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Chapter 5: Problem 8

Bestimme mit dem Schmidtschen Verfahren eine Orthonormalbasis des folgenden Untervektorraums des \(\mathbb{R}^{5}\) : $$ \operatorname{span}\left(\left(\begin{array}{l} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{array}\right),\left(\begin{array}{l} 1 \\ 0 \\ 1 \\ 0 \\ 0 \end{array}\right),\left(\begin{array}{l} 1 \\ 1 \\ 1 \\ 0 \\ 2 \end{array}\right),\left(\begin{array}{l} 2 \\ 1 \\ 0 \\ 2 \\ 3 \end{array}\right)\right) $$

Short Answer

Expert verified
The orthonormal basis of the vector subspace of \(\mathbb{R}^{5}\) is given by the set of vectors \( \{ u_{1}, u_{2}, u_{3}, u_{4} \} \). If \( u_{4} \) is not defined because \( v_{4}' \) is a zero vector, then the orthonormal basis is given by the set of vectors \( \{ u_{1}, u_{2}, u_{3} \} \). Calculation will provide the exact vectors in this set.

Step by step solution

01

Initialization

Start with the first vector \( v_{1} = (1,0,0,0,0)^T \). Normalize it and obtain the first basis vector \( u_{1} = v_{1} / ||v_{1}|| \) where \( ||v_{1}|| \) represents the norm of \( v_{1} \).
02

First step of Schmidt process

Remove the component of the 2nd vector \( v_{2} = (1,0,1,0,0)^T \) that is in the direction of \( u_{1} \). The resulting vector is orthogonal to \( u_{1} \), \( v_{2}' = v_{2} - (v_{2}\cdot u_{1})u_{1} \), where \( \cdot \) indicates the dot product. Normalize \( v_{2}' \) to get the 2nd basis vector \( u_{2} = v_{2}' / ||v_{2}'|| \).
03

Second step of Schmidt process

Remove the component of the 3rd vector \( v_{3} = (1,1,1,0,2)^T \) in the directions of \( u_{1} \) and \( u_{2} \). The resulting vector is orthogonal to \( u_{1} \) and \( u_{2} \), \( v_{3}' = v_{3} - (v_{3}\cdot u_{1})u_{1} - (v_{3}\cdot u_{2})u_{2} \). Normalize \( v_{3}' \) to get the 3rd basis vector \( u_{3} = v_{3}' / ||v_{3}'|| \).
04

Third step of Schmidt process

Remove the component of the 4th vector \( v_{4} = (2,1,0,2,3)^T \) in the directions of \( u_{1} \), \( u_{2} \) and \( u_{3} \). The resulting vector is orthogonal to \( u_{1} \), \( u_{2} \) and \( u_{3} \), \( v_{4}' = v_{4} - (v_{4}\cdot u_{1})u_{1} - (v_{4}\cdot u_{2})u_{2} - (v_{4}\cdot u_{3})u_{3} \). Normalize \( v_{4}' \) to get the 4th basis vector \( u_{4} = v_{4}' / ||v_{4}'|| \). If \( v_{4}' \) is a zero vector, then \( v_{4} \) is linearly dependent on \( u_{1} \), \( u_{2} \), and \( u_{3} \), and we do not need a fourth basis vector.

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Key Concepts

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

Schmidt Process
The Schmidt process is a method used to transform a set of linearly independent vectors into an orthonormal basis. This is achieved by making vectors orthogonal to each other and then normalizing them. Normalization means adjusting the length of each vector to be one. This is essential when dealing with vector spaces to simplify calculations and applications.
Start with an initial set of vectors and initialize by taking the first vector as it is. The second step involves removing the components of subsequent vectors that are in the same direction as the previously processed vectors. By doing so, we ensure orthogonality. Once a vector is orthogonalized, it is normalized to a unit vector to retain the orthonormal property. This iterative method results in an orthonormal set that spans the same subspace as the original vectors.
Span of Vectors
The span of a set of vectors is the collection of all possible linear combinations of these vectors. In simple terms, if you have vectors in a space such as \(\mathbb{R}^5\), the span covers every vector you could create by scaling and adding those vectors together. For example, if you have vectors \(v_1, v_2, \text{and } v_3\), every vector in their span can be written in the form \(c_1 v_1 + c_2 v_2 + c_3 v_3\), where \(c_1, c_2,\text{and } c_3\) are scalars.
This concept is crucial when discussing the coverage a set of vectors provides, especially when determining if they can form a basis for a particular vector space. In practical terms, knowing the span helps us understand the reach and limitations of a vector set in representing or approximating other vectors.
Gram-Schmidt Orthogonalization
Gram-Schmidt Orthogonalization is a step-by-step process of converting a set of linearly independent vectors into an orthogonal or orthonormal set. This technique is particularly helpful in solving problems like finding orthonormal bases in vector spaces.
The process involves several steps:
  • Start with a vector and normalize it, creating the first orthonormal vector.
  • For each subsequent vector, subtract its projection onto all previous orthonormal vectors, ensuring orthogonality.
  • Normalize each new orthogonal vector to maintain the orthonormal property.
This method is performed iteratively for every vector in the set and is vital in fields such as numerical analysis, simplifying computations related to orthogonal projections and orthonormal bases.
Vector Spaces in R5
Vector spaces in \(\mathbb{R}^5\) refer to the structure where vectors have five components, each corresponding to a dimension in the real number field. Such spaces offer a context for representing and manipulating vectors mathematically.
Each vector in \(\mathbb{R}^5\) can be visualized as a point or a directed line segment in a five-dimensional space. These vector components can be manipulated through vector addition and scalar multiplication, all while adhering to the axioms that define a vector space.
Within \(\mathbb{R}^5\), a basis could be formed by any set of linearly independent vectors that span the space. The convenience of working within \(\mathbb{R}^5\) lies in its structure, providing a robust framework for computations and problem-solving in multi-dimensional settings.

