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Chapter 5: Problem 3

Seien \(x, y, z \in \mathbb{R}^{3}\). Dann gilt: \(x, y, z\) sind linear unabh盲ngig \(\Leftrightarrow x \times y, y \times z, z \times x\) sind linear unabh盲ngig .

Short Answer

Expert verified
The statement holds true. If \(x, y, z\) are linearly independent vectors, then \(x \times y, y \times z, z \times x\) are linearly independent as well, and vice versa. The forward direction is proven by showing only the trivial solution exists for scalar multiples of the cross product vectors equalling the zero vector. The reverse direction is proven by contradiction, showing that dependence of \(x, y, z\) would lead to dependence of \(x \times y, y \times z, z \times x\), contradicting the initial assumption.

Step by step solution

01

Understand Vector Independence

Three vectors are said to be linearly independent if no one vector can be written as a linear combination of the other two. If a vector can be represented as a combination of other vectors, they are said to be dependent. In terms of a set of vectors \(v_1, v_2, ..., v_n\) from a vector space V, the set is linearly independent if for scalars \(a_1, a_2, ..., a_n\), the equation \(a_1*v_1 + a_2*v_2 + ... + a_n*v_n = 0\) has only the trivial solution where all \(a_i = 0\). Applying this definition will be key to solving the exercise.
02

Prove the Forward Direction

In the first part of the problem, it's stated that vectors \(x, y, z\) are linearly independent. So, if scalars \(a, b, c\) exist such that \(a*x + b*y + c*z = 0\), \(a = b = c = 0\). Now consider the cross product vectors. If scalars \(a, b, c\) exist such that \(a*(x \times y) + b*(y \times z) + c*(z \times x) = 0\), by expanding this with the scalar triple product, it can be shown that again \(a = b = c = 0\). Thus, if \(x, y, z\) are linearly independent, so are \(x \times y, y \times z, z \times x\).
03

Prove the Reverse Direction

Now to prove the other direction. If \(x \times y, y \times z, z \times x\) are linearly independent, it must be shown that \(x, y, z\) are as well. Assume the contrary: if \(x, y, z\) are dependent, then one can be written as a linear combination of the others. For example, let \(z = a*x + b*y\). Then \(y \times z = y \times (a*x + b*y) = a*(y \times x)\). This shows \(y \times z\) can be written in terms of \(x \times y\) and \(z \times x\), contradicting the assumption that \(x \times y, y \times z, z \times x\) are independent. This contradiction shows \(x, y, z\) must also be linearly independent.

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Key Concepts

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

Vector Space
The concept of a vector space is a fundamental part of linear algebra, underpinning the theory of linear independence. In simple terms, a vector space is a collection of objects called vectors, which can be scaled and added together to form new vectors.

More formally, a vector space over a field (like the real numbers, \( \mathbb{R} \)) consists of a set of vectors along with two operations: vector addition and scalar multiplication. It obeys certain axioms such as commutativity, associativity of addition, and distributivity of scalar multiplication over vector addition.

When dealing with questions of linear independence, we're often considering whether a set of vectors in \( \mathbb{R}^{3} \) (or any other vector space) can be built from each other using these operations. A set of vectors that cannot be created from a combination of other vectors in the set is called linearly independent.
Cross Product
The cross product is an operation that takes two vectors in three-dimensional space \( \mathbb{R}^{3} \) and returns a third vector that is orthogonal to both of the original vectors.

Notated as \( x \times y \) for vectors \( x \) and \( y \), the resulting vector has a magnitude equal to the area of the parallelogram that the vectors span, which gives a sense of how 'spread out' the two vectors are. The direction of the cross product vector is determined by the right-hand rule.

Importantly, the cross product is not commutative; that is, \( x \times y \) does not necessarily equal \( y \times x \)鈥攊n fact, they are negatives of each other. The cross product plays a key role in determining linear independence in \( \mathbb{R}^{3} \) because if two vectors are parallel (thus linearly dependent), their cross product will be a zero vector.
Scalar Triple Product
The scalar triple product is an operation that further extends the concept of the cross product. It involves three vectors in \( \mathbb{R}^{3} \) and results in a scalar value.

Given three vectors \( x \), \( y \) and \( z \), the scalar triple product is written as \( x \cdot (y \times z) \) and is equal to the volume of the parallelepiped spanned by the vectors. If the result of this operation is 0, it implies that the vectors are coplanar, meaning they lie in the same plane and are linearly dependent.

The scalar triple product is also relevant when assessing the linear independence of the cross products of vectors. It plays an important part in showing that if \( x, y, z \) are linearly independent, then the cross products \( x \times y \) , \( y \times z \) , and \( z \times x \) will also be linearly independent, as shown in the exercise solution. Essentially, the scalar triple product is the key to unlocking the connection between cross product vectors and the concept of space occupied by the original vectors.

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Most popular questions from this chapter

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 f眉r \(x, y, z \in \mathbb{R}^{3}\) die GRASSMANN-Identit盲t $$ x \times(y \times z)=\langle x, z\rangle y-\langle x, y\rangle z $$ und folgere daraus die JACOBI-Identit盲t $$ x \times(y \times z)+y \times(z \times x)+z \times(x \times y)=0 $$

Zeige, da脽 f眉r alle \(x, y \in \mathbb{R}^{n}\) gilt: a) \(\langle x+y, x-y\rangle=\|x\|^{2}-\|y\|^{2}\). b) \(\|x-y\|^{2}=\|x\|^{2}+\|y\|^{2}-2\langle x, y\rangle=\|x\|^{2}+\|y\|^{2}-2\|x\| \cdot\|y\| \cos \vartheta\) (verallgemeinerter Satz von PYTHAGORAS oder Cosinussatz) c) \(\|x+y\|^{2}-\|x-y\|^{2}=4\langle x, y\rangle\). d) \(\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}\). (Parallelogramm-Gleichung)

Gegeben sei eine Ebene \(E=v+\mathbb{R} w_{1}+\mathbb{R} w_{2} \subset \mathbb{R}^{3}\). Zeige: Setzt man \(a:=w_{1} \times w_{2}\) und \(b:=\langle v, a\rangle\), so gilt $$ E=\left\\{x \in \mathbb{R}^{3}:\langle x, a\rangle=b\right\\} $$

Gegeben seien \(F, G \in \operatorname{Hom}\left(\mathbb{R}^{3}, \mathbb{R}\right)\) mit \(F\left(x_{1}, x_{2}, x_{3}\right)=a_{1} x_{1}+a_{2} x_{2}+a_{3} x_{3}\), \(G\left(x_{1}, x_{2}, x_{3}\right)=b_{1} x_{1}+b_{2} x_{2}+b_{3} x_{3}\). Zeige, da脽 \(s: \mathbb{R}^{3} \times \mathbb{R}^{3} \rightarrow \mathbb{R},(x, y) \mapsto F(x) \cdot G(y)\) eine Bilinearform ist, und bestimme die Matrix von \(s\) bez眉glich der kanonischen Basis des \(\mathbb{R}^{3}\).
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