Problem 22 - 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) Vectors are linearly independent. (b) \( \mathbf{a} = 7\mathbf{u}_1 - 12\mathbf{u}_2 + 8\mathbf{u}_3 \).

Step 1: Understand Linear Independence

Vectors are linearly independent if the equation \( c_1 \mathbf{u}_1 + c_2 \mathbf{u}_2 + c_3 \mathbf{u}_3 = \mathbf{0} \) has only the trivial solution \( c_1 = c_2 = c_3 = 0 \).

Step 2: Setup the Linear Independence Equation

Substitute the given vectors into the equation \( c_1 \langle 1, 0, 0 \rangle + c_2 \langle 1, 1, 0 \rangle + c_3 \langle 1, 1, 1 \rangle = \langle 0, 0, 0 \rangle \). This gives the system: \( c_1 + c_2 + c_3 = 0 \), \( c_2 + c_3 = 0 \), \( c_3 = 0 \).

Step 3: Solve the System for Trivial Solution

From the third equation \( c_3 = 0 \). Substitute into the second equation to get \( c_2 = 0 \). Substitute both into the first equation to get \( c_1 = 0 \). Thus, the only solution is the trivial one, showing linear independence.

Step 4: Setup the Linear Combination Equation

To express \( \mathbf{a} = \langle 3, -4, 8 \rangle \) as a linear combination, solve \( c_1 \langle 1, 0, 0 \rangle + c_2 \langle 1, 1, 0 \rangle + c_3 \langle 1, 1, 1 \rangle = \langle 3, -4, 8 \rangle \). This gives the system: \( c_1 + c_2 + c_3 = 3 \), \( c_2 + c_3 = -4 \), \( c_3 = 8 \).

Step 5: Solve for the Coefficients

From \( c_3 = 8 \), substitute into \( c_2 + 8 = -4 \) to find \( c_2 = -12 \). Substitute \( c_2 \) and \( c_3 \) into \( c_1 + c_2 + c_3 = 3 \) to find \( c_1 = 7 \).

Step 6: Write the Linear Combination

The vector \( \mathbf{a} \) can be expressed as \( 7\mathbf{u}_1 - 12\mathbf{u}_2 + 8\mathbf{u}_3 \).

Key Concepts

Linear Independence

In linear algebra, determining if vectors are linearly independent is essential to understanding the structure of a vector space. Vectors are said to be linearly independent if the only way to express the zero vector as a linear combination of these vectors is by setting all coefficients to zero. For example, consider the vectors \(\mathbf{u}_1 = \langle 1, 0, 0 \rangle\), \(\mathbf{u}_2 = \langle 1, 1, 0 \rangle\), and \(\mathbf{u}_3 = \langle 1, 1, 1 \rangle\). These vectors are linearly independent if the equation

• \(c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + c_3\mathbf{u}_3 = \mathbf{0}\)

has only the trivial solution \(c_1 = c_2 = c_3 = 0\).
This concept is fundamental because if vectors in a vector space can be rewritten in terms of each other, they don't provide new dimensions to the space. Linear independence ensures that each vector contributes a unique direction, enhancing the vector space's dimension.

Linear Combination

The concept of a linear combination involves creating a new vector by scaling and adding other vectors. If you have vectors \(\mathbf{u}_1\), \(\mathbf{u}_2\), and \(\mathbf{u}_3\), a linear combination would look like

• \(c_1\mathbf{u}_1 + c_2\mathbf{u}_2 + c_3\mathbf{u}_3\).

Here, \(c_1, c_2,\) and \(c_3\) are scalars and can be any real numbers. To express a vector \(\mathbf{a}\) as a linear combination of other vectors means finding these scalars such that the equation holds true. Let's examine the vector \(\mathbf{a} = \langle 3, -4, 8 \rangle\) and express it using the vectors given. We solve the system:

• \(c_1 + c_2 + c_3 = 3\)
• \(c_2 + c_3 = -4\)
• \(c_3 = 8\)

By solving, we find \(c_1 = 7\), \(c_2 = -12\), and \(c_3 = 8\). So, \(\mathbf{a}\) can be written as \(7\mathbf{u}_1 - 12\mathbf{u}_2 + 8\mathbf{u}_3\). This illustrates how combining vectors can reproduce another vector.

Vector Space

A vector space is a collection of vectors where vector addition and scalar multiplication are defined and follow specific rules. Key properties of vector spaces include:

• Closure under addition and scalar multiplication
• Existence of a zero vector
• Existence of additive inverses

In a vector space like \(\mathbb{R}^3\), any vector can be described using a combination of basis vectors. For instance, the vectors \(\mathbf{u}_1\), \(\mathbf{u}_2\), and \(\mathbf{u}_3\) span the entire space, providing full coverage of all points in \(\mathbb{R}^3\). This means any vector you choose within \(\mathbb{R}^3\) can be described as a linear combination of those basis vectors, which is a powerful tool for understanding and manipulating spaces when tackling complex problems.

Basis of a Vector Space

The basis of a vector space refers to a set of linearly independent vectors that span the entire space. These vectors are like building blocks for the vector space. For \(\mathbb{R}^3\), a set of three linearly independent vectors forms a basis because it covers the three-dimensional space.
Consider the basis vectors \(\mathbf{u}_1 = \langle 1, 0, 0 \rangle\), \(\mathbf{u}_2 = \langle 1, 1, 0 \rangle\), and \(\mathbf{u}_3 = \langle 1, 1, 1 \rangle\). They form a basis because:

• They are linearly independent, as shown by the trivial solution \(c_1 = c_2 = c_3 = 0\) for the zero vector.
• They can create any vector in \(\mathbb{R}^3\) through linear combinations, providing all necessary directions needed.

The basis is crucial for simplifying problem-solving in vector spaces because it provides a manageable yet comprehensive set of vectors to describe every element in the space.