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Most popular questions from this chapter

$$ \text { Eine symmetrische Matrix } A \in \mathrm{M}(n \times n ; \mathbb{R}) \text { hei脽t negativ definit, wenn } $$ $$ { }^{t} x A x<0 $$ f眉r jedes \(0 \neq x \in \mathbb{R}^{n}\). Zeige: \(A\) negativ definit \(\Leftrightarrow-A\) positiv definit.

F眉r \(x, x^{\prime}, y, y^{\prime} \in \mathbb{R}^{3}\) gilt: a) \((x \times y) \times\left(x^{\prime} \times y^{\prime}\right)=x^{\prime} \cdot \operatorname{det}\left(\begin{array}{lll}x_{1} & y_{1} & y_{1}^{\prime} \\\ x_{2} & y_{2} & y_{2}^{\prime} \\ x_{3} & y_{3} & y_{3}^{\prime}\end{array}\right)-y^{\prime} \cdot \operatorname{det}\left(\begin{array}{lll}x_{1} & y_{1} & x_{1}^{\prime} \\\ x_{2} & y_{2} & x_{2}^{\prime} \\ x_{3} & y_{3} & x_{3}^{\prime}\end{array}\right)\). b) \(\left\langle x \times y, x^{\prime} \times y^{\prime}\right\rangle=\left\langle x, x^{\prime}\right\rangle\left\langle y, y^{\prime}\right\rangle-\left\langle y, x^{\prime}\right\rangle\left\langle x, y^{\prime}\right\rangle\)

Zeige, da脽 f眉r einen \(\mathbb{R}\)-Vektorraum \(V\) der folgende Zusammenhang zwischen Normen und Skalarprodukten gilt: a) Ist \((,\), ein Skalarprodukt auf \(V\) mit zugeh枚riger Norm \(\|v\|=\sqrt{\langle v, v}\rangle\), so gilt die Parallelogramm-Gleichung $$ \|v+w\|^{2}+\|v-w\|^{2}=2\|v\|^{2}+2\|w\|^{2} $$ b) \(^{*}\) Ist umgekehrt \|\| eine Norm auf \(V\), die die Parallelogrammgleichung erf眉llt, so existiert ein Skalarprodukt \(\langle,\),\(rangle auf V\) mit \(\|v\|=\sqrt{\langle v, v\rangle}\).

Wir wollen mit Hilfe des Vektorproduktes eine Parameterdarstellung der Schnittgeraden zweier nichtparalleler Ebenen im \(\mathbb{R}^{3}\) bestimmen. Sind zwei Ebenen \(E=v+\mathbb{R} w_{1}+\mathbb{R} w_{2}, E^{\prime}=v^{\prime}+\mathbb{R} w_{1}^{\prime}+\mathbb{R} w_{2}^{\prime} \subset \mathbb{R}^{3}\) gegeben, so sei \(W=\mathbb{R} w_{1}+\mathbb{R} w_{2}\) \(W^{\prime}=\mathbb{R} w_{1}^{\prime}+\mathbb{R} w_{2}^{\prime} .\) Da die beiden Ebenen nicht parallel sind, ist \(W \neq W^{\prime}\), und dar hat \(U=W \cap W^{\prime}\) die Dimension 1 . Zeige: a) Ist \(L=E \cap E^{\prime}\) und \(u \in L\), so ist \(L=u+U\). b) Seien \(s=w_{1} \times w_{2}, s^{\prime}=w_{1}^{\prime} \times w_{2}^{\prime}\) und \(w=s \times s^{\prime}\). Dann gilt \(U=\mathbb{R} w\). Bestimme nach diesem Verfahren eine Parameterdarstellung von \(E \cap E^{\prime}\), wobei $$ \begin{aligned} &E=(0,2,3)+\mathbb{R}(3,6,5)+\mathbb{R}(1,7,-1) \\ &E^{\prime}=(-1,3,2)+\mathbb{R}(8,2,3)+\mathbb{R}(2,-1,-2) \end{aligned} $$

