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Chapter 6: Problem 22

Determine whether the function is a linear transformation. $$T: P_{2} \rightarrow P_{2}, T\left(a_{0}+a_{1} x+a_{2} x^{2}\right)=a_{1}+2 a_{2} x$$

Short Answer

Expert verified
Yes, the function \( T\left(a_{0}+a_{1} x+a_{2} x^{2}\right)=a_{1}+2 a_{2} x \) is a linear transformation, because it satisfies the two conditions for a transformation to be linear.

Step by step solution

01

Write down the function.

We first write the function being studied: \( T\left(a_{0}+a_{1} x+a_{2} x^{2}\right)=a_{1}+2 a_{2} x \)
02

Check if it satisfies the condition \( T(u + v) = T(u) + T(v) \)

Let's use two general vectors for polynomials of degree 2 for \( u = b_{0}+b_{1} x+b_{2} x^{2} \) and \( v = c_{0}+c_{1} x+c_{2} x^{2} \). The sum of these polynomials \( u + v = (b_{0}+c_{0})+(b_{1}+c_{1}) x+(b_{2}+c_{2}) x^{2} \). If we apply the transformation to u + v: \( T(u + v) = (b_{1}+c_{1})+2 (b_{2}+c_{2}) x \) and adding the transformations of u and v individually, \( T(u) + T(v) = (b_{1}+2b_{2} x) + (c_{1}+2c_{2} x) = (b_{1}+c_{1})+2 (b_{2}+c_{2}) x \). Hence, the function T satisfies the first condition of a linear transformation.
03

Check if it satisfies the condition \( T(cu) = cT(u) \)

For scalar c, \( cu = c(b_{0}+b_{1} x+b_{2} x^{2}) = (cb_{0})+(cb_{1}) x+ (cb_{2}) x^{2} \). Applying the transformation, \( T(cu) = cb_{1}+2 cb_{2} x \). On the other hand, \( cT(u) = c(b_{1}+2b_{2} x) = cb_{1}+2 cb_{2} x \). Thus, the function T satisfies the second condition of a linear transformation.
04

Conclusion

Since both conditions for a transformation to be linear are fulfilled (i.e., \( T(u + v) = T(u) + T(v) \) and \( T(cu) = cT(u) \)), we conclude that the function \( T\left(a_{0}+a_{1} x+a_{2} x^{2}\right)=a_{1}+2 a_{2} x \) is indeed a linear transformation.

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Key Concepts

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

Vectors in Polynomial Form
When dealing with linear transformations, particularly in polynomial spaces such as \( P_2 \), it's essential to understand what vectors look like. In this context, vectors are polynomials of a certain form. For \( P_2 \), these are polynomials of degree 2. This means they can be expressed as \( a_0 + a_1 x + a_2 x^2 \). Here, \( a_0, a_1, \) and \( a_2 \) are coefficients, and \( x \) is the variable. Imagine each polynomial as a point or vector in a space formed by these coefficients. Think of the coefficients \( a_0 \), \( a_1 \), and \( a_2 \) as coordinates in a 3-dimensional space. Just like how a vector \([x, y, z]\) is a point in 3D space, \([a_0, a_1, a_2]\) is a polynomial in this space. This perspective helps in understanding operations on polynomials in terms of linear transformations.
Function Properties
Let's delve into what it means for a function to be a linear transformation. A function or transformation \( T \) is considered linear when it satisfies two primary properties:
  • Additivity: This means that for any two vectors \( u \) and \( v \), \( T(u + v) = T(u) + T(v) \). This property ensures that the transformation distributes over addition.
  • Homogeneity: For any scalar \( c \) and vector \( u \), the condition \( T(cu) = cT(u) \) holds. It ensures that scaling a vector before transformation yields the same result as transforming and then scaling.
In our exercise, these two conditions are applied to polynomials in \( P_2 \). For a transformation like \( T(a_0 + a_1 x + a_2 x^2) = a_1 + 2a_2 x \), checking these properties confirms if \( T \) truly behaves linearly.
Polynomial Degree
Polynomials are classified by their degree, which tells you the highest power of \( x \) present in the polynomial. In this exercise, we focus on polynomials of degree 2, which means the highest power of \( x \) is \( x^2 \). Understanding this concept is crucial when dealing with transformations, as the degree determines the space the polynomial vector belongs to, like \( P_2 \) in this case. Each polynomial in \( P_2 \) can be seen as a vector in a space determined by three coefficients \( a_0, a_1, \) and \( a_2 \). Linear transformations often alter these coefficients or combine them in different ways. But, they should not increase the degree of a polynomial. So, knowing the degree ensures that the transformation stays within its designated space without complicating the polynomial with higher degree terms.

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Most popular questions from this chapter

Let \(T\) be a linear transformation from \(P\) into \(R\) such that $$T(p)=\int_{0}^{1} p(x) d x$$ Find \((a) T\left(-2+3 x^{2}\right),(b) T\left(x^{3}-x^{5}\right),\) and \((c) T(-6+4 x)\)

Let the matrix \(A\) represent the linear transformation \(T: R^{3} \rightarrow R^{3}\). Describe the orthogonal projection to which \(T\) maps every vector in \(R^{3}\). $$A=\left[\begin{array}{lll}0 & 0 & 0 \\\0 & 1 & 0 \\\0 & 0 & 1\end{array}\right]$$

True or False? In Exercises 53 and \(54,\) determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text. Guided Proof Let \(T_{1}: V \rightarrow V\) and \(T_{2}: V \rightarrow V\) be one- to-one linear transformations. Prove that the composition \(T=T_{2} \circ T_{1}\) is one-to-one and that \(T^{-1}\) exists and is equal to \(T_{1}^{-1} \circ T_{2}^{-1}\) Getting Started: To show that \(T\) is one-to-one, use the definition of a one-to-one transformation and show that \(T(\mathbf{u})=T(\mathbf{v})\) implies \(\mathbf{u}=\mathbf{v} .\) For the Nsecond statement you first need to use Theorems 6.8 and 6.12 to show that \(T\) is invertible, and then show that \(T \circ\left(T_{1}^{-1} \circ T_{2}^{-1}\right)\) and \(\left(T_{1}^{-1} \circ T_{2}^{-1}\right) \circ T\) are identity transformations.(i) \(\operatorname{Let} T(\mathbf{u})=T(\mathbf{v}) .\) Recall that \(\left(T_{2} \circ T_{1}\right)(\mathbf{v})=T_{2}\left(T_{1}(\mathbf{v})\right)\) for all vectors v. Now use the fact that \(T_{2}\) and \(T_{1}\) are one-to-one to conclude that \(\mathbf{u}=\mathbf{v}\) (ii) Use Theorems 6.8 and 6.12 to show that \(T_{1}, T_{2}\) and \(T\) are all invertible transformations. So, \(T_{1}^{-1}\) and \(T_{2}^{-1}\) exist. (iii) Form the composition \(T^{\prime}=T_{1}^{-1} \circ T_{2}^{-1} .\) It is a linear transformation from \(V\) into \(V .\) To show that it is the inverse of \(T,\) you need to determine whether the composition of \(T\) with \(T^{\prime}\) on both sides gives an identity transformation.

Proof Prove that if \(A\) and \(B\) are similar matrices and \(A\) is nonsingular, then \(B\) is also nonsingular and \(A^{-1}\) and \(B^{-1}\) are similar matrices.

Sketch the image of the unit square [a square with vertices at \((0,0),(1,0),(1,1), \text { and }(0,1)]\) under the specified transformation. \(T\) is the contraction represented by \(T(x, y)=(x / 2, y)\)
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