Problem 33 - Differential Equations and Linear Algebra

Question

Let \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}\right\\}\) be linearly independent vectors in an inner product space \(V,\) and suppose that \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal. Define the vector \(\mathbf{u}_{3}\) in \(V\) by $$ \mathbf{u}_{3}=\mathbf{v}+\lambda \mathbf{u}_{1}+\mu \mathbf{u}_{2} $$ where \(\lambda, \mu\) are scalars. Derive the values of \(\lambda\) and \(\mu\) such that \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) is an orthogonal basis for the subspace of \(V\) spanned by \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}\right\\}\)

Step-by-Step Solution

Answer

The values for \(\lambda\) and \(\mu\) that make \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) an orthogonal basis for the subspace of \(V\) spanned by \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}\right\\}\) are: \[\lambda = -\frac{\langle \mathbf{u}_{1},\mathbf{v} \rangle}{\langle \mathbf{u}_{1},\mathbf{u}_{1} \rangle}\] and \[\mu = -\frac{\langle \mathbf{u}_{2},\mathbf{v} \rangle}{\langle \mathbf{u}_{2},\mathbf{u}_{2} \rangle}\]

Step 1: Write down the orthogonality conditions

Since we want the new basis \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) to be orthogonal, the orthogonality conditions are: \[\langle \mathbf{u}_{1},\mathbf{u}_{3} \rangle = 0\] \[\langle \mathbf{u}_{2},\mathbf{u}_{3} \rangle = 0\]

Step 2: Substitute the given expression for \(\mathbf{u}_{3}\) in the orthogonality conditions

Substituting the definition of \(\mathbf{u}_{3}=\mathbf{v}+\lambda \mathbf{u}_{1}+\mu \mathbf{u}_{2}\) into the orthogonality conditions, we get the following equations: \[\langle \mathbf{u}_{1},\mathbf{v}+\lambda \mathbf{u}_{1}+\mu \mathbf{u}_{2} \rangle = 0\] \[\langle \mathbf{u}_{2},\mathbf{v}+\lambda \mathbf{u}_{1}+\mu \mathbf{u}_{2} \rangle = 0\]

Step 3: Simplify the orthogonality equations

Using the bilinearity of the inner product, we can rewrite the orthogonality equations as follows: \[\langle \mathbf{u}_{1},\mathbf{v} \rangle + \lambda \langle \mathbf{u}_{1},\mathbf{u}_{1} \rangle + \mu \langle \mathbf{u}_{1},\mathbf{u}_{2} \rangle = 0\] \[\langle \mathbf{u}_{2},\mathbf{v} \rangle + \lambda \langle \mathbf{u}_{2},\mathbf{u}_{1} \rangle + \mu \langle \mathbf{u}_{2},\mathbf{u}_{2} \rangle = 0\] Since \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal, we know that: \[\langle \mathbf{u}_{1},\mathbf{u}_{2} \rangle = \langle \mathbf{u}_{2},\mathbf{u}_{1} \rangle = 0\] Therefore, our orthogonality equations become: \[\langle \mathbf{u}_{1},\mathbf{v} \rangle + \lambda \langle \mathbf{u}_{1},\mathbf{u}_{1} \rangle = 0\] \[\langle \mathbf{u}_{2},\mathbf{v} \rangle + \mu \langle \mathbf{u}_{2},\mathbf{u}_{2} \rangle = 0\]

Step 4: Solve for \(\lambda\) and \(\mu\)

Isolate \(\lambda\) and \(\mu\) in the equations: \[\lambda = -\frac{\langle \mathbf{u}_{1},\mathbf{v} \rangle}{\langle \mathbf{u}_{1},\mathbf{u}_{1} \rangle}\] \[\mu = -\frac{\langle \mathbf{u}_{2},\mathbf{v} \rangle}{\langle \mathbf{u}_{2},\mathbf{u}_{2} \rangle}\] So the values for \(\lambda\) and \(\mu\) that make \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) an orthogonal basis for the subspace of \(V\) spanned by \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}\right\\}\) are: \[\lambda = -\frac{\langle \mathbf{u}_{1},\mathbf{v} \rangle}{\langle \mathbf{u}_{1},\mathbf{u}_{1} \rangle}\] and \[\mu = -\frac{\langle \mathbf{u}_{2},\mathbf{v} \rangle}{\langle \mathbf{u}_{2},\mathbf{u}_{2} \rangle}\]

Key Concepts

Inner Product Space

An inner product space is a type of vector space that comes equipped with an additional structure, allowing us to define angles and lengths. This structure is called the 'inner product'. The inner product is a function that combines two vectors to produce a scalar. It is typically denoted by angle brackets, such as \( \langle \mathbf{u}, \mathbf{v} \rangle \).
In an inner product space, the inner product must satisfy certain properties:

• Linearity: It is linear in both arguments. For example, \( \langle a\mathbf{u} + b\mathbf{v}, \mathbf{w} \rangle = a \langle \mathbf{u}, \mathbf{w} \rangle + b \langle \mathbf{v}, \mathbf{w} \rangle \).
• Symmetry: The product of \( \mathbf{u} \) and \( \mathbf{v} \) is the same as \( \mathbf{v} \) and \( \mathbf{u} \); \( \langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v}, \mathbf{u} \rangle \).
• Positive-Definiteness: The inner product of a vector with itself is always non-negative and zero only for the zero vector; \( \langle \mathbf{u}, \mathbf{u} \rangle \geq 0 \), and \( \langle \mathbf{u}, \mathbf{u} \rangle = 0 \) implies \( \mathbf{u} = \mathbf{0} \).

These properties enable calculations and definitions, such as orthogonality and norm, which are foundational in geometry and algebra. Understanding these is crucial to working with concepts like orthogonal vectors and spaces.

Linearly Independent Vectors

Linearly independent vectors are a set of vectors in a vector space that do not depend on each other. This means no vector in the set can be written as a linear combination of the others. Understanding linear independence is essential for determining the dimensions of a vector space and defining the concept of a basis.
Mathematically, a set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, ..., \mathbf{v}_n \} \) is linearly independent if the equation \( a_1\mathbf{v}_1 + a_2\mathbf{v}_2 + ... + a_n\mathbf{v}_n = \mathbf{0} \) only holds when all coefficients \( a_1, a_2, ..., a_n \) are zero.
This property is invaluable because:

• It means the set can uniquely span a particular subspace.
• It ensures that no vector in the set is redundant.

In the exercise, we assess linear independence to confirm that \( \{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{v}\} \) can indeed constitute a basis for a subspace they span, devoid of redundancy.

Orthogonal Basis

An orthogonal basis of a vector space is a basis where all vectors are orthogonal to each other. In other words, each pair of distinct vectors in the set has an inner product of zero. Orthogonal bases are especially useful because they simplify many calculations, such as finding the coordinates of vectors and computing projections.
To highlight:

• When vectors in a basis are mutually orthogonal, any vector in the space can be expressed as a linear combination of those basis vectors.
• The coefficients of this linear combination are easily found by computing the inner product of a vector with each basis vector, aiding in clear and straightforward decomposition.

In the exercise, we are given a basis \( \{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\} \) and need to ensure it is orthogonal. The conditions for orthogonality lead us to derive the specific values of \( \lambda \) and \( \mu \) that will maintain orthogonality in the presence of an initial vector \( \mathbf{v} \). These values are critical in constructing an orthogonal basis for the subspace of \( V \) that is conveniently managed.