Problem 20 - Differential Equations and Linear Algebra

Question

Determine the matrix representation \([T]_{B}^{C}\) for the given linear transformation \(T\) and ordered bases \(B\) and \(C,\) and for the given vector \(\mathbf{v}\) in \(V,\) use Theorem 6.5 .4 to compute \(T(\mathbf{v})\) $$\begin{aligned} &\begin{aligned} T: P_{3}(\mathbb{R}) \rightarrow & \mathbb{R}^{2} \text { given by } \\ T(p(x)) &=(p(0), p(1)) \end{aligned}\\\ &\text { with } B=\left\\{1, x, x^{2}, x^{3}\right\\}, C=\\{(1,0),(0,1)\\}, \text { and }\\\ &\mathbf{v}=2-x^{2}+2 x^{3} \end{aligned}$$

Step-by-Step Solution

Answer

Solution: Step 1: Apply T to elements of B: \(T(1) = (0, 1)\), \(T(x) = (0, 1)\), \(T(x^2) = (0, 1)\), \(T(x^3) = (0, 1)\). Step 2: Find coordinate vectors: \(T(1)_C = (0, 1)\), \(T(x)_C = (0, 1)\), \(T(x^2)_C = (0, 1)\), \(T(x^3)_C = (0, 1)\). Step 3: Create \([T]_{B}^{C}\): \(\begin{pmatrix}0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\end{pmatrix}\). Step 4: Express v with respect to B: \([v]_B = \begin{pmatrix}2\\0\\-1\\2\end{pmatrix}\). Step 5: Compute T(v) using Theorem 6.5.4: \(T(\mathbf{v})=\begin{pmatrix}0\\4\end{pmatrix}\). Final answer: The image of the vector v under the linear transformation T is (0, 4).

Step 1: Apply the transformation T to the basis elements of B

Apply the linear transformation T to each element in the ordered basis B, i.e. compute T(1), T(x), T(x^2), and T(x^3): \(T(1) = (1(0), 1(1)) = (0, 1)\) \(T(x) = (x(0), x(1)) = (0, 1)\) \(T(x^2) = (x^2(0), x^2(1)) = (0, 1)\) \(T(x^3) = (x^3(0), x^3(1)) = (0, 1)\)

Step 2: Find the coordinate vectors of the images

Express the images T(1), T(x), T(x^2), and T(x^3) as coordinate vectors with respect to the ordered basis C: \(T(1)_C = (0, 1) = 0(1, 0) + 1(0, 1) \Rightarrow (0, 1)\) \(T(x)_C = (0, 1) = 0(1, 0) + 1(0, 1) \Rightarrow (0, 1)\) \(T(x^2)_C = (0, 1) = 0(1, 0) + 1(0, 1) \Rightarrow (0, 1)\) \(T(x^3)_C = (0, 1) = 0(1, 0) + 1(0, 1) \Rightarrow (0, 1)\)

Step 3: Create the matrix representation [T]_B^C

In this step, we create the matrix representation of the linear transformation T with respect to the ordered bases B and C by using the coordinate vectors found in Step 2 as columns: \([[T]_{B}^{C} = \begin{pmatrix}0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\end{pmatrix}\)

Step 4: Find the coordinate vector of v with respect to B

In this step, we express the given vector v with respect to the ordered basis B: \(\mathbf{v} = 2-x^2+2x^3 = 2(1) + 0(x) - 1(x^2) + 2(x^3) \Rightarrow [v]_B = \begin{pmatrix}2\\0\\-1\\2\end{pmatrix}\)

Step 5: Compute T(v) using Theorem 6.5.4

Now, we apply Theorem 6.5.4 (change of basis formula) to compute the image of the vector v under the transformation T: \(T(\mathbf{v})=[T]_{B}^{C} [v]_{B}\) Using the matrix and the coordinate vector that we found in Steps 3 and 4: \( T(\mathbf{v})=\begin{pmatrix}0 & 0 & 0 & 0\\ 1 & 1 & 1 & 1\end{pmatrix} \begin{pmatrix}2\\0\\-1\\2\end{pmatrix} \) After performing the matrix-vector multiplication, we obtain: \(T(\mathbf{v})=\begin{pmatrix}0\\4\end{pmatrix}\) Therefore, the image of the vector v under the linear transformation T is (0, 4).

Key Concepts

Change of Basis

The concept of a change of basis is central in linear algebra. It involves transforming vectors from one coordinate system to another. Consider two different bases, B and C, in the same vector space. To change from basis B to basis C, we need to express each vector in terms of the new basis. This action is analogous to converting measurements from inches to centimeters in the physical world; it's the same 'length' but expressed in a different 'language'.

The matrix that converts coordinate vectors from basis B to basis C is known as the change of basis matrix. In the given exercise, the matrix representation \[T\]_B^C embodies this idea, representing how the linear transformation T changes as we move from the domain basis B to the codomain basis C.

Coordinate Vectors

Coordinate vectors are representations of vectors within a specific basis. Think of it as the address of a point in a city grid: depending on your reference (which corner you start from), the same point can have different coordinates. In linear algebra, for any vector \(v\) and basis B, the coordinate vector \[v\]_B provides the coordinates of \(v\) with respect to B.

Importantly, when basis B changes to basis C, the corresponding coordinate vectors change as well. In our exercise, the vector \(\mathbf{v}\) in the polynomial vector space had its coordinates reevaluated from the standard basis to basis B, a necessary step before applying the transformation T to find T(\mathbf{v}).

Linear Algebra

Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between these spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces. The linear transformation T in our exercise is an example of a function that takes vectors from one space and maps them into another, preserving vector addition and scalar multiplication.

Linear algebra provides the language for solving systems of equations, describing geometrical shapes, and operationally, it underpins various scientific computations. The steps taken in the exercise demonstrate the practicality of linear algebra in transforming and interpreting vectors in diverse coordinate systems.

Polynomial Vector Space

A polynomial vector space, such as \(P_3(\mathbb{R})\), consists of all polynomials with a degree less than or equal to 3. These polynomials form a vector space because they can be added together or multiplied by scalars to produce another polynomial of the same type.

In the given exercise, the vector space is mapped to \(\mathbb{R}^2\) by the linear transformation T. This demonstrates an important aspect of polynomial vector spaces: they can be subject to linear transformations just like vectors in \(\mathbb{R}^n\), and their interactions can be interpreted within the framework of linear algebra. Polynomial vector spaces are fundamental examples in linear algebra, showing the versatility and power of abstract vectors beyond mere points in space.