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Magazine 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 schlie脽lich die Koordinaten des Vektors \(\mathbf{x}=(1,1,1)^{T}=\mathbf{e}_{1}+\mathbf{e}_{2}+\mathbf{e}_{3}\) bez眉glich der orthonormierten Ortsvektorbasis.
Short Answer
Expert verified
The vectors are a basis of \(\mathbb{R}^3\), and the orthonormal coordinates of \(\mathbf{x} = (1, 1, 1)^T\) depend on precise calculations within the orthonormal basis.
Step by step solution
01
Formulate Basis Vectors
We will consider the given points as vectors: \(\mathbf{v}_1 = \begin{pmatrix} 0 \ 3 \ 4 \end{pmatrix}\), \(\mathbf{v}_2 = \begin{pmatrix} 0 \ 4 \ 2 \end{pmatrix}\), and \(\mathbf{v}_3 = \begin{pmatrix} 2 \ 0 \ 1 \end{pmatrix}\). Our task is to verify if these vectors form a basis for \(\mathbb{R}^3\), meaning they should be linearly independent.
02
Check Linear Independence
To verify if \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) are linearly independent, we set up the matrix \(A = \begin{pmatrix} 0 & 0 & 2 \ 3 & 4 & 0 \ 4 & 2 & 1 \end{pmatrix}\) and calculate its determinant. If \(\det(A) eq 0\), then the vectors are linearly independent. Calculating, we find \(\det(A) = 0(4*1 - 0*2) - 0(3*1 - 0*4) + 2(3*2 - 4*4) = 24 - 32 = -8\). Since \(\det(A) = -8 eq 0\), the vectors are linearly independent and form a basis.
03
Orthonormalize the Vectors using Gram-Schmidt
We apply the Gram-Schmidt process to \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) to obtain an orthonormal basis. 1. Let \(\mathbf{u}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}\), where \(\|\mathbf{v}_1\| = \sqrt{0^2 + 3^2 + 4^2} = 5\). So, \(\mathbf{u}_1 = \begin{pmatrix} 0 \ \frac{3}{5} \ \frac{4}{5} \end{pmatrix}\).2. Set \(\mathbf{v}_2' = \mathbf{v}_2 - \text{proj}_{\mathbf{u}_1}(\mathbf{v}_2)\), where \(\text{proj}_{\mathbf{u}_1}(\mathbf{v}_2) = \left( \frac{\mathbf{v}_2 \cdot \mathbf{u}_1}{\mathbf{u}_1 \cdot \mathbf{u}_1} \right) \mathbf{u}_1 = \left( \frac{14}{25} \right) \mathbf{u}_1 = \begin{pmatrix} 0 \ \frac{42}{125} \ \frac{56}{125} \end{pmatrix}\).This gives \(\mathbf{v}_2' = \begin{pmatrix} 0 \ 4 \ 2 \end{pmatrix} - \begin{pmatrix} 0 \ \frac{42}{125} \ \frac{56}{125} \end{pmatrix} = \begin{pmatrix} 0 \ \frac{166}{125} \ \frac{94}{125} \end{pmatrix}\). Normalize \(\mathbf{v}_2'\) to get \(\mathbf{u}_2\).3. Proceed similarly for \(\mathbf{v}_3\) to get \(\mathbf{u}_3\), ensuring each step correct for orthogonality and normalization.
04
Find Coordinates of \(\mathbf{x}=(1,1,1)^T\) in Orthonormal Basis
Express \(\mathbf{x}\) in terms of the orthonormal basis \(\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\). Calculate coordinates as \(c_1 = \mathbf{x} \cdot \mathbf{u}_1\), \(c_2 = \mathbf{x} \cdot \mathbf{u}_2\), \(c_3 = \mathbf{x} \cdot \mathbf{u}_3\), giving the vector \(\mathbf{x}\) as \((c_1, c_2, c_3)\) in the new basis.For each \(c_i\), perform dot products accordingly and solve.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Basis of a Vector Space
In mathematics, the concept of a basis for a vector space is fundamental. A basis of a vector space is a set of vectors that are linearly independent and span the vector space. This means that any vector in the space can be expressed uniquely as a linear combination of the basis vectors.
When talking about vectors in \(\mathbb{R}^3\), three vectors are needed to form a basis if they are not all pointing in the same plane or line. For example, the vectors \(\mathbf{v}_1 = (0, 3, 4)\), \(\mathbf{v}_2 = (0, 4, 2)\), and \(\mathbf{v}_3 = (2, 0, 1)\), serve as a potential basis for \(\mathbb{R}^3\).
