Question

Let \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3}\) be defined by \(T(p)=(p(0), p(1), p(2))\) for all \(p\) in \(\mathbf{P}_{2}\). Let \(B=\left\\{1, x, x^{2}\right\\}\) and \(D=\\{(1,0,0),(0,1,0),(0,0,1)\\}\) . a. Show that \(M_{D B}(T)=\left[\begin{array}{lll}1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 2 & 4\end{array}\right]\) and conclude that \(T\) is an isomorphism. b. Generalize to \(T: \mathbf{P}_{n} \rightarrow \mathbb{R}^{n+1}\) where \(T(p)=\left(p\left(a_{0}\right), \quad p\left(a_{1}\right), \ldots, \quad p\left(a_{n}\right)\right) \quad\) and \(a_{0}, a_{1}, \ldots, a_{n}\) are distinct real numbers.

Short answer

\(T\) is an isomorphism since its matrix is invertible. Generalization involves using a Vandermonde matrix.

Step-by-step solution

Identify the Transformation

The transformation \(T\) maps a polynomial \(p\) in \(\mathbf{P}_2\) to a vector in \(\mathbb{R}^3\) by evaluating the polynomial at 0, 1, and 2. This gives the transformation \(T(p) = (p(0), p(1), p(2))\).

Apply Basis \(B\)

The basis \(B\) for \(\mathbf{P}_2\) is \(\{1, x, x^2\}\). We evaluate each basis polynomial at 0, 1, and 2 to find their images under \(T\):- \(T(1) = (1, 1, 1)\)- \(T(x) = (0, 1, 2)\)- \(T(x^2) = (0, 1, 4)\).

Form Matrix \(M_{DB}(T)\)

To form the matrix \(M_{DB}(T)\), we arrange the image vectors of the basis elements as columns:\[M_{DB}(T) = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 2 & 4 \end{bmatrix}\]

Verify Isomorphism

To check if \(T\) is an isomorphism, we need to determine if \(M_{DB}(T)\) is invertible. This holds if its determinant is non-zero. Calculate the determinant of \(M_{DB}(T)\):\[\det \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 2 & 4 \end{bmatrix} = 1\cdot(1\cdot4 - 1\cdot2) - 0 + 0 = 2\]Since the determinant is non-zero, \(T\) is an isomorphism.

Generalize for \(T: \mathbf{P}_n \to \mathbb{R}^{n+1}\)

For \(T: \mathbf{P}_n \to \mathbb{R}^{n+1}\), we define \(T(p) = (p(a_0), p(a_1), \ldots, p(a_n))\) where each \(a_i\) is distinct. The resulting matrix \(M_{DB}(T)\) will have entries from evaluating polynomials at those distinct points, creating a Vandermonde matrix. A Vandermonde matrix is invertible (non-singular) if and only if its rows (or columns) are distinct, which is guaranteed by the distinct \(a_i\)'s.

Key concepts

Polynomial Transformation

A polynomial transformation is a function that maps polynomials from one space to another by evaluating them at specific points. Consider the transformation \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3}\), which takes a polynomial \(p(x)\) and maps it to a vector in \(\mathbb{R}^{3}\) by evaluating the polynomial at three points: 0, 1, and 2. This means that the transformation is defined as \(T(p) = (p(0), p(1), p(2))\).
This simple evaluation method effectively transforms the polynomial into a vector, capturing the essence of the polynomial's behavior at the specified points. This process is powerful in applications such as signal processing and numerical analysis, where understanding polynomial behavior at particular instances is crucial.

Matrix Representation

Matrix representation is a method of representing linear transformations between vector spaces as matrices. In linear algebra, every linear transformation can be associated with a matrix. For the transformation \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}^{3}\), we seek to find its matrix representation relative to the bases \(B = \{1, x, x^2\}\) and \(D = \{(1,0,0), (0,1,0), (0,0,1)\}\).
We first evaluate the basis polynomials at the given points of the transformation:

• \(T(1) = (1, 1, 1)\)
• \(T(x) = (0, 1, 2)\)
• \(T(x^2) = (0, 1, 4)\)

Next, we form the matrix \(M_{DB}(T)\) using these evaluated vectors as its columns: \[M_{DB}(T) = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 2 & 4 \end{bmatrix}\]This matrix encapsulates the transformation and can be used for further computational purposes like determining invertibility.

Isomorphism

Isomorphism in the context of linear algebra refers to a bijective (one-to-one and onto) linear transformation between vector spaces. Determining if a transformation is an isomorphism often involves checking if its matrix representation is invertible.
For our transformation \(T\), its matrix \(M_{DB}(T)\) is:\[\begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 2 & 4 \end{bmatrix}\]We calculate its determinant to evaluate invertibility:\[\det \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 2 & 4 \end{bmatrix} = 1 \cdot (1 \cdot 4 - 1 \cdot 2) = 2\]Since the determinant is non-zero, the matrix is invertible, confirming that the transformation \(T\) is an isomorphism. This means \(T\) not only maps \(\mathbf{P}_2\) onto \(\mathbb{R}^3\) perfectly, but it can also be reversed, providing a robust mathematical equivalence between these spaces.

Vandermonde Matrix

The Vandermonde matrix plays a crucial role when generalizing polynomial transformations. When extending the transformation \(T\) to \(\mathbf{P}_{n} \rightarrow \mathbb{R}^{n+1}\), with \(T(p) = (p(a_0), p(a_1), \ldots, p(a_n))\), the matrix that represents this transformation is often a Vandermonde matrix.
A Vandermonde matrix for distinct values \(a_0, a_1, \ldots, a_n\) is structured as follows:\[\begin{bmatrix}1 & a_0 & a_0^2 & \cdots & a_0^n \1 & a_1 & a_1^2 & \cdots & a_1^n \\vdots & \vdots & \vdots & & \vdots \1 & a_n & a_n^2 & \cdots & a_n^n\end{bmatrix}\]The key property of a Vandermonde matrix is that it is invertible if and only if the values \(a_i\) are distinct. This unique property ensures that the transformation remains an isomorphism, preserving the integrity and independent mappings within the polynomial space. Vandermonde matrices are instrumental in numerical analysis, interpolation, and systems of equations, thanks to their unique structure and invertibility properties.