Chapter 6: Problem 3
Let \(T: \mathscr{P}_{2} \rightarrow \mathbb{R}^{2}\) be the linear transformation defined by \\[ T\left(a+b x+c x^{2}\right)=\left[\begin{array}{l} a-b \\ b+c \end{array}\right] \\] (a) Which, if any, of the following polynomials are in \(\operatorname{ker}(T) ?\) (i) \(1+x\) (ii) \(x-x^{2}\) (iii) \(1+x-x^{2}\) (b) Which, if any, of the following vectors are in \(\operatorname{range}(T) ?\) (i) \(\left[\begin{array}{l}0 \\ 0\end{array}\right]\) (ii) \(\left[\begin{array}{l}1 \\ 0\end{array}\right]\) (iii) \(\left[\begin{array}{l}0 \\ 1\end{array}\right]\) (c) Describe ker \((T)\) and range \((T)\)
Short Answer
Step by step solution
Define Transformation
Finding Ker(T)
Checking Ker(T) Candidates
Finding Range(T)
Checking Range(T) Candidates
Describing Ker(T) and Range(T)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kernel of a Linear Transformation
- \( a - b = 0 \)
- \( b + c = 0 \)
Range of a Linear Transformation
- Selecting \( a = 0, b = 0, c = 0 \) yields \( \begin{bmatrix} 0 \ 0 \end{bmatrix} \).
- Choosing \( a = 1, b = 1, c = 0 \) results in \( \begin{bmatrix} 0 \ 1 \end{bmatrix} \).
- With \( a = 1, b = 0, c = -1 \), we get \( \begin{bmatrix} 1 \ -1 \end{bmatrix} \).
Polynomial Space
Vector Space
- Closure under addition: Sum of two vectors results in a vector within the space.
- Closure under scalar multiplication: Multiplying a vector by a scalar produces a vector within the space.
- Presence of zero vector: A vector exists that, when added to any vector, returns the same vector.
- Existence of additive inverses: Each vector has a corresponding negative vector that sums to the zero vector.
Systems of Equations
- \( a - b = 0 \)
- \( b + c = 0 \)