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Chapter 10: Problem 2

Suppose \(\left\\{W_{i}\right\\}\) is a collection of \(T\) -invariant subspaces of a vector space \(V\). Show that the intersection \(W=\cap_{i} W_{i}\) is also \(T\) -invariant.

Short Answer

Expert verified
To show that the intersection \(W = \cap_{i} W_{i}\) of a collection of T-invariant subspaces \(\left\{W_{i}\right\}\) of vector space V is also T-invariant, we follow these steps: 1. Define a vector in W, say \(v \in W\). This means \(v \in W_i\) for all \(i\). 2. Recall that each subspace \(W_i\) is T-invariant, so if \(v \in W_i\), then \(T(v) \in W_i\). 3. Find \(T(v)\) in each subspace \(W_i\), which means \(T(v) \in W_i\) for all \(i\). 4. Since \(T(v)\) belongs to every \(W_i\), it must be in their intersection, so \(T(v) \in W\). 5. As this holds for any vector \(v \in W\), the intersection W is T-invariant.

Step by step solution

01

Define a vector in W

Let \(v \in W\). This means that the vector \(v\) belongs to every subspace \(W_i\) in the collection. In other words, \(v \in W_i\) for all \(i\).
02

Recall that each subspace \(W_i\) is T-invariant

Since \(W_i\) is a T-invariant subspace for all \(i\), it means that if \(v \in W_i\), then \(T(v) \in W_i\).
03

Find T(v) in each subspace W_i

Since \(v \in W_i\) for all \(i\), then applying the linear transformation T on vector v, we get \(T(v) \in W_i\) for all \(i\).
04

Conclude that T(v) is in W

Because \(T(v)\) belongs to every subspace \(W_i\) in the collection, it is in the intersection of all the T-invariant subspaces. Therefore, \(T(v) \in W\).
05

Prove that W is T-invariant

We showed that for any vector \(v \in W\), \(T(v) \in W\). This satisfies the definition of T-invariant subspaces. Therefore, the intersection \(W = \cap_{i}W_i\) is also a T-invariant subspace of V.

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Key Concepts

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

Linear Transformation
A linear transformation is a foundational concept in linear algebra. It essentially refers to a function between two vector spaces that preserves the operations of vector addition and scalar multiplication.
  • If we have two vector spaces, say, \( V \) and \( W \), a linear transformation \( T: V \to W \) maps vectors from \( V \) to vectors in \( W \).
  • The function is "linear" because it satisfies two important properties: (1) Additivity: \( T(u + v) = T(u) + T(v) \) for any vectors \( u, v \) in \( V \). (2) Homogeneity: \( T(c \cdot v) = c \cdot T(v) \) for any scalar \( c \) and any vector \( v \) in \( V \).
These properties ensure that the transformation does not "distort" the structure of the vector space: vectors remain vectors, and the rules for handling them stay the same.
In the context of our exercise, a linear transformation \( T \) has an additional nuance. It requires working within subspaces that are invariant under \( T \), meaning that applying \( T \) results in a vector that stays within the same subspace it started in.
Vector Space
To grasp what a vector space is, imagine a vast system of "vectors" arranged in space. A vector space is a collection of vectors that can be added together and multiplied by numbers, called scalars.
Here are the main features of a vector space:
  • Vector Addition: Adding two vectors in the space yields another vector in the same space.
  • Scalar Multiplication: Multiplying a vector by a scalar results in another vector in the space.
  • Contains a Zero Vector: There's always a vector named the "zero vector" which acts as an identity for vector addition, meaning adding it to any vector leaves the vector unchanged.
Vector spaces provide the framework needed for many areas of mathematics, starting from solving systems of linear equations to advanced fields like quantum physics.
In the exercise provided, the concept of a vector space serves as the backdrop where various subspaces (special subsets of vectors) exist and interact under the operation of a linear transformation \( T \).
Intersection of Subspaces
The intersection of subspaces conceptually resembles defining a "common area" between several subsets of a larger vector space.
  • Consider a collection of subspaces \( \{W_i\} \) within a vector space \( V \). The intersection \( W = \cap_i W_i \) comprises all vectors that exist in every subspace \( W_i \).
  • In simpler terms, if a vector \( v \) belongs to the intersection \( W \), then it means that \( v \) is part of every individual subspace \( W_i \) included in the collection.
Understanding intersections is vital for establishing limits and conditions in vector spaces.
In the given exercise, proving that the intersection of \( T \)-invariant subspaces is itself \( T \)-invariant involves confirming that the transformation \( T(v) \) of any vector \( v \) in the intersection remains within the intersection. This holds true as shown in the original solution where applying \( T \) ensures \( T(v) \) is in every subspace \( W_i \), thus in their intersection \( W \).

