Problem 23 - Advanced Engineering Mathematics

Question

The vectors \(\mathbf{u}_{1}=\langle 1,0,0\rangle, \mathbf{u}_{2}=\langle 1,1,0\rangle\), and \(\mathbf{u}_{3}=\langle 1,1,1\rangle\) form a basis for the vector space \(R^{3}\). (a) Show that \(\mathbf{u}_{1}, \mathbf{u}_{2}\), and \(\mathbf{u}_{3}\) are linearly independent. (b) Express the vector \(\mathbf{a}=\langle 3,-4,8\rangle\) as a linear combination of \(\mathbf{u}_{1}, \mathbf{u}_{2}\), and \(\mathbf{u}_{3}\).

Step-by-Step Solution

Answer

(a) The vectors are linearly independent. (b) \( \mathbf{a} = 7 \mathbf{u}_1 - 12 \mathbf{u}_2 + 8 \mathbf{u}_3 \).

Step 1: Define Linear Independence

Vectors are linearly independent if there is no nontrivial combination such that the sum equals the zero vector. We consider a linear combination: \[ c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3 = \langle 0,0,0 \rangle \]which translates to the system of equations:1. \( c_1 + c_2 + c_3 = 0 \)2. \( c_2 + c_3 = 0 \)3. \( c_3 = 0 \)

Step 2: Solve System of Equations for Linear Independence

From equation (3), we see that \( c_3 = 0 \). Substituting \( c_3 = 0 \) into equation (2) gives \( c_2 = 0 \). Finally, substituting \( c_2 = 0 \) and \( c_3 = 0 \) into equation (1) gives \( c_1 = 0 \). Hence, \( c_1 = c_2 = c_3 = 0 \) is the only solution, proving the vectors are linearly independent.

Step 3: Write Vector \(\mathbf{a}\) as a Linear Combination

To express \( \mathbf{a} \) as a linear combination of \( \mathbf{u}_1, \mathbf{u}_2, \mathbf{u}_3 \), we have:\[ \mathbf{a} = a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2 + a_3 \mathbf{u}_3 \]which leads to the equation:\[ \langle 3, -4, 8 \rangle = a_1 \langle 1, 0, 0 \rangle + a_2 \langle 1, 1, 0 \rangle + a_3 \langle 1, 1, 1 \rangle \]

Step 4: Set Up and Solve the Equations from Combination

From the equation \[ \langle 3, -4, 8 \rangle = \langle a_1 + a_2 + a_3, a_2 + a_3, a_3 \rangle \]we get the system:1. \( a_1 + a_2 + a_3 = 3 \)2. \( a_2 + a_3 = -4 \)3. \( a_3 = 8 \)

Step 5: Solve System to Find Coefficients

From equation (3), \( a_3 = 8 \). Substituting into equation (2) gives \( a_2 + 8 = -4 \) or \( a_2 = -12 \). Substituting \( a_2 \) and \( a_3 \) into equation (1) yields:\[ a_1 - 12 + 8 = 3 \] which simplifies to \( a_1 = 7 \). Thus, the vector \( \mathbf{a} \) can be expressed as:\[ \mathbf{a} = 7 \mathbf{u}_1 - 12 \mathbf{u}_2 + 8 \mathbf{u}_3 \]

Key Concepts

Vector Space

In linear algebra, a vector space is a fundamental concept that acts as a framework for understanding vectors and their combinations. Think of a vector space as a collection of vectors that can be added together and multiplied by scalars (numbers) to produce another vector in the same space.
For example, the vectors \[ \mathbf{u}_{1}=\langle 1,0,0 \rangle, \mathbf{u}_{2}=\langle 1,1,0 \rangle, \text{ and } \mathbf{u}_{3}=\langle 1,1,1 \rangle \]
create a vector space when any number can be used to scale these vectors and also added together, forming new vectors like \( \mathbf{v} = 2\mathbf{u}_{1} + 3\mathbf{u}_{2} - \mathbf{u}_{3} \).
Each new vector that we form still stays within the same vector space. It adheres to rules like:

• Addition: Sum of two vectors is still in the vector space.
• Scalar Multiplication: Scaling a vector by a number keeps it in the same vector space.

This system creates a structure in which we can perform complex calculations using simple linear transformations.

Linear Independence

Linear independence is a crucial concept that comes into play when checking if a list of vectors spans a vector space uniquely. Vectors are considered linearly independent when no vector in the list can be represented as a combination of the others.
In our exercise, we were checking if the vectors\[\mathbf{u}_1=\langle 1,0,0\rangle, \mathbf{u}_2=\langle 1,1,0\rangle, \text{ and } \mathbf{u}_3=\langle 1,1,1\rangle\]
are linearly independent by solving the equation:
\[ c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3 = \langle 0,0,0 \rangle \]
If the only solution is \(c_1 = c_2 = c_3 = 0\), then these vectors are independent.

• This means that no single vector can be created by scaling and adding the others.
• In such cases, they provide a unique set of directions or dimensions in the vector space.

Linear independence ensures that the vectors set forms a basis for the space without redundancy.

Linear Combination

A linear combination in linear algebra refers to an expression where vectors are scaled and added together using coefficients. This forms a new vector within the same vector space.
To express a vector \( \mathbf{a} = \langle 3, -4, 8 \rangle \) as a linear combination of our basis vectors\[\mathbf{u}_1=\langle 1,0,0\rangle, \mathbf{u}_2=\langle 1,1,0\rangle, \text{ and } \mathbf{u}_3=\langle 1,1,1\rangle\],
we need to find coefficients \(a_1, a_2, \text{ and } a_3\) such that:
\[ \mathbf{a} = a_1 \mathbf{u}_1 + a_2 \mathbf{u}_2 + a_3 \mathbf{u}_3 \]Solving this requires substituting the given vector into a set of equations:
\[\langle 3, -4, 8 \rangle = \langle a_1 + a_2 + a_3, a_2 + a_3, a_3 \rangle\]
By solving these, we determine:

• \(a_1 = 7\)
• \(a_2 = -12\)
• \(a_3 = 8\)

Therefore, the vector \( \mathbf{a} \) can be expressed uniquely using a linear combination of these basis vectors, highlighting the structure and flexibility provided by vector spaces.