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Determine whether the linear transformation T is (a) one-to-one and ( \(b\) ) onto. \(T: \mathscr{P}_{2} \rightarrow \mathbb{R}^{2}\) defined by \(T(p(x))=\left[\begin{array}{l}p(0) \\ p(1)\end{array}\right]\)

Short Answer

Expert verified
The transformation \( T \) is not one-to-one but is onto.

Step by step solution

01

Understand the Transformation

The given transformation \( T : \mathscr{P}_{2} \rightarrow \mathbb{R}^{2} \) takes a polynomial \( p(x) = ax^2 + bx + c \) and maps it to a vector in \( \mathbb{R}^2 \) such that: \( T(p(x)) = \begin{bmatrix} p(0) \ p(1) \end{bmatrix} \), where \( p(0) = c \) and \( p(1) = a + b + c \).
02

Check for One-to-Oneness

To determine if \( T \) is one-to-one, we must show that if \( T(p(x)) = T(q(x)) \) implies \( p(x) = q(x) \). Set \( T(p(x)) = T(q(x)) \), which gives \( \begin{bmatrix} c_p \ a_p + b_p + c_p \end{bmatrix} = \begin{bmatrix} c_q \ a_q + b_q + c_q \end{bmatrix} \). This leads to the system \( c_p = c_q \) and \( a_p + b_p + c_p = a_q + b_q + c_q \). Substitute \( c_p = c_q \) into the second equation: \( a_p + b_p = a_q + b_q \). There are infinite solutions to \( a_p, b_p, a_q, \) and \( b_q \) that satisfy this equality for non-equal polynomials, hence, \( T \) is not one-to-one.
03

Check for Onto Property

To determine if \( T \) is onto, every vector \( \begin{bmatrix} y_1 \ y_2 \end{bmatrix} \) in \( \mathbb{R}^2 \) must be image of some polynomial \( p(x) = ax^2 + bx + c \). From \( T(p(x)) = \begin{bmatrix} c \ a + b + c \end{bmatrix} = \begin{bmatrix} y_1 \ y_2 \end{bmatrix} \), we get \( c = y_1 \) and \( a + b + c = y_2 \). Substituting \( c = y_1 \) into the second equation gives \( a + b = y_2 - y_1 \). The system of equations has a solution for every \( y_1 \) and \( y_2 \), demonstrating that \( T \) maps onto all of \( \mathbb{R}^2 \). Thus, \( T \) is onto.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-to-One Mapping
In the realm of linear transformations like our function \( T \), understanding "one-to-one mapping" is essential. A transformation is considered one-to-one, or injective, if it assigns distinct outputs to distinct inputs. In simpler terms, if \( T(p(x)) = T(q(x)) \) implies that \( p(x) = q(x) \), then \( T \) is one-to-one.
In our exercise, we set the outputs equal, i.e., \( T(p(x)) = T(q(x)) \), and ended up with equations requiring \( c_p = c_q \) and \( a_p + b_p + c_p = a_q + b_q + c_q \). Even after simplifying, we found infinite possibilities of \( a_p \), \( b_p \), \( a_q \), and \( b_q \) such that the polynomials can still be distinct. This leads to the conclusion that \( T \) isn't one-to-one as different inputs can yield the same result.
Onto Mapping
Onto mapping, or surjectivity, is about ensuring the function covers every possible output in its target space. A transformation \( T \) is onto if, for every vector in the target space, there is a polynomial that maps to it.
In our case, we consider any vector \( \begin{bmatrix} y_1 \ y_2 \end{bmatrix} \) in \( \mathbb{R}^2 \) and pose the question: is there a polynomial \( p(x) = ax^2 + bx + c \) such that \( T(p(x)) = \begin{bmatrix} y_1 \ y_2 \end{bmatrix} \)? We see through the equations \( c = y_1 \) and \( a + b + c = y_2 \) that solutions can be found for any choice of \( y_1 \) and \( y_2 \). Thus, \( T \) is onto, demonstrating that we can reach any point in \( \mathbb{R}^2 \) using our polynomials.
Polynomial Functions
Polynomial functions form the backbone of the transformation in our example. Simply put, a polynomial is a mathematical expression involving sums and powers of variables, typically \( x \), with coefficients. In \( \mathscr{P}_2 \), we specifically look at quadratic polynomials \( p(x) = ax^2 + bx + c \).
The transformation \( T \) cleverly uses these polynomials: it evaluates them at specific points—in this case \( x = 0 \) and \( x = 1 \)—to form a vector in \( \mathbb{R}^2 \). With polynomials of degree up to 2, their behavior and structure allow us to explore concepts like one-to-one and onto mappings, as seen in our example. By examining outputs at particular points, we capture essential information about the polynomial's properties.
System of Equations
In analyzing linear transformations, systems of equations become crucial tools. They help us decode relationships between inputs and outputs of the transformation. Here, we encountered two specific equations: \( c_p = c_q \) and \( a_p + b_p + c_p = a_q + b_q + c_q \).
By resolving these, we checked the injectivity of \( T \)—whether different inputs could lead to different outputs. The system showed multiple solutions for \( a \) and \( b \), signaling that \( T \) isn't one-to-one. Similarly, the system \( c = y_1 \) and \( a + b + c = y_2 \) confirmed the surjectivity, or onto nature, of \( T \), revealing that all output vectors in \( \mathbb{R}^2 \) have a corresponding input. Understanding these systems is key to unpacking the behavior of transformations.

