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Chapter 5: Problem 8

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

Short Answer

Expert verified
The orthonormal basis is \( \{ \mathbf{q}_1, \mathbf{q}_2, \mathbf{q}_3, \mathbf{q}_4 \} \).

Step by step solution

01

Select the First Basis Vector

Start with the first vector \( \mathbf{v}_1 = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \).This vector will be our first orthonormal vector, so we denote it as \( \mathbf{q}_1 \). Since it is already unit length, \( \mathbf{q}_1 = \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} \).
02

Orthogonalize the Second Vector

Take the second vector \( \mathbf{v}_2 = \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} \)and subtract the projection onto \( \mathbf{q}_1 \):\[ \mathbf{u}_2 = \mathbf{v}_2 - (\mathbf{v}_2 \cdot \mathbf{q}_1) \mathbf{q}_1 = \begin{pmatrix} 1 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} - 1 \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} \].
03

Normalize the Second Vector

Normalize \(\mathbf{u}_2\) to get \(\mathbf{q}_2\):\[ \mathbf{q}_2 = \frac{\mathbf{u}_2}{||\mathbf{u}_2||} = \frac{1}{1} \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix}.\]
04

Orthogonalize the Third Vector

Take the third vector \( \mathbf{v}_3 = \begin{pmatrix} 1 \ 1 \ 1 \ 0 \ 2 \end{pmatrix} \) and subtract the projections onto \( \mathbf{q}_1 \) and \( \mathbf{q}_2 \):\[ \mathbf{u}_3 = \mathbf{v}_3 - (\mathbf{v}_3 \cdot \mathbf{q}_1) \mathbf{q}_1 - (\mathbf{v}_3 \cdot \mathbf{q}_2) \mathbf{q}_2 = \begin{pmatrix} 1 \ 1 \ 1 \ 0 \ 2 \end{pmatrix} - 1 \begin{pmatrix} 1 \ 0 \ 0 \ 0 \ 0 \end{pmatrix} - 1 \begin{pmatrix} 0 \ 0 \ 1 \ 0 \ 0 \end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \ 2 \end{pmatrix}.\]
05

Normalize the Third Vector

Normalize \(\mathbf{u}_3\) to get \(\mathbf{q}_3\):\[ ||\mathbf{u}_3|| = \sqrt{0^2 + 1^2 + 0^2 + 0^2 + 2^2} = \sqrt{5}, \quad \mathbf{q}_3 = \frac{\mathbf{u}_3}{||\mathbf{u}_3||} = \begin{pmatrix} 0 \ \frac{1}{\sqrt{5}} \ 0 \ 0 \ \frac{2}{\sqrt{5}} \end{pmatrix}. \]
06

Orthogonalize the Fourth Vector

Take the fourth vector \( \mathbf{v}_4 = \begin{pmatrix} 2 \ 1 \ 0 \ 2 \ 3 \end{pmatrix} \) and subtract the projections onto \( \mathbf{q}_1 \), \( \mathbf{q}_2 \), and \( \mathbf{q}_3 \).Calculate the projections:\[ \mathbf{v}_4 \cdot \mathbf{q}_1 = 2, \; \mathbf{v}_4 \cdot \mathbf{q}_2 = 0, \; \mathbf{v}_4 \cdot \mathbf{q}_3 = \frac{7}{\sqrt{5}}. \]Thus,\[ \mathbf{u}_4 = \mathbf{v}_4 - 2\mathbf{q}_1 - \frac{7}{\sqrt{5}}\mathbf{q}_3 = \begin{pmatrix} 0 \ -\frac{2}{\sqrt{5}} \ 0 \ 2 \ \frac{-1}{\sqrt{5}} \end{pmatrix}.\]
07

Normalize the Fourth Vector

Normalize \(\mathbf{u}_4\) to get \(\mathbf{q}_4\):\[ ||\mathbf{u}_4|| = \sqrt{\left(-\frac{2}{\sqrt{5}}\right)^2 + 2^2 + \left(\frac{-1}{\sqrt{5}}\right)^2} = \sqrt{5}, \quad \mathbf{q}_4 = \begin{pmatrix} 0 \ \frac{-2}{5} \ 0 \ \frac{2}{\sqrt{5}} \ \frac{-1}{5} \end{pmatrix}. \]

