-2 t+5\right)(t-3)^{3} .\) Find all possible rational canonical forms \(M\) of \(T\) Because \(\operatorname{dim} V=7\), there are only two possible characteristic polynomials, \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)^{2}\) \((t-3)^{3}\) or \(\Delta_{1}(t)=\left(t^{2}-2 t+5\right)(t-3)^{5} .\) Moreover, the sum of the orders of the companion matrices must add up to 7. Also, one companion matrix must be \(C\left(t^{2}-2 t+5\right)\) and one must be \(C\left((t-3)^{3}\right)=\) \(C\left(t^{3}-9 t^{2}+27 t-27\right)\). Thus, \(M\) must be one of the following block diagonal matrices: (a) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right]\right)\) (b) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],\left[\begin{array}{rr}0 & -9 \\ 1 & 6\end{array}\right]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{rr}0 & -5 \\ 1 & 2\end{array}\right],\left[\begin{array}{rrr}0 & 0 & 27 \\ 1 & 0 & -27 \\ 0 & 1 & 9\end{array}\right],[3],[3]\right)\)(b) We have that $$ w_{k}=0+\cdots+0+w_{k}+0+\cdots+0 $$ is the unique sum corresponding to \(w_{k} \in W_{k}\); hence, \(E\left(w_{k}\right)=w_{k} .\) Then, for any \(v \in V\), $$ E^{2}(v)=E(E(v))=E\left(w_{k}\right)=w_{k}=E(v) $$ Thus, \(E^{2}=E\), as required.

Prove Theorem 10.8: In Theorem 10.7 (Problem 10.9 ), if \(f(t)\) is the minimal polynomial of \(T\) (and \(g(t)\) and \(h(t)\) are monic), then \(g(t)\) is the minimal polynomial of the restriction \(T_{1}\) of \(T\) to \(U\) and \(h(t)\) is the minimal polynomial of the restriction \(T_{2}\) of \(T\) to \(W\).

Let \(W\) be the solution space of the homogeneous equation \(2 x+3 y+4 z=0\). Describe the cosets of \(W\) in \(\mathbf{R}^{3}\).

Prove Theorem 10.1: Let \(T: V \rightarrow V\) be a linear operator whose characteristic polynomial factors into linear polynomials. Then \(V\) has a basis in which \(T\) is represented by a triangular matrix. The proof is by induction on the dimension of \(V\). If \(\operatorname{dim} V=1\), then every matrix representation of \(T\) is a \(1 \times 1\) matrix, which is triangular. Now suppose \(\operatorname{dim} V=n>1\) and that the theorem holds for spaces of dimension less than \(n .\) Because the characteristic polynomial of \(T\) factors into linear polynomials, \(T\) has at least one eigenvalue and so at least one nonzero eigenvector \(v\), say \(T(v)=a_{11} v .\) Let \(W\) be the one- dimensional subspace spanned by \(v\). Set \(\bar{V}=V / W .\) Then (Problem 10.26) \(\operatorname{dim} \bar{V}=\operatorname{dim} V-\operatorname{dim} W=n-1 .\) Note also that \(W\) is invariant under \(T .\) By Theorem \(10.16, T\) induces a linear operator \(\bar{T}\) on \(\bar{V}\) whose minimal polynomial divides the minimal polynomial of \(T\). Because the characteristic polynomial of \(T\) is a product of linear polynomials, so is its minimal polynomial, and hence, so are the minimal and characteristic polynomials of \(\bar{T}\). Thus, \(\bar{V}\) and \(\bar{T}\) satisfy the hypothesis of the theorem. Hence, by induction, there exists a basis \(\left\\{\bar{v}_{2}, \ldots, \bar{v}_{n}\right\\}\) of \(\bar{V}\) such that $$ \begin{aligned} &\bar{T}\left(\bar{v}_{2}\right)=a_{22} \bar{v}_{2} \\ &\bar{T}\left(\bar{v}_{3}\right)=a_{32} \bar{v}_{2}+a_{33} \bar{v}_{3} \\ &\cdots \ldots \ldots \cdots \\ &\bar{T}\left(\bar{v}_{n}\right)=a_{n 2} \bar{v}_{n}+a_{n 3} \bar{v}_{3}+\cdots+a_{n n} \bar{v}_{n} \end{aligned} $$Now let \(v_{2}, \ldots, v_{n}\) be elements of \(V\) that belong to the cosets \(v_{2}, \ldots, v_{n}\), respectively. Then \(\left\\{v, v_{2}, \ldots, v_{n}\right\\}\) is a basis of \(V\) (Problem 10.26). Because \(\bar{T}\left(v_{2}\right)=a_{22} \bar{v}_{2}\), we have $$ \bar{T}\left(\bar{v}_{2}\right)-a_{22} \bar{v}_{22}=0, \quad \text { and so } \quad T\left(v_{2}\right)-a_{22} v_{2} \in W $$ But \(W\) is spanned by \(v\); hence, \(T\left(v_{2}\right)-a_{22} v_{2}\) is a multiple of \(v\), say, $$ T\left(v_{2}\right)-a_{22} v_{2}=a_{21} v, \quad \text { and so } \quad T\left(v_{2}\right)=a_{21} v+a_{22} v_{2} $$ Similarly, for \(i=3, \ldots, n\) $$ T\left(v_{i}\right)-a_{i 2} v_{2}-a_{i 3} v_{3}-\cdots-a_{i i} v_{i} \in W, \quad \text { and so } \quad T\left(v_{i}\right)=a_{i 1} v+a_{i 2} v_{2}+\cdots+a_{i i} v_{i} $$ Thus, $$ \begin{aligned} &T(v)=a_{11} v \\ &T\left(v_{2}\right)=a_{21} v+a_{22} v_{2} \\ &\ldots \ldots \ldots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ &T\left(v_{n}\right)=a_{n 1} v+a_{n 2} v_{2}+\cdots+a_{n n} v_{n} \end{aligned} $$ and hence the matrix of \(T\) in this basis is triangular.

Suppose \(W\) is a subspace of a vector space \(V\). Show that the operations in Theorem \(10.15\) are well defined; namely, show that if \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W\), then (a) \((u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W \quad\) and (b) \(k u+W=k u^{\prime}+W \quad\) for any \(k \in K\) (a) Because \(u+W=u^{\prime}+W\) and \(v+W=v^{\prime}+W\), both \(u-u^{\prime}\) and \(v-v^{\prime}\) belong to \(W\). But then \((u+v)-\left(u^{\prime}+v^{\prime}\right)=\left(u-u^{\prime}\right)+\left(v-v^{\prime}\right) \in W .\) Hence, \((u+v)+W=\left(u^{\prime}+v^{\prime}\right)+W .\) (b) Also, because \(u-u^{\prime} \in W\) implies \(k\left(u-u^{\prime}\right) \in W\), then \(k u-k u^{\prime}=k\left(u-u^{\prime}\right) \in W\); accordingly, \(k u+W=k u^{\prime}+W\)
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