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Most popular questions from this chapter

Table 6.2 gives the population of the United States at 10-year intervals for the years \(1900-2000\) (a) Assuming an exponential growth model, use the data for 1900 and 1910 to find a formula for \(p(t)\) the population in year \(t .\) ( Hint: Let \(t=0\) be 1900 and let \(t=1\) be \(1910 .\) ) How accurately does your formula calculate the U.S. population in 2000 ? (b) Repeat part (a), but use the data for the years 1970 and 1980 to solve for \(p(t) .\) Does this approach give a better approximation for the year \(2000 ?\) (c) What can you conclude about U.S. population growth? $$\begin{array}{lc} \text { Year } & \text { (Population in millions) } \\ \hline 1900 & 76 \\ 1910 & 92 \\ 1920 & 106 \\ 1930 & 123 \\ 1940 & 131 \\ 1950 & 150 \\ 1960 & 179 \\ 1970 & 203 \\ 1980 & 227 \\ 1990 & 250 \\ 2000 & 281 \end{array}$$

The set of all linear transformations from a vector space \(V\) to a vector space \(W\) is denoted by \(\mathscr{L}(V, W)\). If \(S\) and \(T\) are in \(\mathscr{L}(V, W),\) we can define the sum \(S+T\) of \(\operatorname{Sand} T b y\) \\[(S+T)(\mathbf{v})=S(\mathbf{v})+T(\mathbf{v})\\] for all v in \(V\). If \(c\) is a scalar, we define the scalar multiple \(c T\) of T by c to be \\[(c T)(\mathbf{v})=c T(\mathbf{v})\\] for all \(\mathbf{v}\) in \(V\). Then \(S+T\) and \(c\) T are both transformations from \(V\) to \(W\). Let \(R, S\), and \(T\) be linear transformations such that the following operations make sense. Prove that: (a) \(R \circ(S+T)=R \circ S+R \circ T\) (b) \(c(R \circ S)=(c R) \circ S=R \circ(c S)\) for any scalar \(c\)

Show that the solution set \(S\) of the second-order differential equation \(y^{\prime \prime}+a y^{\prime}+b y=0\) is a subspace of \(\mathscr{F}\)

Let \(T: V \rightarrow W\) be a linear transformation between finite-dimensional vector spaces \(V\) and \(W\). Let \(\mathcal{B}\) and \(\mathcal{C}\) be bases for \(V\) and \(W\), respectively, and let \(A=[T]_{C+B}\). Show that nullity \((T)=\) nullity \((A)\)

Verify that \(S\) and \(T\) are inverses. $$\begin{aligned}&S: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \text { defined by } S\left[\begin{array}{l}x \\\y\end{array}\right]=\left[\begin{array}{l}4 x+y \\\3 x+y\end{array}\right] \text { and } T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\\\ &\text { defined by } T\left[\begin{array}{l}x \\\y\end{array}\right]=\left[\begin{array}{c}x-y \\\\-3 x+4 y\end{array}\right]\end{aligned}$$

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