Zwei Vektoren \(x, y \in \mathbb{R}^{n}\) hei脽en orthogonal (in Zeichen \(x \perp y\) ), wenn \(\langle x, y\rangle=0\). Sind \(x, y \neq 0\), so gilt offenbar $$ x \perp y \Leftrightarrow \angle(x, y)=\frac{\pi}{2} $$ Ist \(L=v+\mathbb{R} w \subset \mathbb{R}^{n}\) eine Gerade, so hei脽t \(s \in \mathbb{R}^{n}\) orthogonal zu \(L\), wenn \(\langle s, x-y\rangle=0\) f眉r alle \(x, y \in L\). Zeige: a) Ist \(L=v+\mathbb{R} w \subset \mathbb{R}^{n}\) eine Gerade und \(s \in \mathbb{R}^{n}\), so gilt: \(s\) ist orthogonal zu \(L \Leftrightarrow s \perp w\). b) Ist \(L=\left\\{\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2}: a_{1} x_{1}+a_{2} x_{2}=b\right\\}\) eine Gerade im \(\mathbb{R}^{2}\), so ist \(\left(a_{1}, a_{2}\right)\) orthogonal zu \(L\). Zu einer Geraden orthogonale Vektoren kann man benutzen, um den k眉rzesten Abstand zwischen einem Punkt und einer Geraden zu bestimmen. Ist \(L=v+\mathbb{R} w \subset \mathbb{R}^{n}\) eine Gerade und \(u \in \mathbb{R}^{n}\), so ist der Abstand zwischen \(u\) und \(L\) definiert als $$ d(u, L):=\min \\{\|x-u\|: x \in L\\} $$ Zeige, \(\mathrm{da} \beta\) f眉r den Abstand zwischen \(u\) und \(L\) gilt: c) Es gibt ein eindeutig bestimmtes \(x \in L\), so da脽 \((x-u)\) orthogonal zu \(L\) ist. F眉r \(x\) gilt \(d(u, L)=\|x-u\|\) (d. h. der senkrechte Abstand ist der k眉rzeste). F眉r Geraden im \(\mathbb{R}^{2}\) kann man den Abstand von einem Punkt noch einfacher beschreiben. Es gilt: d) Ist \(L \subset \mathbb{R}^{2}\) eine Gerade, \(s \in \mathbb{R}^{2} \backslash\\{0\\}\) orthogonal zu \(L\) und \(v \in L\) beliebig, so ist $$ L=\left\\{x \in \mathbb{R}^{2}:\langle s, x-v\rangle=0\right\\} $$ Ist \(u \in \mathbb{R}^{2}\), so folgt aus c), da脽 $$ d(u, L)=\frac{|\langle s, u-v\rangle|}{\|s\|} $$ Ist speziell \(L=\left\\{\left(x_{1}, x_{2}\right) \in \mathbb{R}^{2}: a_{1} x_{1}+a_{2} x_{2}=b\right\\}\) und \(u=\left(u_{1}, u_{2}\right)\), so ergibt sich $$ d(u, L)=\frac{\left|a_{1} u_{1}+a_{2} u_{2}-b\right|}{\sqrt{a_{1}^{2}+a_{2}^{2}}} $$ Mit Hilfe von d) k枚nnen wir nun f眉r Gleichungen von Geraden im \(\mathbb{R}^{2}\) die sogenannte Hessesche Normalform herleiten: Ist \(s \in \mathbb{R}^{2} \backslash\\{0\\}\) orthogonal zur Geraden \(L \subset \mathbb{R}^{2}\), so sei \(n:=\frac{1}{\|s\|} \cdot s\). Dann ist \(\|n\|=1\). Man nennt \(n\) einen Normalenvektor zu \(L\); nach d) gilt f眉r beliebiges \(v \in L, \mathrm{da} 脽\) $$ L=\left\\{x \in \mathbb{R}^{2}:\langle n, x-v\rangle=0\right\\} $$ F眉r jedes \(u \in \mathbb{R}^{2}\) gilt dann \(d(u, L)=|\langle n, u-v\rangle|\), die Funktion \(\langle n, u-v\rangle\) mi脽t also mit Vorzeichen den Abstand von \(u\) zu \(L\).
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