Question

\(V\) denotes a finite dimensional inner product space. Define \(T: \mathbf{P} \rightarrow \mathbf{P}\) by \(T(f)=x f(x)\) for all \(f \in \mathbf{P},\) and define an inner product on \(\mathbf{P}\) as follows: If \(f=a_{0}+a_{1} x+a_{2} x^{2}+\cdots\) and \(g=b_{0}+b_{1} x+b_{2} x^{2}+\cdots\) are in \(\mathbf{P},\) define \(\langle f, g\rangle=a_{0} b_{0}+a_{1} b_{1}+a_{2} b_{2}+\cdots\) a. Show that \(\langle,\rangle\) is an inner product on \(\mathbf{P}\). b. Show that \(T\) is an isometry of \(\mathbf{P}\). c. Show that \(T\) is one-to-one but not onto.

Short answer

a. It is an inner product; b. \(T\) is an isometry; c. \(T\) is one-to-one but not onto.

Step-by-step solution

Show that \(\langle,\rangle\) is an inner product

To prove this, we need to verify the four properties of inner products: linearity in the first argument, conjugate symmetry, positivity, and positive definiteness. 1. **Linearity in the first argument**: For polynomials \(f = a_0 + a_1x + a_2x^2 + \cdots\) and \(h = c_0 + c_1x + c_2x^2 + \cdots\), and scalar \(\alpha\), we have: \[\langle \alpha f + h, g \rangle = \sum (\alpha a_i + c_i)b_i = \alpha \sum a_ib_i + \sum c_ib_i = \alpha \langle f, g \rangle + \langle h, g \rangle. \]2. **Conjugate symmetry**: Since we are dealing with real coefficients, we have \(\langle f, g \rangle = \langle g, f \rangle.\)3. **Positivity**: \(\langle f, f \rangle = a_0^2 + a_1^2 + a_2^2 + \cdots \geq 0\).4. **Positive definiteness**: \(\langle f, f \rangle = 0\) implies each \(a_i = 0\), hence \(f = 0\).All properties are satisfied; hence, \(\langle,\rangle\) is an inner product on \(\mathbf{P}\).

Prove that \(T\) is an isometry

To show \(T\) is an isometry, we need \(\langle T(f), T(g) \rangle = \langle f, g \rangle\) for all \(f, g \in \mathbf{P}\).If \(T(f) = xf\) and \(T(g) = xg\), then:\[\begin{align*}\langle T(f), T(g) \rangle &= \langle xf, xg \rangle \&= \sum (a_{i+1}x^{i+1})(b_{i+1}x^{i+1}) \&= \sum a_{i+1}b_{i+1} \quad (\text{which corresponds to the same indices as the inner products}).\end{align*}\]Upon inspection, the inner products remain unchanged, hence \(\langle T(f), T(f) \rangle = \langle f, f \rangle\) showing that \(T\) is an isometry.

Demonstrate that \(T\) is one-to-one

\(T\) is one-to-one if \(T(f) = T(g)\) implies \(f = g\). Suppose \(T(f) = T(g)\). Then \(xf = xg\), which implies \(f = g\) since cancelling \(x\) (considering polynomial degrees) occurs when coefficients for corresponding terms are equal. Thus, \(T\) is injective, so \(T\) is one-to-one.

Show that \(T\) is not onto

\(T\) is onto if for every polynomial \(h(x) = c_0 + c_1x + c_2x^2 + \cdots\) there is a polynomial \(f\) such that \(T(f) = xf = h(x)\). However, notice the constant term \(c_0\) cannot be obtained as it's annihilated in \(xf(x)\). Thus, constant non-zero polynomials cannot be in the range of \(T\), demonstrating that \(T\) is not onto.

Key concepts

Isometry

In the realm of mathematics, particularly when exploring inner product spaces, the concept of isometry shines through. An isometry is simply a transformation that preserves distance. In our context, when we say a transformation is an isometry, it means that distances between all points after the transformation remain the same as they were before.
In the given problem, the mapping \( T(f) = x f(x) \) is an isometry of the polynomial function space \( \mathbf{P} \). Why is this true? Because when we transform two polynomials using \( T \), the inner product remains unchanged. The inner product is a form of 'distance' in polynomial space.
So, if you think of the inner product as a special kind of measurement between two polynomial functions, an isometry ensures that these measurements stay constant. This is a vital property that helps us understand how transformations behave in inner product spaces. The concept of isometry is not just confined to polynomial spaces; it's a broader principle applicable across various mathematical structures that involve symmetry and distance.

Finite Dimensional Space

A finite dimensional space is a space that can be spanned by a finite number of basis vectors. This is an important concept because it tells us that every vector in the space can be expressed as a combination of the basis vectors.
For example, consider the space of polynomials \( \mathbf{P} \). If we restrict ourselves to polynomials of degree \( n \), then \( \mathbf{P} \) is finite dimensional. It can be represented by the basis \( \{1, x, x^2, \ldots, x^n\} \). Each polynomial in \( \mathbf{P} \) can be written as a linear combination of these basis elements.
Understanding finite dimensionality is crucial because it simplifies many computations and proves useful in theoretical mathematics. It means we can apply matrix theory and other linear algebra tools effectively, solving problems in an organized manner. Finite dimensional spaces are much easier to work with compared to infinite dimensional spaces, where basis cannot be defined so straightforwardly.

Polynomial Function

The polynomial function is a cornerstone in mathematics, especially when it comes to vector spaces and transformations. A polynomial function of degree \( n \) takes the form \( f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \). Each term in the polynomial is a product of a coefficient and a power of \( x \).
In the given exercise, the polynomial function defines the elements of the space \( \mathbf{P} \), and each of these functions can be combined, transformed, and analyzed using operations like addition, multiplication, and the inner product.
The beauty of polynomial functions lies in their versatility and simplicity. They can approximate other functions, provide solutions to equations, and model a wide range of phenomena in engineering, science, and economics. In the exercise, polynomial functions are manipulated using the transformation \( T(f) = xf(x) \), showing their role in expressing isometries and examining mappings in finite dimensional spaces.