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Key Concepts

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

Schmidtsches Verfahren
The Schmidtsches Verfahren, known as the Gram-Schmidt process in English, is a method used to generate an orthonormal basis from a set of linearly independent vectors. The goal of this process is to form orthogonal vectors (perpendicular to each other) and then normalize these vectors to have a unit length.
This process is essential for simplifying problems in linear algebra, particularly in vector spaces.
Let's break down the steps:
  • **Select a Basis Vector:** Start with the first vector. Since it does not depend on any previous vectors, it remains unchanged and already forms part of the orthonormal basis.
  • **Orthogonalization:** For each subsequent vector, subtract the component that lies in the direction of all previous orthonormal vectors. This requires computing the projection of the vector onto each of these orthogonal vectors and subtracting them from the original vector.
  • **Normalization:** Once orthogonal vectors are found, you normalize them by dividing each vector by its magnitude. This ensures each vector in the set has a length or norm of 1.

Using these steps allows us to transform any set of vectors into an orthonormal set which is very useful for various applications like simplifying matrix operations and solving systems of equations.
Untervektorraum
An Untervektorraum, or subspace, is a concept in linear algebra where a subset of a vector space still satisfies the conditions required of a vector space. This means it is closed under vector addition and scalar multiplication.
In simpler terms, if you choose any vectors from this smaller space (the subspace), adding them together or multiplying them by a number will still keep you in the subspace.
Let's look at an example to make this more tangible:
  • **Closure Under Addition:** Taking two vectors from the subspace, their sum also lies within the same subspace.
  • **Closure Under Scalar Multiplication:** Multiplying any vector in the subspace by a scalar results in a new vector that still belongs to the subspace.

This property is essential when working with vector spaces as it ensures that any operation within the subspace doesn't exit its boundaries, which makes calculations and transformations cohesive and manageable.
R^5
The R^5 space is a five-dimensional real vector space, generally represented by vectors with five components. Each element in this vector space is a vector of the form (\( x_1, x_2, x_3, x_4, x_5 \)), where each \( x_i \)is a real number.
In any such space, operations like addition and scalar multiplication follow standard rules, allowing for transformations and operations essential in mathematics and physics.
Important properties of R^5 include:
  • **Vector Addition:** Adding two vectors in R^5 is straightforward, adding component-wise, ensuring the result is also in R^5.
  • **Scalar Multiplication:** Each component of a vector can be multiplied by a real number, resulting again in a vector of R^5.

Understanding higher dimensions like R^5 is crucial for complex computations in disciplines like computer graphics and systems modeling, where multi-dimensional data are often involved.

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Most popular questions from this chapter

Ein symplektischer Vektorraum ist ein \(\mathbb{R}\)-Vektorraum \(V\) mit einer schiefsymmetrischen Bilinearform \(\omega\), die nicht-entartet ist (d.h. dass aus \(\omega(v, w)=0\) f眉r alle \(w \in V\) stets \(v=0\) folgt). Eine Basis \(\left(v_{1}, \ldots, v_{n}, w_{1}, \ldots, w_{n}\right)\) von \(V\) hei脽t Darboux-Basis, wenn gilt: \(\omega\left(v_{i}, v_{j}\right)=\omega\left(w_{i}, w_{j}\right)=0\) und \(\omega\left(v_{i}, w_{j}\right)=\delta_{i j}\) f眉r alle \(i, j .\) Zeigen Sie, dass jeder endlichdimensionale symplektische Vektorraum eine Darboux-Basis besitzt (und damit insbesondere gerade Dimension hat).

Zeigen Sie f眉r \(x, y, z \in \mathbb{R}^{3}\) die Grassmann-Identit?t $$ x \times(y \times z)=\langle x, z\rangle y-\langle x, y\rangle z $$ und folgern daraus die Jacobi-Identit盲t $$ x \times(y \times z)+y \times(z \times x)+z \times(x \times y)=0 $$

Zeigen Sie, dass f眉r \(F \in O(3)\) gilt: \(F(x) \times F(y)=(\operatorname{det} F) \cdot F(x \times y)\).

