Chapter 4: Problem 6
Weisen Sie nach, dass die Ortsvektoren der Punkte \(P_{1}=(0,3,4), P_{2}=(0,4,2)\) \(P_{3}=(2,0,1)\) eine Basis des \(\mathbb{R}^{3}\) bilden. Orthonormieren Sie diese Basis. Berechnen Sie schließlich die Koordinaten des Vektors \(\mathbf{x}=(1,1,1)^{T}=\mathbf{e}_{1}+\mathbf{e}_{2}+\mathbf{e}_{3}\) bezüglich der orthonormierten Ortsvektorbasis.
Short Answer
Step by step solution
Check Linear Independence
Apply Gram-Schmidt Process
Process Second Vector
Normalize Second Vector
Process Third Vector
Normalize Third Vector
Express Vector in Orthonormal Basis
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gram-Schmidt Process
To begin the process, we start with a set of vectors, for example, \( \vec{P}_1, \vec{P}_2, \vec{P}_3 \) in \( \mathbb{R}^3 \). The first vector in the orthonormal set, \( \vec{u}_1 \), is simply the normalized version of \( \vec{P}_1 \). This involves dividing \( \vec{P}_1 \) by its magnitude.
The second vector in the set, \( \vec{u}_2 \), is found by subtracting the projection of \( \vec{P}_2 \) onto \( \vec{u}_1 \) from \( \vec{P}_2 \), and then normalizing the result. This ensures the new vector is orthogonal to the first.
Continuing this process, the third vector \( \vec{u}_3 \) is computed by removing the components of \( \vec{P}_3 \) that lie in the directions of \( \vec{u}_1 \) and \( \vec{u}_2 \). After subtracting these projections, the remaining vector is normalized to form \( \vec{u}_3 \).
By following these steps, we construct a set of orthonormal vectors from a set of linearly independent vectors. This transformation facilitates easier calculations, like those involving projections or transformations in vector spaces.
Orthonormal Bases
An orthonormal set \( \{ \vec{u}_1, \vec{u}_2, \ldots, \vec{u}_n \} \) must satisfy two main conditions:
- Orthogonality: Each pair of vectors is orthogonal, meaning \( \vec{u}_i \cdot \vec{u}_j = 0 \) for all \( i eq j \).
- Normalization: Each vector has a unit length, meaning \( \|\vec{u}_i\| = 1 \) for all \( i \).
When a vector \( \mathbf{x} \) is expressed in terms of an orthonormal basis \( \{\vec{u}_1, \vec{u}_2, \vec{u}_3\} \), the coefficients \( c_i \) in the linear combination \( \mathbf{x} = c_1 \vec{u}_1 + c_2 \vec{u}_2 + c_3 \vec{u}_3 \) are easily computed as \( c_i = \mathbf{x} \cdot \vec{u}_i \). This is due to the normalization property which ensures each basis vector \( \vec{u}_i \) impacts one coefficient directly, simplifying projections and other vector operations.
Linear Independence
To verify the linear independence of vectors \( \vec{P}_1, \vec{P}_2, \vec{P}_3 \), we need to solve the equation: \( a\vec{P}_1 + b\vec{P}_2 + c\vec{P}_3 = \vec{0} \). For the vectors to be linearly independent, the only solution for \( a, b, \) and \( c \) must be \( a = b = c = 0 \).
In practical terms, checking linear independence often involves setting up and solving a system of linear equations or constructing a matrix and inspecting its determinant or rank. If the matrix formed by placing the vectors as its columns has a non-zero determinant, the vectors are linearly independent.
Linear independence is vital because it ensures vectors can form a basis for a vector space, allowing every vector in the space to be uniquely represented as a linear combination of the basis vectors.