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Most popular questions from this chapter

Prove that two \(3 \times 3\) matrices with the same minimal and characteristic polynomials are similar.

Suppose \(E: V \rightarrow V\) is linear and \(E^{2}=E .\) Show that (a) \(E(u)=u\) for any \(u \in \operatorname{Im} E\) (i.e., the restriction of \(E\) to its image is the identity mapping); (b) \(V\) is the direct sum of the image and kernel of \(E: V=\operatorname{Im} E \oplus \operatorname{Ker} E ;\) (c) \(E\) is the projection of \(V\) into \(\operatorname{Im} E\), its image. Thus, by the preceding problem, a linear mapping \(T: V \rightarrow V\) is a projection if and only if \(T^{2}=T ;\) this characterization of a projection is frequently used as its definition. (a) If \(u \in \operatorname{Im} E\), then there exists \(v \in V\) for which \(E(v)=u\); hence, as required, $$ E(u)=E(E(v))=E^{2}(v)=E(v)=u $$ (b) Let \(v \in V\). We can write \(v\) in the form \(v=E(v)+v-E(v)\). Now \(E(v) \in \operatorname{Im} E\) and, because $$ E(v-E(v))=E(v)-E^{2}(v)=E(v)-E(v)=0 $$ \(v-E(v) \in \operatorname{Ker} E .\) Accordingly, \(V=\operatorname{Im} E+\operatorname{Ker} E\) Now suppose \(w \in \operatorname{Im} E \cap \operatorname{Ker} E .\) By \((i), E(w)=w\) because \(w \in \operatorname{Im} E .\) On the other hand, \(E(w)=0\) because \(w \in \operatorname{Ker} E\). Thus, \(w=0\), and so \(\operatorname{Im} E \cap \operatorname{Ker} E=\\{0\\} .\) These two conditions imply that \(V\) is the direct sum of the image and kernel of \(E\). (c) Let \(v \in V\) and suppose \(v=u+w\), where \(u \in \operatorname{Im} E\) and \(w \in\) Ker \(E\). Note that \(E(u)=u\) by (i), and \(E(w)=0\) because \(w \in \operatorname{Ker} E\). Hence, $$ E(v)=E(u+w)=E(u)+E(w)=u+0=u $$ That is, \(E\) is the projection of \(V\) into its image.

Find all invariant subspaces of \(A=\left[\begin{array}{ll}2 & -5 \\ 1 & -2\end{array}\right]\) viewed as an operator on \(\mathbf{R}^{2}\).

Let \(\hat{T}\) denote the restriction of an operator \(T\) to an invariant subspace \(W\). Prove (a) For any polynomial \(f(t), f(\hat{T})(w)=f(T)(w)\). (b) The minimal polynomial of \(T\) divides the minimal polynomial of \(T\).

Let \(W=Z(v, T),\) and suppose the \(T\) -annihilator of \(v\) is \(f(t)^{n},\) where \(f(t)\) is a monic irreducible polynomial of degree \(d\). Show that \(f(T)^{3}(W)\) is a cyclic subspace generated by \(f(T)^{2}(v)\) and that it has dimension \(d(n-s)\) if \(n>s\) and dimension 0 if \(n \leq s\).
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