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Chapter 10: Problem 15

Let \(\left.T: V \text { be linear. Let } X=\operatorname{Ker} T^{i-2}, Y=\operatorname{Ker} T^{i-1}, Z=\operatorname{Ker} T^{i} . \text { Therefore (Problem } 10.14\right)\) \(X \subseteq Y \subseteq Z .\) Suppose \\[ \left\\{u_{1}, \ldots, u_{r}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}\right\\}, \quad\left\\{u_{1}, \ldots, u_{r}, v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{t}\right\\} \\] are bases of \(X, Y, Z,\) respectively.

Short Answer

Expert verified
In short, given bases for subspaces X, Y, and Z as \(\{u_1, \ldots, u_r\}, \{u_1, \ldots, u_r, v_1, \ldots, v_s\}, \{u_1, \ldots, u_r, v_1, \ldots, v_s, w_1, \ldots, w_t\}\), respectively, with X 鈯� Y 鈯� Z, we have: 1. Applying the linear transformation T on the basis elements of X gives elements in Ker(T^(i-1)). 2. Applying T on the additional basis elements of Y gives elements in Ker(T^i). 3. Applying T on the additional basis elements of Z gives elements in T(V). Thus, dimensions of the subspaces X, Y, and Z are r, r+s, and r+s+t, respectively.

Step by step solution

01

Apply T on the basis of X

Since all the elements of X are in Y, we will analyze what happens when we apply T to the basis elements of X in the context of Y. We have: \[T(u_1), \dots, T(u_r)\]
02

Apply T on the additional basis elements of Y

Now we will analyze the transformation on the additional basis elements of Y that are not in X. Specifically, we will analyze T applied to the vectors: \[T(v_1), \dots, T(v_s)\]
03

Apply T on the additional basis elements of Z

Finally, we will analyze the transformation on the additional basis elements of Z that are not in Y. We will apply T on the vectors: \[T(w_1), \dots, T(w_t)\]
04

Determine the subspace membership of the transformed basis elements

Let's determine the subspace memberships of the transformed basis elements of X, Y, and Z: - T(u_1), ..., T(u_r) 鈭� Ker(T^(i-1)) since applying T to the elements of Ker(T^(i-2)) will result in elements of Ker(T^(i-1)). - T(v_1), ..., T(v_s) 鈭� Ker(T^(i)) since applying T to the elements of Ker(T^(i-1)) will result in elements of Ker(T^i). - T(w_1), ..., T(w_t) 鈭� T(V) since applying T to the elements of Ker(T^i) will result in elements in the image of T.
05

Analyze the dimensions of the subspaces

Now we will analyze the dimensions of subspaces X, Y, and Z: - dim(X) = r since the basis of X has r elements. - dim(Y) = r + s since the basis of Y has r + s elements. - dim(Z) = r + s + t since the basis of Z has r + s + t elements. In conclusion, the dimension of X is r, the dimension of Y is r+s, and the dimension of Z is r+s+t. By applying the transformation T on the bases of X, Y, and Z, we obtain elements of subspaces Ker(T^(i-1)), Ker(T^i), and T(V).

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Key Concepts

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

Basis of a Vector Space
When working with vector spaces, the concept of a **basis** is fundamental. A basis of a vector space is a set of vectors that satisfy two main criteria:
  • Linearly independent: No vector in the set can be written as a combination of the others.
  • Spans the vector space: Any vector in the space can be expressed as a combination of the basis vectors.
In essence, a basis provides a way to uniquely represent every element of the vector space using a finite combination of the basis vectors.
In the context of the exercise, we deal with three bases corresponding to subspaces \(X\), \(Y\), and \(Z\).

These bases are given by:
\[ \{u_1, \ldots, u_r\} \text{ for } X, \quad \{u_1, \ldots, u_r, v_1, \ldots, v_s\} \text{ for } Y, \quad \{u_1, \ldots, u_r, v_1, \ldots, v_s, w_1, \ldots, w_t\} \text{ for } Z \]
The different elements in these sets show how each step adds more vectors to span a larger subspace.
Dimension of a Subspace
The **dimension** of a subspace is a measure of its size, defined by the number of vectors in any of its bases.
Each subspace of a vector space will have a dimension equal to the number of vectors in its basis.

For the subspaces \(X\), \(Y\), and \(Z\) in the exercise, we have the following dimension calculations based on their bases:
  • \(\text{dim}(X) = r\), as \(X\) is spanned by \(r\) vectors \(\{u_1, \ldots, u_r\}\).
  • \(\text{dim}(Y) = r + s\), since \(Y\) includes all vectors of \(X\) plus additional \(s\) vectors \(\{v_1, \ldots, v_s\}\).
  • \(\text{dim}(Z) = r + s + t\), encompassing \(Y\) and further extending it with \(t\) more vectors \(\{w_1, \ldots, w_t\}\).
These dimensions indicate how the kernel of transformations changes as the power \(i\) increases, enlarging the subspace.
Linear Transformation Properties
A **linear transformation** is a function between two vector spaces that preserves vector addition and scalar multiplication.
This means that if you have two vectors \(\mathbf{u}\) and \(\mathbf{v}\) in vector space \(V\) and a scalar \(c\), then a linear transformation \(T\) satisfies:
\[ T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \]
\[ T(c \mathbf{u}) = c T(\mathbf{u}) \]

In our problem, we specifically look at the kernel of such transformations at different stages (for example, \(\operatorname{Ker} T^{i}\)).
This affects the basis and consequently the dimension of subspaces as described above. By applying \(T\) to vectors within these kernels, we understand how they map within the transformations:\
  • Applying \(T\) to basis elements of \(X\) maps them to \(\operatorname{Ker} T^{i-1}\).
  • Applying \(T\) to basis elements in \(Y\) maps them to \(\operatorname{Ker} T^{i}\).
  • Applying \(T\) to \(W\) vectors extends mapping into the full image, \(T(V)\).
Understanding these properties helps us track how dimensions evolve and how transformations impact vector spaces.

