Problem 23 - Advanced Engineering Mathematics

Question

The set of vectors \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\), where $$ \mathbf{u}_{1}=\langle 1,1,3\rangle, \mathbf{u}_{2}=\langle 1,4,1\rangle, \text { and } \mathbf{u}_{3}=\langle 1,10,-3\rangle $$ is linearly dependent in \(R^{3}\) since \(\mathbf{u}_{3}=-2 \mathbf{u}_{1}+3 \mathbf{u}_{2} .\) Discuss what you would expect when the Gram-Schmidt process in (4) is applied to these vectors. Then carry out the orthogonalization process.

Step-by-Step Solution

Answer

The Gram-Schmidt process results in one zero vector due to linear dependence.

Step 1: Understanding Linear Dependence

When a set of vectors is linearly dependent, at least one vector can be expressed as a linear combination of the others. In this case, \(\mathbf{u}_3 = -2\mathbf{u}_1 + 3\mathbf{u}_2\). This means when applying the Gram-Schmidt process, the third vector will become zero after orthogonalization because it does not add any new direction to the span of \(\mathbf{u}_1\) and \(\mathbf{u}_2\).

Step 2: Apply Gram-Schmidt Process for Orthogonalization

To start the Gram-Schmidt process, we first take \(\mathbf{v}_1 = \mathbf{u}_1 = \langle 1, 1, 3 \rangle\) as our first orthogonal vector. Next, we take \(\mathbf{v}_2 = \mathbf{u}_2 - \text{proj}_{\mathbf{v}_1}(\mathbf{u}_2)\), where the projection formula is \(\text{proj}_{\mathbf{v}_1}(\mathbf{u}_2) = \frac{\mathbf{u}_2 \cdot \mathbf{v}_1}{\mathbf{v}_1 \cdot \mathbf{v}_1} \mathbf{v}_1\). Calculate this to find \(\mathbf{v}_2\). Finally, since \(\mathbf{u}_3\) is linearly dependent on the first two vectors, \(\mathbf{v}_3\) will result in the zero vector when we calculate \(\mathbf{v}_3 = \mathbf{u}_3 - \text{proj}_{\mathbf{v}_1}(\mathbf{u}_3) - \text{proj}_{\mathbf{v}_2}(\mathbf{u}_3)\).

Step 3: Orthogonalization Calculation

First, compute the projection of \(\mathbf{u}_2\) onto \(\mathbf{v}_1\): - \(\mathbf{u}_2 \cdot \mathbf{v}_1 = 1 \cdot 1 + 4 \cdot 1 + 1 \cdot 3 = 8\) - \(\mathbf{v}_1 \cdot \mathbf{v}_1 = 1^2 + 1^2 + 3^2 = 11\) - \(\text{proj}_{\mathbf{v}_1}(\mathbf{u}_2) = \frac{8}{11} \langle 1, 1, 3 \rangle = \langle \frac{8}{11}, \frac{8}{11}, \frac{24}{11} \rangle\)Now, \(\mathbf{v}_2 = \mathbf{u}_2 - \text{proj}_{\mathbf{v}_1}(\mathbf{u}_2) = \langle 1, 4, 1 \rangle - \langle \frac{8}{11}, \frac{8}{11}, \frac{24}{11} \rangle = \langle \frac{3}{11}, \frac{36}{11}, -\frac{13}{11} \rangle\). Now calculate \(\mathbf{v}_3:\)a) \(\text{proj}_{\mathbf{v}_1}(\mathbf{u}_3):\) Use similar calculations.b) \(\text{proj}_{\mathbf{v}_2}(\mathbf{u}_3):\) Use similar projection technique.Finally, \(\mathbf{v}_3 = 0\) because of the vector dependency.

Step 4: Result for the Gram-Schmidt Process

After following the orthogonalization process, we find:- \(\mathbf{v}_1 = \langle 1, 1, 3 \rangle\)- \(\mathbf{v}_2 = \langle \frac{3}{11}, \frac{36}{11}, -\frac{13}{11} \rangle\)- \(\mathbf{v}_3 = 0\)This result indicates that only two orthogonal directions exist in the span of \(\{\mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3\}\), corresponding to \(\mathbf{v}_1\) and \(\mathbf{v}_2\). The third vector becomes zero due to linear dependence.

Key Concepts

Linear Dependence

When discussing linear dependence, it's vital to understand what makes a group of vectors dependent. If vectors are linearly dependent, it means that at least one of them can be constructed as a linear combination of others. In simple terms, they are not all pointing in new, unique directions. For the vectors provided in the problem:

• e.g.,
  \(\mathbf{u}_3 = -2\mathbf{u}_1 + 3\mathbf{u}_2\)

This equation reveals that \(\mathbf{u}_3\) does not add a new direction to the space spanned by \(\mathbf{u}_1\) and \(\mathbf{u}_2\). Because of this, \(\mathbf{u}_3\) is not bringing any additional information about the space.

• In the context of linear algebra, dependent vectors reduce the dimensionality of the vector space they are in.

Examine other problems where vectors might be dependent. Try expressing one vector as a combination of others, to check if new directions are being introduced or not.

Orthogonalization

Orthogonalization is a method used to convert a set of vectors into a set of orthogonal vectors. This is particularly important because orthogonal vectors simplify computations in vector spaces. The Gram-Schmidt process is a popular method to achieve orthogonalization. Here's how it generally works:

• Start with the first vector of the set. This remains the same as it's already orthogonal to nothing.
• For each subsequent vector, remove the components that are in the direction of the already processed orthogonal vectors by using projections.

The equations for projections might look complex:

• For vector \(\mathbf{u}_{2}\), calculate the projection on \(\mathbf{v}_{1}\).
• Subtract this projection from \(\mathbf{u}_{2}\) to obtain your second orthogonal vector \(\mathbf{v}_{2}\).

In the provided exercise, the projections ensure that \(\mathbf{u}_{3}\) becomes a zero vector after orthogonalization since it is linearly dependent on \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\). Thus, orthogonalization effectively reveals independence and dependence amongst vectors.

Vector Spaces

Vector spaces are fundamental in understanding linear transformations and dependencies. A vector space is an assembly of vectors where two operations, vector addition and scalar multiplication, are defined. These spaces extend across multiple dimensions:

• Vectors in a space can be scaled by numbers (scalars) and added together.
• Vector spaces often help solve linear equations and understand geometric concepts in higher dimensions.

In \(R^3\), vectors possess three components and can be visualized as arrows in three-dimensional space. The exercise deals with the space \( \mathbb{R}^3 \), and asks us to analyze a set of vectors within it.

• When vectors are linearly dependent, as seen with \( \mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \), the problem becomes about understanding how they relate spatially.
• Gram-Schmidt in this context is visually separating non-contributing (dependent) parts of the vector space.

Understanding vector spaces gives rise to grasping concepts like basis, dimensions, and subspaces. They serve as a canvas for mathematical modeling and solving real-world problems.