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Chapter 2: Problem 4

Sei \(F: V \rightarrow V\) linear mit \(F^{2}=F\). Zeige, da脽 es Untervektorr盲ume \(U, W\) von \(V\) gibt mit \(V=U \oplus W\) und \(F(W)=0, F(u)=u\) f眉r alle \(u \in U\)

Short Answer

Expert verified
Yes, it can be shown by correctly interpreting the properties of the linear map. First, we defined subspaces \(U\) and \(W\) as the range and kernel of \(F\) respectively. Then we proved that each vector in \(V\) can be written as a sum of vectors in \(U\) and \(W\), and that these subspaces only intersect at the zero vector. With these subspaces, \(F(W) = 0\) holds by definition, and \(F(u) = u\) for all \(u \in U\) follows from \(F^2 = F\). Thus, the given conditions are satisfied.

Step by step solution

01

Defining the Subspaces

First, let's define the subspaces \(U\) and \(W\) of \(V\). Grounded on the properties of linear maps, we can define \(U\) as the range of \(F\) (also called the image of \(F\)), i.e., \(U = F(V)\), and \(W\) as the kernel of \(F\), i.e., \(W = ker(F)\).
02

Showing that V = U + W

We need to show that every vector \(v \in V\) can be written as a sum of a vector from \(U\) and a vector from \(W\). Since \(U = F(V)\), we have \(v = F(v) + v - F(v)\), where \(v - F(v)\) is in the kernel of \(F\) and hence, is in \(W\). Therefore, each vector in \(V\) can be written as a sum of vectors in \(U\) and \(W\).
03

Showing that U and W are disjoint

For the direct sum, \(U\) and \(W\) should also be disjoint. This means that they only meet at the zero vector, i.e., \(U \cap W = {0}\). Let \(v\) be in \(U \cap W\). Then \(v = F(u)\) for some \(u \in V\), and since \(v \in W = ker(F)\), we have that \(F(v) = 0\). But \(F(u) = v\), so applying \(F\) to both sides gives \(F^2(u) = 0\), and since \(F^2 = F\), we obtain \(F(u) = u = 0\). So, indeed, \(U \cap W = {0}\).
04

Showing F(W) = 0

The kernel of a linear map consists of all vectors that map to zero, so by definition of our subspace \(W = ker(F)\), we have that \(F(W) = 0\).
05

Showing F(U) = U

Finally, we have to show that \(F(u) = u\) for all \(u \in U = F(V)\). But this is simply the consequence of \(F^2 = F\), which says that applying \(F\) to any vector already in the image of \(F\) leaves it unchanged. Hence, \(F(u) = u\).

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Key Concepts

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

Linear Maps
A linear map, or linear transformation, is a foundational concept in linear algebra with critical importance in various mathematical and real-world applications. At its core, a linear map is a function between two vector spaces that preserves the operations of addition and scalar multiplication. This means that for a linear map \( F: V \rightarrow W \), if we take any vectors \( v_1 \) and \( v_2 \) in \( V \) and any scalar \( c \) , the following properties hold true:
  • \( F(v_1 + v_2) = F(v_1) + F(v_2) \) (additivity)
  • \( F(cv_1) = cF(v_1) \) (homogeneity).

In the given exercise, we encounter a special type of linear map where \( F^2 = F \) . This property indicates that once the map is applied, applying it again does not change the result. This attribute is characteristic of projection maps, which reduce a vector space to a subspace by 'projecting' vectors onto it. Such maps have direct implications on the structure of the vector space, leading to the concept of direct sum decomposition, which is explored in the solution step by step.
Subspaces of Vector Spaces
In the realm of linear algebra, a subspace is a subset of a vector space that is itself a vector space under the same operations of addition and scalar multiplication. Subspaces arise naturally in the study of linear maps as the range, also known as the image, and the kernel of a linear map. Specifically, the image includes all vectors that can be obtained from applying the linear map to vectors in the original vector space, indicating what the map 'covers'. Contrarily, the kernel comprises all vectors that map to the zero vector, revealing the 'blind spots' of the map.

The exercise illustrates utilizing the properties of a linear map \( F \) to define two essential subspaces:
  • \( U \) as the image \( F(V) \), where \( V \) is the original vector space,
  • and \( W \) as the kernel \( ker(F) \), embodying the idea that there is a portion of \( V \) that is essentially 'flattened out' or nullified by \( F \) .

These subspaces become instrumental in partitioning the vector space into simpler, manageable pieces.
Direct Sum Decomposition
The direct sum decomposition is an elegant concept that takes center stage in the discussion of vector spaces. It is a way to decompose a vector space into a direct sum of its subspaces in such a manner that every vector in the space can be uniquely represented as the sum of vectors from these subspaces. The decomposition \( V = U \oplus W \) presented in our exercise signifies that any vector \( v \) in \( V \) can be expressed uniquely as \( v = u + w \) where \( u \) belongs to the subspace \( U \) and \( w \) belongs to \( W \). Furthermore, for the sum to be 'direct', \( U \) and \( W \) must intersect only at the zero vector.

