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Chapter 9: Problem 1

Let \(T: \mathbb{P}_{2} \rightarrow \mathbb{R}\) be a linear transformation such that $$ T\left(x^{2}\right)=1 ; T\left(x^{2}+x\right)=5 ; T\left(x^{2}+x+1\right)=-1 $$ Find \(T\left(a x^{2}+b x+c\right)\)

Short Answer

Expert verified
T(ax^2 + bx + c) = a + 4b - 6c

Step by step solution

01

- Understand the Linear Transformation

Given that T is a linear transformation, recall that for any polynomials P(x) and Q(x) and scalars a and b, the following property holds: T(aP(x) + bQ(x)) = aT(P(x)) + bT(Q(x)).
02

- Express the Polynomial

The polynomial given is ax^2 + bx + c. This can be broken down as follows: ax^2 + bx + c = a(x^2) + b(x) + c(1). By linearity, we need to determine T(ax^2 + bx + c).
03

- Use Given Transformations

Using the given transformations, T(x^2) = 1, T(x^2 + x) = 5, and T(x^2 + x + 1) = -1, express T(x) and T(1) in terms of these. First, we compute T(1) and T(x):
04

- Compute T(1)

We know from T(x^2 + x + 1) = -1 and the previous transformations that T(x^2) + T(x) + T(1) = -1. Substitute T(x^2) = 1 and T(x^2 + x) = 5, solve for T(1):
05

- Solve for T(1)

\[ 1 + T(x) + T(1) = -1 \]
06

- Simplify Equation

We also know that T(x^2 + x) = T(x^2) + T(x) = 5. Thus, T(x) = 4. Substitute T(x) to solve for T(1):
07

- Solve for T(1)

1 + 4 + T(1) = -1. Simplifying gives T(1) = -6.
08

- Apply Linearity

Now, apply linearity to find T(ax^2 + bx + c):
09

- Compute Final Expression

T(ax^2 + bx + c) = aT(x^2) + bT(x) + cT(1). Substitute the known values:
10

- Simplify Final Expression

\[ T(ax^2 + bx + c) = a(1) + b(4) + c(-6) \]
11

- Final Result

Simplify to get the final answer: \[ T(ax^2 + bx + c) = a + 4b - 6c \]

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Key Concepts

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

Polynomial Expression
A polynomial expression is a mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In this exercise, we deal with a polynomial of the form (ax^2 + bx + c). Here:
  • 'a', 'b', and 'c' are the coefficients.
  • 'x' is the variable, and
  • The highest power of 'x' is 2, making this a second-degree polynomial.
Polynomials like ax^2 + bx + c can be broken down based on the degree of each term. For example:
  • The term 'ax^2' is a degree-2 term.
  • The term 'bx' is a degree-1 term.
  • 'c' is a constant (degree-0 term).
When analyzing such polynomials, we often consider each term individually, which is a helpful step when applying transformations.
Linearity Property
The linearity property of a transformation is crucial. It tells us how to deal with sums and scalar multiples of functions. A transformation T is linear if for any two polynomials P(x) and Q(x) and scalars a and b, the following holds: \[T(aP(x) + bQ(x)) = aT(P(x)) + bT(Q(x))\] This property allows us to handle complex polynomials systematically by breaking them into simpler parts. For the given problem, this means we express the polynomial ax^2 + bx + c as a combination of simpler polynomials:
  • \[ax^2\]
  • \[bx\]
  • \[c\]
Thanks to linearity, we can transform each part independently and then sum the results:
  • \(T(ax^2 + bx + c) = aT(x^2) + bT(x) + cT(1)\)
Transformation Computation
Using the given transformations and applying linearity, we can compute the necessary values to transform any polynomial of the form ax^2 + bx + c. Here's how we proceed:

Given transformations:
  • \(T(x^2) = 1\)
  • \(T(x^2 + x) = 5\)
  • \(T(x^2 + x + 1) = -1\)
Steps:
  • Identify relationships from the given transformations.
  • Calculate unknown values like T(x) and T(1).
From the given equations:
\[1 + T(x) + T(1) = -1\] Solving step-by-step, we find:
\[T(x) = 4\]
\[1 + 4 + T(1) = -1\] so, ### Final Expression for Polynomial Transformation With the values of T(x) and T(1) identified, we use linearity to find:
\[T(ax^2 + bx + c) = aT(x^2) + bT(x) + cT(1)\]
By substituting known values:
\left( \[T(ax^2 + bx + c) = a(1) + b(4) + c(-6)\] \right) Slowing down for simplification, we get:
\[T(ax^2 + bx + c) = a + 4b - 6c\]
So, the transformed polynomial is T(ax^2 + bx + c), in the form: \[a + 4b - 6c\]

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Most popular questions from this chapter

Let \(V\) and \(W\) be subspaces of \(\mathbb{R}^{n}\) and \(\mathbb{R}^{m}\) respectively and let \(T: V \rightarrow W\) be a linear transformation. Show that if \(T\) is onto \(W\) and if \(\left\\{\vec{v}_{1}, \cdots, \vec{v}_{r}\right\\}\) is a basis for \(V,\) then span \(\left\\{T \vec{v}_{1}, \cdots, T \vec{v}_{r}\right\\}=\) \(W\)

Find the matrix for \(T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})\) where \(\vec{v}=\left[\begin{array}{lll}1 & -2 & 3\end{array}\right]^{T}\).

Find a basis in \(\mathbb{P}_{3}\) for the subspace $$ \operatorname{span}\left\\{x^{3}-2 x^{2}+x+2,3 x^{3}-x^{2}+2 x+2,7 x^{3}+x^{2}+4 x+2,5 x^{3}+3 x+2\right\\} $$ If the above three vectors do not yield a basis, exhibit one of them as a linear combination of the others.

Let $$ V=\operatorname{span}\left\\{\left[\begin{array}{l} 1 \\ 1 \\ 2 \\ 0 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 1 \\ 1 \end{array}\right],\left[\begin{array}{l} 1 \\ 1 \\ 0 \\ 1 \end{array}\right]\right\\} $$ Let \(T \vec{x}=A \vec{x}\) where \(A\) is the matrix $$ \left[\begin{array}{llll} 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 1 & 1 & 2 \end{array}\right] $$ Give a basis for im \((T)\).

Determine if the following set is linearly independent. If it is linearly dependent, write one vector as a linear combination of the other vectors in the set. $$ \left\\{\left[\begin{array}{ll} 1 & 2 \\ 0 & 1 \end{array}\right],\left[\begin{array}{ll} -7 & 2 \\ -2 & -3 \end{array}\right],\left[\begin{array}{ll} 4 & 0 \\ 1 & 2 \end{array}\right]\right\\} $$
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