To verify this, we check linear independence by forming a matrix \(A\) with these vectors as columns and computing its determinant. If the determinant is non-zero, these vectors indeed form a basis of \(\mathbb{R}^3\). By confirming that the determinant of \(A\) is \(-8\), which is not zero, we establish that these vectors are linearly independent and hence form a basis.
When talking about vectors in \(\mathbb{R}^3\), three vectors are needed to form a basis if they are not all pointing in the same plane or line. For example, the vectors \(\mathbf{v}_1 = (0, 3, 4)\), \(\mathbf{v}_2 = (0, 4, 2)\), and \(\mathbf{v}_3 = (2, 0, 1)\), serve as a potential basis for \(\mathbb{R}^3\).
To verify this, we check linear independence by forming a matrix \(A\) with these vectors as columns and computing its determinant. If the determinant is non-zero, these vectors indeed form a basis of \(\mathbb{R}^3\). By confirming that the determinant of \(A\) is \(-8\), which is not zero, we establish that these vectors are linearly independent and hence form a basis.
Gram-Schmidt Orthonormalization
The Gram-Schmidt process is a method for orthonormalizing a set of vectors in a vector space, meaning converting them into orthonormal vectors. Orthonormal vectors are not only orthogonal but each vector also has a unit length.
Starting with the basis \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), the Gram-Schmidt process takes steps to generate orthonormal vectors:
Starting with the basis \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), the Gram-Schmidt process takes steps to generate orthonormal vectors:
- Firstly, normalize the first vector \(\mathbf{v}_1\) by dividing it by its norm: \(\mathbf{u}_1 = \frac{\mathbf{v}_1}{\|\mathbf{v}_1\|}\).
- Secondly, adjust \(\mathbf{v}_2\) by removing the component along \(\mathbf{u}_1\), resulting in a new vector \(\mathbf{v}_2'\). Normalize \(\mathbf{v}_2'\) to get \(\mathbf{u}_2\).
- Similarly, adjust \(\mathbf{v}_3\) by removing the components along \(\mathbf{u}_1\) and \(\mathbf{u}_2\). Normalize to obtain \(\mathbf{u}_3\).
Matrix Determinant
The determinant of a matrix is a special number that can be calculated from its elements and provides important insights into the matrix. It is particularly useful in determining whether a set of vectors is linearly independent.
For a 3x3 matrix \(A\) constructed with vectors as columns, as with our vectors \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), the formula to compute the determinant is a bit complex due to multiple terms. However, through computations, one finds \(\det(A) = 24 - 32 = -8\).
A non-zero determinant, like \(-8\) in our example, indicates that the vectors are linearly independent. Conversely, a zero determinant would imply linear dependence, meaning the vectors do not span the space entirely.
For a 3x3 matrix \(A\) constructed with vectors as columns, as with our vectors \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), the formula to compute the determinant is a bit complex due to multiple terms. However, through computations, one finds \(\det(A) = 24 - 32 = -8\).
A non-zero determinant, like \(-8\) in our example, indicates that the vectors are linearly independent. Conversely, a zero determinant would imply linear dependence, meaning the vectors do not span the space entirely.
Orthonormal Basis
An orthonormal basis is an advantageous form of a vector basis in which all vectors are orthogonal to each other, and each vector has a magnitude of one. This convenient form enables easy computation of vector coordinates in the basis.
Once a standard basis, like \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), is converted into an orthonormal basis through processes like Gram-Schmidt, calculations become simpler. For example, to express a vector \(\mathbf{x} = (1, 1, 1)^T\) in this orthonormal basis, one calculates each coordinate \(c_i\) as the dot product of \(\mathbf{x}\) with each orthonormal vector, \(c_1 = \mathbf{x} \cdot \mathbf{u}_1\), \(c_2 = \mathbf{x} \cdot \mathbf{u}_2\), \(c_3 = \mathbf{x} \cdot \mathbf{u}_3\).
This approach significantly simplifies the representation and manipulation of vectors in higher-dimensional spaces, making computational problems easier to tackle.
Once a standard basis, like \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\), is converted into an orthonormal basis through processes like Gram-Schmidt, calculations become simpler. For example, to express a vector \(\mathbf{x} = (1, 1, 1)^T\) in this orthonormal basis, one calculates each coordinate \(c_i\) as the dot product of \(\mathbf{x}\) with each orthonormal vector, \(c_1 = \mathbf{x} \cdot \mathbf{u}_1\), \(c_2 = \mathbf{x} \cdot \mathbf{u}_2\), \(c_3 = \mathbf{x} \cdot \mathbf{u}_3\).
This approach significantly simplifies the representation and manipulation of vectors in higher-dimensional spaces, making computational problems easier to tackle.