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Most popular questions from this chapter

Suppose dim \(V=n\). Show that \(T: V \rightarrow V\) has a triangular matrix representation if and only if there exist \(T\) -invariant subspaces \(W_{1} \subset W_{2} \subset \cdots \subset W_{n}=V\) for which \(\operatorname{dim} W_{k}=k, k=1, \ldots, n\).

Prove the Primary Decomposition Theorem 10.6: Let \(T: V \rightarrow V\) be a linear operator with minimal polynomial \\[ m(t)=f_{1}(t)^{n_{1}} f_{2}(t)^{n_{2}} \ldots f_{r}(t)^{n_{r}} \\] where the \(f_{i}(t)\) are distinct monic irreducible polynomials. Then \(V\) is the direct sum of \(T\) -invariant subspaces \(W_{1}, \ldots, W_{r}\) where \(W_{i}\) is the kernel of \(f_{i}(T)^{n_{i}}\). Moreover, \(f_{i}(t)^{n_{i}}\) is the minimal polynomial of the restriction of \(T\) to \(W_{i}\).

Determine all possible Jordan canonical forms for a linear operator \(T: V \rightarrow V\) whose characteristic polynomial \(\Delta(t)=(t-2)^{3}(t-5)^{2} .\) In each case, find the minimal polynomial \(m(t)\) Because \(t-2\) has exponent 3 in \(\Delta(t), 2\) must appear three times on the diagonal. Similarly, 5 must appear twice. Thus, there are six possibilities: (a) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right],\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\), (b) \(\operatorname{diag}\left(\left[\begin{array}{lll}2 & 1 & \\ & 2 & 1 \\\ & & 2\end{array}\right],[5],[5]\right)\) (c) \(\operatorname{diag}\left(\left[\begin{array}{rr}2 & 1 \\ & 2\end{array}\right],[2],\left[\begin{array}{ll}5 & 1 \\ & 5\end{array}\right]\right)\) (d) \(\operatorname{diag}\left(\left[\begin{array}{ll}2 & 1 \\ & 2\end{array}\right],[2],[5],[5]\right)\), (e) \(\operatorname{diag}\left([2],[2],[2],\left[\begin{array}{ll}5 & 1 \\\ 5\end{array}\right]\right)\), (f) \(\operatorname{diag}([2],[2],[2],[5],[5])\) The exponent in the minimal polynomial \(m(t)\) is equal to the size of the largest block. Thus, (a) \(m(t)=(t-2)^{3}(t-5)^{2}\), (b) \(m(t)=(t-2)^{3}(t-5)\), (c) \(m(t)=(t-2)^{2}(t-5)^{2}\), (d) \(m(t)=(t-2)^{2}(t-5)\), (e) \(m(t)=(t-2)(t-5)^{2}\), (f) \(m(t)=(t-2)(t-5)\)

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 \(m_{1}(t)\) and \(m_{2}(t)\) be the minimal polynomials of \(T_{1}\) and \(T_{2}\), respectively. Note that \(g\left(T_{1}\right)=0\) and \(h\left(T_{2}\right)=0\) because \(U=\) Ker \(g(T)\) and \(W=\) Ker \(h(T)\). Thus, $$ m_{1}(t) \text { divides } g(t) \quad \text { and } \quad m_{2}(t) \text { divides } h(t) $$ By Problem 10.9, \(f(t)\) is the least common multiple of \(m_{1}(t)\) and \(m_{2}(t) .\) But \(m_{1}(t)\) and \(m_{2}(t)\) are relatively prime because \(g(t)\) and \(h(t)\) are relatively prime. Accordingly, \(f(t)=m_{1}(t) m_{2}(t)\). We also have that \(f(t)=g(t) h(t)\). These two equations together with (1) and the fact that all the polynomials are monic imply that \(g(t)=m_{1}(t)\) and \(h(t)=m_{2}(t)\), as required.

Let \(T: V \rightarrow V^{\prime \prime}\) be linear with kernel \(W\) and image \(U .\) Show that the quotient space \(V / W\) is isomorphic to \(U\) under the mapping \(\theta: V / W \rightarrow U\) defined by \(\theta(v+W)=T(v) .\) Furthermore, show that \(T=i \circ \theta \circ \eta,\) where \(\eta: V \rightarrow V / W\) is the natural mapping of \(V\) into \(V / W\) (i.e., \(\eta(v)=v+W\), and \(i: U \hookrightarrow V^{\prime}\) is the inclusion mapping (i.e., \(i(u)=u\) ). (See diagram.)
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