Problem 25 - Differential Equations and Linear Algebra

Question

Let \(V\) be a real inner product space, and let \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) be fixed (nonzero) vectors in \(V .\) Define \(T: V \rightarrow \mathbb{R}^{2}\) by $$T(\mathbf{v})=\left(\left\langle\mathbf{u}_{1}, \mathbf{v}\right\rangle,\left\langle\mathbf{u}_{2}, \mathbf{v}\right\rangle\right)$$. Use properties of the inner product to show that \(T\) is a linear transformation.

Step-by-Step Solution

Answer

We can show that $T$ is a linear transformation by verifying the two properties of linearity using properties of the inner product. Property 1 states that $T(a \mathbf{v} + b \mathbf{w}) = a T(\mathbf{v}) + b T(\mathbf{w})$. Applying inner product properties gives $(a\langle\mathbf{u}_{1},\mathbf{v}\rangle+b\langle\mathbf{u}_{1},\mathbf{w}\rangle, a\langle\mathbf{u}_{2},\mathbf{v}\rangle+b\langle\mathbf{u}_{2},\mathbf{w}\rangle)$, which matches $a T(\mathbf{v}) + b T(\mathbf{w})$. Property 2 states that $T(\mathbf{v}+\mathbf{w}) = T(\mathbf{v}) + T(\mathbf{w})$. Again, applying inner product properties gives $(\langle\mathbf{u}_{1},\mathbf{v}\rangle+\langle\mathbf{u}_{1},\mathbf{w}\rangle, \langle\mathbf{u}_{2},\mathbf{v}\rangle+\langle\mathbf{u}_{2},\mathbf{w}\rangle)$, which matches $T(\mathbf{v})+T(\mathbf{w})$. Thus, as both properties hold, we can conclude that $T$ is a linear transformation.

Step 1: Property 1: Scalar multiplication and addition

We want to show that T(a * v + b * w) = a * T(v) + b * T(w) for all vectors v, w in V and scalars a, b in R. Given, \[T(\mathbf{v}) = (\langle\mathbf{u}_{1}, \mathbf{v}\rangle, \langle\mathbf{u}_{2}, \mathbf{v}\rangle)\] Let's calculate \(T(a * v + b * w)\) and apply the inner product properties: \[T(a\mathbf{v}+b\mathbf{w})=(\langle\mathbf{u}_{1},a\mathbf{v}+b\mathbf{w}\rangle, \langle\mathbf{u}_{2},a\mathbf{v}+b\mathbf{w}\rangle)\] Using the distributive property of the inner product, \[(a\langle\mathbf{u}_{1},\mathbf{v}\rangle+b\langle\mathbf{u}_{1},\mathbf{w}\rangle, a\langle\mathbf{u}_{2},\mathbf{v}\rangle+b\langle\mathbf{u}_{2},\mathbf{w}\rangle)\] Now, we know that \(T(\mathbf{v})=(\langle\mathbf{u}_{1},\mathbf{v}\rangle, \langle\mathbf{u}_{2},\mathbf{v}\rangle)\) and \(T(\mathbf{w})=(\langle\mathbf{u}_{1},\mathbf{w}\rangle, \langle\mathbf{u}_{2},\mathbf{w}\rangle)\) So, \[a\cdot T(\mathbf{v})+b\cdot T(\mathbf{w})=(a\langle\mathbf{u}_{1},\mathbf{v}\rangle+b\langle\mathbf{u}_{1},\mathbf{w}\rangle, a\langle\mathbf{u}_{2},\mathbf{v}\rangle+b\langle\mathbf{u}_{2},\mathbf{w}\rangle)\] Thus, T(a * v + b * w) = a * T(v) + b * T(w), which verifies Property 1.