This decomposition is critical in understanding the structure of vector spaces, as it allows one to break down the space into simpler components that are easier to analyze. In the context of the exercise, \( U \) and \( W \) represent the 'functional' and 'non-functional' parts of the vector space under the map \( F \) respectively, illustrating the powerful interplay between linear maps and vector space structure.

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Most popular questions from this chapter

Zeige mit Hilfe der universellen Eigenschaft des Quotientenvektorraumes, \(\mathrm{da} B\) f眉r Vektorr盲ume \(V, W\) sowie einen Untervektorraum \(U \subset V\) die lineare Abbildung $$ \\{F \in \operatorname{Hom}(V, W): F \mid U=0\\} \rightarrow \operatorname{Hom}(V / U, W) \quad \text { mit } \quad F \mapsto \bar{F} $$ (vgl. Satz 2.2.7) ein Isomorphismus von Vektorr盲umen ist.

$$ \text { Sei } I \subset \mathbb{R} \text { ein Intervall und } $$ $$ d: \mathcal{D}(I ; \mathbb{R}) \rightarrow \mathcal{D}(I ; \mathbb{R}), \quad f \mapsto f^{\prime} $$ Zeige, \(\operatorname{da} 脽 d\) eine \(\mathbb{R}\)-lineare Abbildung ist, und gib eine Basis von Ker \(d\) an. Wie sieht Ker \(d\) aus im Fall, da脽 I disjunkte Vereinigung von Intervallen ist?

Beweise: a) F眉r \(A \in \mathrm{M}(n \times n ; K)\) und \(m \in \mathbb{N}\) gilt: $$ E_{n}-A^{m}=\left(E_{n}-A\right)\left(\sum_{i=0}^{m-1} A^{i}\right)=\left(\sum_{i=0}^{m-1} A^{i}\right)\left(E_{n}-A\right) $$ (Dabei sei \(A^{0}:=E_{n}\).) b) Ist \(A \in \mathrm{M}(n \times n ; K)\) eine Matrix, f眉r die ein \(m \in \mathbb{N}\) existiert mit \(A^{m}=0\), so ist \(E_{n}-A\) invertierbar. Wie sieht die inverse Matrix aus?

\(V\) sei ein \(\mathbb{R}\)-Vektorraum mit Basis \(\mathcal{A}=\left(v_{1}, \ldots, v_{4}\right), W\) sei ein \(\mathbb{R}\)-Vektorraum mit Basis \(\mathcal{B}=\left(w_{1}, \ldots, w_{5}\right) . F: V \rightarrow W\) sei die lineare Abbildung, die gegeben ist durch $$ M_{\mathcal{B}}^{\mathcal{A}}(F)=\left(\begin{array}{rrrr} 3 & 1 & -2 & 2 \\ -2 & -2 & 7 & -3 \\ 4 & 0 & 3 & 1 \\ 1 & 3 & 12 & 4 \\ 0 & 4 & -17 & 5 \end{array}\right) $$ Schlie脽lich seien \(\mathcal{A}^{\prime}=\left(v_{1}^{\prime}, \ldots, v_{4}^{\prime}\right)\) mit \(v_{1}^{\prime}=v_{1}+v_{2}, v_{2}^{\prime}=v_{2}+v_{3}, v_{3}^{\prime}=v_{3}+v_{4}\) \(v_{4}^{\prime}=v_{4}\) und \(\mathcal{B}^{\prime}=\left(w_{1}^{\prime}, \ldots, w_{5}^{\prime}\right)\) mit \(w_{1}^{\prime}=w_{1}, w_{2}^{\prime}=w_{1}+w_{2}, w_{3}^{\prime}=-w_{1}+w_{3}\) \(w_{4}^{\prime}=w_{1}+w_{4}, w_{5}^{\prime}=w_{1}+w_{5}\) a) Zeige, \(\mathrm{daB} \mathcal{A}^{\prime}\) eine Basis von \(V\) und \(\mathcal{B}^{\prime}\) eine Basis von \(W\) ist. b) Berechne \(M_{\mathcal{B}}^{\mathcal{A}^{\prime}}(F), M_{\mathcal{B}^{\prime}}^{\mathcal{A}}(F)\) und \(M_{\mathcal{B}^{\prime}}^{\mathcal{A}^{\prime}}(F)\). c) Bestimme \(F^{-1}\left(\operatorname{span}\left(w_{1}, w_{2}, w_{3}\right)\right)\).

Im \(\mathbb{R}^{3}\) seien die Basen $$ \mathcal{A}=((1,-1,2),(2,3,7),(2,3,6)) \text { und } \mathcal{B}=((1,2,2),(-1,3,3),(-2,7,6)) $$ gegeben. a) Berechne die Transformationsmatrix \(T_{\mathcal{B}}^{\mathcal{A}}\). b) Bestimme die Koordinaten des Vektors $$ v=2 \cdot(1,-1,2)+9 \cdot(2,3,7)-8 \cdot(2,3,6) $$ bez眉glich der Basis \(\mathcal{B}\).
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