Sei \(V\) ein 3-dimensionaler \(\mathbb{R}\)-Vektorraum, \(\mathcal{A}=\left(v_{1}, v_{2}, v_{3}\right)\) eine Basis von \(V\) und \(s\) eine Bilinearform auf \(V\) mit $$ M_{\mathcal{A}}(s)=\left(\begin{array}{lll} 1 & 1 & 2 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array}\right) $$ Zeigen Sie, dass \(\mathcal{B}=\left(v_{1}+v_{2}, v_{2}+v_{3}, v_{2}\right)\) eine Basis von \(V\) ist, und berechnen Sie \(M_{\mathcal{B}}(s)\)

Aufgabe 4 l盲sst sich leicht verallgemeinern, um den Abstand zwischen einem Punkt und einer Hyperebene im \(\mathbf{R}^{n}\) zu bestimmen; eine Teilmenge \(H\) des \(\mathbb{R}^{n}\) hei脽t dabei \(H y\) perebene, falls \(H\) ein affiner Unterraum der Dimension \((n-1)\) ist, d. h. es existiert ein \(v \in \mathbf{R}^{n}\) und ein Untervektorraum \(W \subset \mathbb{R}^{n}\) der Dimension \((n-1)\), so dass \(H=v+W\). Ist \(H=v+\operatorname{span}\left(w_{1}, \ldots, w_{n-1}\right) \subset \mathbb{R}^{n}\) eine Hyperebene, so hei脽t \(s \in \mathbb{R}^{n}\) orthogonal zu \(H\), wenn \(\langle s, x-y\rangle=0\) f眉r alle \(x, y \in H .\) Zeigen Sie: a) \(s\) ist orthogonal zu \(H \Leftrightarrow s \perp w_{i}\) f眉r \(i=1, \ldots, n-1\). b) Ist die Hyperebene gegeben durch \(H=\left\\{\left(x_{1}, \ldots, x_{n}\right) \in \mathbf{R}^{n}: a_{1} x_{1}+\ldots+a_{n} x_{n}=b\right\\}\), so ist \(\left(a_{1}, \ldots, a_{n}\right)\) orthogonal zu \(H\).Ist die Hyperebene \(H\) also durch eine Gleichung gegeben, so findet man leicht einen zu \(H\) orthogonalen Vektor. Was man tun kann, falls die Ebene in Parameterdarstellung gegeben ist, wird in Aufgabe \(6 \mathrm{zu} 5.2\) gezeigt. Ist \(H \subset \mathbb{R}^{n}\) eine Hyperebene und \(u \in \mathbb{R}^{n}\), so ist der Abstand zwischen \(u\) und \(H\) erkl盲rt durch $$ d(u, H):=\min \\{\|x-u\|: x \in H\\} $$ Beweisen Sie: c) Es gibt ein eindeutig bestimmtes \(x \in H\), so dass \((x-u)\) orthogonal zu \(H\) ist. Es gilt \(d(u, H)=\|x-u\|\) (d. h. der senkrechte Abstand ist der k眉rzeste). d) Ist \(H=\left\\{\left(x_{1}, \ldots, x_{n}\right) \in \mathbb{R}^{n}: a_{1} x_{1}+\ldots+a_{n} x_{n}=b\right\\}\) und \(u=\left(u_{1}, \ldots, u_{n}\right) \in \mathbb{R}^{n}\) so gilt $$ d(u, H)=\frac{\left|a_{1} u_{1}+\ldots+a_{n} u_{n}-b\right|}{\sqrt{a_{1}^{2}+\ldots+a_{n}^{2}}} $$ Ist \(N\) orthogonal zu \(H\) mit \(\|N\|=1\) und \(v \in H\) beliebig, so leitet man wie in Aufgabe 4 die Hessesche Normalform der Gleichung der Hyperebene ab: $$ H=\left\\{x \in \mathbb{R}^{n}:\langle N, x-v\rangle=0\right\\} $$
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