Step 2: Property 2: Vector addition

We want to show that T(v + w) = T(v) + T(w) for all vectors v, w in V. Given, \[T(\mathbf{v}) = (\langle\mathbf{u}_{1}, \mathbf{v}\rangle, \langle\mathbf{u}_{2}, \mathbf{v}\rangle)\] Let's calculate \(T(v + w)\) and apply the inner product properties: \[T(\mathbf{v}+\mathbf{w})=(\langle\mathbf{u}_{1},\mathbf{v}+\mathbf{w}\rangle, \langle\mathbf{u}_{2},\mathbf{v}+\mathbf{w}\rangle)\] Using the distributive property of the inner product, \[(\langle\mathbf{u}_{1},\mathbf{v}\rangle+\langle\mathbf{u}_{1},\mathbf{w}\rangle, \langle\mathbf{u}_{2},\mathbf{v}\rangle+\langle\mathbf{u}_{2},\mathbf{w}\rangle)\] Now, we know that \(T(\mathbf{v})=(\langle\mathbf{u}_{1},\mathbf{v}\rangle, \langle\mathbf{u}_{2},\mathbf{v}\rangle)\) and \(T(\mathbf{w})=(\langle\mathbf{u}_{1},\mathbf{w}\rangle, \langle\mathbf{u}_{2},\mathbf{w}\rangle)\) So, \[T(\mathbf{v})+T(\mathbf{w})=(\langle\mathbf{u}_{1},\mathbf{v}\rangle+\langle\mathbf{u}_{1},\mathbf{w}\rangle, \langle\mathbf{u}_{2},\mathbf{v}\rangle+\langle\mathbf{u}_{2},\mathbf{w}\rangle)\] Thus, T(v + w) = T(v) + T(w), which verifies Property 2. Since both properties are verified, we can conclude that T is a linear transformation.

Key Concepts

Inner Product Space

An inner product space is a vector space equipped with a special operation called the inner product. This inner product allows us to measure angles and lengths within the space, which are essential concepts in understanding geometry and physics. Each pair of vectors in the space can be combined using the inner product to produce a scalar quantity, often representing some form of 'dot' product.

For a real vector space, the inner product \( \langle \mathbf{u}, \mathbf{v} \rangle \) of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) must satisfy several conditions: it must be commutative, linear in the first argument, positive-definite, and satisfy the Cauchy-Schwarz inequality. These properties ensure that the space behaves in an intuitive 'geometric' way, which allows mathematicians and physicists to generalize concepts like projection, orthogonality, and norms from Euclidean space to more abstract settings.

Properties of Inner Product

The properties of an inner product define its true nature and how it interacts with the elements of a vector space. Firstly, it's commutative, which means that \( \langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v}, \mathbf{u} \rangle \) for any vectors \(\mathbf{u}\) and \(\mathbf{v}\) in the space.

Furthermore, the inner product is distributive over vector addition, so \( \langle \mathbf{u}, \mathbf{v} + \mathbf{w} \rangle = \langle \mathbf{u}, \mathbf{v} \rangle + \langle \mathbf{u}, \mathbf{w} \rangle \) for any vectors \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\). It's also homogenous with respect to scalar multiplication, meaning that \( \langle \alpha \mathbf{u}, \mathbf{v} \rangle = \alpha \langle \mathbf{u}, \mathbf{v} \rangle \) for any scalar \(\alpha\) and vectors \(\mathbf{u}, \mathbf{v}\). Lastly, it's positive-definite, which asserts that \( \langle \mathbf{u}, \mathbf{u} \rangle \geq 0 \) with equality if and only if \(\mathbf{u} = \mathbf{0}\), the zero vector.

Vector Addition

Vector addition within an inner product space inherits the intuitive rules from our everyday Euclidean space. Adding two vectors together yields another vector within the same space, and this operation is commutative and associative. So, \(\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}\) and \(\mathbf{u} + (\mathbf{v} + \mathbf{w}) = (\mathbf{u} + \mathbf{v}) + \mathbf{w}\) for any vectors in the space.

These simple rules ensure that vector spaces and consequently inner product spaces are predictable and manageable. When we talk about linear transformations, as in the given exercise, it's these properties that allow us to understand and prove that the transformation preserves these addition operations.

Scalar Multiplication

Scalar multiplication refers to the operation of multiplying a vector by a real number (scalar), scaling it in length without changing its direction. A critical aspect of scalar multiplication in the context of inner product properties is that it is distributive over addition of scalars and vectors, and it is associated with the field of real numbers.

For instance, if \(a\) and \(b\) are scalars and \(\mathbf{u}\), \(\mathbf{v}\) are vectors in the space, we have \(a(\mathbf{u} + \mathbf{v}) = a\mathbf{u} + a\mathbf{v}\) and \((a + b)\mathbf{u} = a\mathbf{u} + b\mathbf{u}\). This interplay between scalar multiplication and vector addition is fundamental to showing the linearity of transformations, just as we've explored in the solutions to the exercise.