Question

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

The vectors form a basis. Orthonormal basis: \( \{ \vec{u}_1, \vec{u}_2, \vec{u}_3 \} \). Coordinates of \( \mathbf{x} \): \( \mathbf{x} = \vec{u}_1 \).

Step-by-step solution

Check Linear Independence

To determine if the vectors \( \vec{P}_1 = (0, 3, 4) \), \( \vec{P}_2 = (0, 4, 2) \), and \( \vec{P}_3 = (2, 0, 1) \) form a basis, we must check if they are linearly independent. Set up the equation for linear independence: \( a\vec{P}_1 + b\vec{P}_2 + c\vec{P}_3 = \vec{0} \). Solve the system:\[\begin{align*}0a + 0b + 2c &= 0, \ 3a + 4b + 0c &= 0, \ 4a + 2b + 1c &= 0 \end{align*}\]Solving these equations shows \( a = 0, b = 0, c = 0 \), thus the vectors are linearly independent and form a basis of \( \mathbb{R}^3 \).

Apply Gram-Schmidt Process

We will now orthonormalize the basis using the Gram-Schmidt process. Start with \( \vec{w}_1 = \vec{P}_1 = (0, 3, 4) \). Normalize to get \( \vec{u}_1 = \frac{\vec{w}_1}{\|\vec{w}_1\|} = \frac{(0, 3, 4)}{5} = (0, \frac{3}{5}, \frac{4}{5}) \).

Process Second Vector

Next, compute \( \vec{w}_2 = \vec{P}_2 - \operatorname{proj}_{\vec{u}_1}(\vec{P}_2) \). The projection is \( \frac{\vec{P}_2 \cdot \vec{u}_1}{\vec{u}_1 \cdot \vec{u}_1} \vec{u}_1 \). Calculate: \( \vec{P}_2 \cdot \vec{u}_1 = 3 \), and \( \|\vec{u}_1\|^2 = 1 \). So, \( \vec{w}_2 = \vec{P}_2 - 3\vec{u}_1 = (0, 4, 2) - 3(0, \frac{3}{5}, \frac{4}{5}) = (0, \frac{7}{5}, -\frac{2}{5}) \).

Normalize Second Vector

Normalize \( \vec{w}_2 \) to get \( \vec{u}_2 = \frac{\vec{w}_2}{\|\vec{w}_2\|} = \frac{(0, \frac{7}{5}, -\frac{2}{5})}{\frac{\sqrt{53}}{5}} = (0, \frac{7}{\sqrt{53}}, -\frac{2}{\sqrt{53}}) \).

Process Third Vector

Find \( \vec{w}_3 = \vec{P}_3 - \operatorname{proj}_{\vec{u}_1}(\vec{P}_3) - \operatorname{proj}_{\vec{u}_2}(\vec{P}_3) \). Compute projections similarly: \( \vec{P}_3 \cdot \vec{u}_1 = \frac{11}{5} \) and \( \vec{P}_3 \cdot \vec{u}_2 = \frac{12}{\sqrt{53}} \). Calculate: \( \vec{w}_3 = (2, 0, 1) - \frac{11}{5}(0, \frac{3}{5}, \frac{4}{5}) - \frac{12}{\sqrt{53}}(0, \frac{7}{\sqrt{53}}, -\frac{2}{\sqrt{53}}) = (2, -\frac{51}{53}, \frac{127}{53}) \).

Normalize Third Vector

Normalize \( \vec{w}_3 \) to get \( \vec{u}_3 = \frac{\vec{w}_3}{\|\vec{w}_3\|} = \left(\frac{2}{\sqrt{(2)^2 + (-\frac{51}{53})^2 + (\frac{127}{53})^2}}, -\frac{51}{53\sqrt{...}}, \frac{127}{53\sqrt{...}}\right) \). Calculate the norm and simplify.

Express Vector in Orthonormal Basis

Now express \( \mathbf{x} = (1, 1, 1) \) in the orthonormal basis \( \{\vec{u}_1, \vec{u}_2, \vec{u}_3\} \). Calculate each coefficient \( c_i = \mathbf{x} \cdot \vec{u}_i \). Substitute and compute: \( c_1 = 1 \), \( c_2 = 0 \), and \( c_3 = 0 \), giving \( \mathbf{x} = \vec{u}_1 \).

Key concepts

Gram-Schmidt Process

The Gram-Schmidt process is a method used to convert a set of linearly independent vectors into an orthonormal basis. This is achieved by orthogonalizing and then normalizing each vector step-by-step.

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 basis in a vector space is a set of vectors that are both orthogonal to each other and each have unit length. These bases are particularly useful because they simplify many vector operations, such as projections and coordinate transformations, thanks to their straightforward properties.

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

Linear independence is a crucial concept in linear algebra. A set of vectors is said to be linearly independent if no vector in the set can be written as a combination of the others.

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.