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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 \(W=\bigcap_{i} W_{i}\) is also \(T\) -invariant. Suppose \(v \in W ;\) then \(v \in W_{i}\) for every \(i .\) Because \(W_{i}\) is \(T\) -invariant, \(T(v) \in W_{i}\) for every \(i .\) Thus, \(T(v) \in W\) and so \(W\) is \(T\) -invariant.

Short Answer

Expert verified
In summary, given a collection of T-invariant subspaces \(\{W_i\}\) of a vector space V, the intersection W is also T-invariant. To prove this, we showed that for any vector \(v \in W\), applying the transformation T results in a vector \(T(v) \in W\), implying that W is T-invariant.

Step by step solution

01

Define the intersection W as the intersection of T-invariant subspaces

We are given a collection of T-invariant subspaces of a vector space V, denoted by \(\{W_i\}\). Their intersection W is defined as: \[ W = \bigcap_{i} W_{i} \] This means that for a vector v to be in W, it must be an element of each of the T-invariant subspaces W_i.
02

Take any vector v from the intersection W

Let's take any vector v that belongs to W: \[ v \in W \] Since W is an intersection of all W_i, it is clear that v also belongs to each W_i, that is: \[ v \in W_{i} \quad \forall i \]
03

Apply the transformation T to the vector v

We apply the linear transformation T to the vector v: \[ T(v) \] Since each subspace W_i is T-invariant and v belongs to all these subspaces, the transformed vector T(v) must also belong to each W_i: \[ T(v) \in W_{i} \quad \forall i \]
04

Show that T(v) is in the intersection W

Because T(v) belongs to each of the T-invariant subspaces W_i, it must also belong to their intersection W: \[ T(v) \in W \]
05

Conclude that W is T-invariant

We have shown that for any vector v in W, the transformed vector T(v) also belongs to W. Therefore, we can conclude that the intersection W is T-invariant.

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Key Concepts

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

Vector Space
In the realm of linear algebra, a vector space is a fundamental concept that underlies many theories and applications. It's essentially a collection of objects called vectors, which can be scaled and added together to form new vectors within the same space. A vector space must satisfy certain axioms that govern addition and scalar multiplication.

Imagine a vector space as a vast landscape where every point is a vector. You can move from one vector to another by either stretching it, flipping it, or combining it with another vector. These transformations play by specific rules鈥攍ike a game where scaling and adding vectors must always result in a vector that also belongs in the same space.

For example, in a three-dimensional vector space, imagine vectors as arrows pointing from the origin to various coordinates in space. Whether you stretch an arrow (scalar multiplication) or combine two arrows (vector addition), the resulting arrow still points to some coordinate in that space.
Linear Transformation
A linear transformation is like a machine that takes a vector from one space and turns it into another vector, possibly in a different space, while keeping the structure of scaling and addition intact. This means that if you put two vectors into this machine, their sum should equal the sum of their individual outputs. Same goes for scaling: a scaled vector should come out the same as if you scaled the output directly.

In mathematical terms, if we have a transformation T and vectors u and v, and a scalar c, we must have:
  • \( T(u + v) = T(u) + T(v) \)
  • \( T(cu) = cT(u) \)
This property is critical because it preserves the 'vector space' behavior even after transformation. It's as if you dressed the vectors in new clothes, but they still walk and talk the same way.
Subspace Intersection
The term subspace intersection might sound complex, but it's really just about finding a common area shared by multiple subspaces. In our vast landscape of vectors, imagine subspace as a flat field or a straight line running through the space. These are areas where certain vectors lie, following the same rules as the larger space.

When we look at where these flat fields overlap, we find an 'intersection'. Much like roads intersecting in a city, the intersection of subspaces is where all the shared vectors of these subspaces can be found. This spot follows the same rules as the initial vector space and each subspace.

In the given exercise, the intersection of T-invariant subspaces鈥攁reas that remain consistent under a certain linear transformation鈥攕till behaves well under the same transformation. It's like an agreement between various clubs (subspaces) that even where their members coincide (intersection), the club rules (transformations) apply.

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Most popular questions from this chapter

Let \(W\) be a subspace of a vector space \(V\). Suppose $\left\\{w_{1}, \ldots, w_{r}\right\\}\( is a basis of \)W$ and the set of \(\operatorname{cosets}\left\\{\bar{v}_{1}, \ldots, \bar{v}_{s}\right\\},\) where \(\bar{v}_{j}=v_{j}+W,\) is a basis of the quotient space. Show that the set of vectors \(B=\left\\{v_{1}, \ldots, v_{s}, w_{1}, \ldots, w_{r}\right\\}\) is a basis of \(V .\) Thus, $\operatorname{dim} V=\operatorname{dim} W+\operatorname{dim}(V / W)$.

Show that \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\) if and only if \(\operatorname{dim}\left[\operatorname{span}\left(W_{i}\right)\right]=\operatorname{dim} W_{1}+\cdots+\operatorname{dim} W_{r}\)

Let \(E_{1}, \ldots, E_{r}\) be linear operators on \(V\) such that (i) \(E_{i}^{2}=E_{i}\) (i.e., the \(E_{i}\) are projections) (ii) \(E_{i} E_{j}=0, i \neq j\) (iii) \(I=E_{1}+\cdots+E_{r}\) Prove that \(V=\operatorname{Im} E_{1} \oplus \cdots \oplus \operatorname{Im} E_{r}\)

The subspaces \(W_{1}, \ldots, W_{r}\) are said to be independent if \(w_{1}+\cdots+w_{r}=0, w_{i} \in W_{i},\) implies that each \(w_{i}=0 .\) Show that \(\operatorname{span}\left(W_{i}\right)=W_{1} \oplus \cdots \oplus W_{r}\) if and only if the \(W_{i}\) are independent. [Here \(\operatorname{span}\left(W_{i}\right)\) denotes the linear span of the \(\left.W_{i} .\right]\)

Let \(C(f(t))\) denote the companion matrix to an arbitrary polynomial \(f(t) .\) Show that \(f(t)\) is the characteristic polynomial of \(C(f(t))\).
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