Question

In each case, assume that \(T\) is a linear transformation. a. If \(T: V \rightarrow \mathbb{R}\) and \(T\left(\mathbf{v}_{1}\right)=1, T\left(\mathbf{v}_{2}\right)=-1,\) find \(T\left(3 \mathbf{v}_{1}-5 \mathbf{v}_{2}\right)\) b. If \(T: V \rightarrow \mathbb{R}\) and \(T\left(\mathbf{v}_{1}\right)=2, T\left(\mathbf{v}_{2}\right)=-3,\) find \(T\left(3 \mathbf{v}_{1}+2 \mathbf{v}_{2}\right)\) c. If \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) and \(T\left[\begin{array}{l}1 \\ 3\end{array}\right]=\left[\begin{array}{l}1 \\\ 1\end{array}\right]\), \(T\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}0 \\\ 1\end{array}\right],\) find \(T\left[\begin{array}{r}-1 \\ 3\end{array}\right]\) d. If \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) and \(T\left[\begin{array}{r}1 \\ -1\end{array}\right]=\left[\begin{array}{l}0 \\\ 1\end{array}\right]\), \(T\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}1 \\\ 0\end{array}\right],\) find \(T\left[\begin{array}{r}1 \\ -7\end{array}\right]\) e. If \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) and \(T(x+1)=x, T(x-1)=1,\) \(T\left(x^{2}\right)=0,\) find \(T\left(2+3 x-x^{2}\right)\) f. If \(T: \mathbf{P}_{2} \rightarrow \mathbb{R}\) and \(T(x+2)=1, T(1)=5\), \(T\left(x^{2}+x\right)=0,\) find \(T\left(2-x+3 x^{2}\right)\)

Short answer

a. 8, b. 0, c. \([1, -1]^T\), d. \([-4, 3]^T\), e. \(2x + 1\), f. 9.

Step-by-step solution

Problem Statement for Part a

Given a linear transformation \(T: V \rightarrow \mathbb{R}\) with \(T(\mathbf{v}_1) = 1\) and \(T(\mathbf{v}_2) = -1\), we need to find \(T(3 \mathbf{v}_1 - 5 \mathbf{v}_2)\).

Apply Linearity to Part a

Using the linearity of \(T\), we have: \[T(3 \mathbf{v}_1 - 5 \mathbf{v}_2) = 3T(\mathbf{v}_1) - 5T(\mathbf{v}_2) \]Substitute the given values: \[T(3 \mathbf{v}_1 - 5 \mathbf{v}_2) = 3(1) - 5(-1) = 3 + 5 = 8\] Thus, \(T(3 \mathbf{v}_1 - 5 \mathbf{v}_2) = 8\).

Problem Statement for Part b

Given a linear transformation \(T: V \rightarrow \mathbb{R}\) with \(T(\mathbf{v}_1) = 2\) and \(T(\mathbf{v}_2) = -3\), find \(T(3 \mathbf{v}_1 + 2 \mathbf{v}_2)\).

Apply Linearity to Part b

Using the linearity of \(T\), we have: \[T(3 \mathbf{v}_1 + 2 \mathbf{v}_2) = 3T(\mathbf{v}_1) + 2T(\mathbf{v}_2)\]Substitute the given values: \[T(3 \mathbf{v}_1 + 2 \mathbf{v}_2) = 3(2) + 2(-3) = 6 - 6 = 0\] Thus, \(T(3 \mathbf{v}_1 + 2 \mathbf{v}_2) = 0\).

Problem Statement for Part c

Given a linear transformation \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) with \(T\left[\begin{array}{c}1 \ 3\end{array}\right] = \left[\begin{array}{c}1 \ 1\end{array}\right]\) and \(T\left[\begin{array}{c}1 \ 1\end{array}\right] = \left[\begin{array}{c}0 \ 1\end{array}\right]\), find \(T\left[\begin{array}{r}-1 \ 3\end{array}\right]\).

Express Vector in Terms of Known Vectors for Part c

Write \(\left[\begin{array}{c}-1 \ 3\end{array}\right]\) as a linear combination of \(\left[\begin{array}{c}1 \ 3\end{array}\right]\) and \(\left[\begin{array}{c}1 \ 1\end{array}\right]\): \[\left[\begin{array}{c}-1 \ 3\end{array}\right] = -2 \left[\begin{array}{c}1 \ 1\end{array}\right] + \left[\begin{array}{c}1 \ 3\end{array}\right]\] Then, apply \(T\) using linearity: \[T\left[-2 \left[\begin{array}{c}1 \ 1\end{array}\right] + \left[\begin{array}{c}1 \ 3\end{array}\right]\right] = -2T\left[\begin{array}{c}1 \ 1\end{array}\right] + T\left[\begin{array}{c}1 \ 3\end{array}\right]\] Substitute the known values: \[-2 \left[\begin{array}{c}0 \ 1\end{array}\right] + \left[\begin{array}{c}1 \ 1\end{array}\right] = \left[\begin{array}{r}1 \ -1\end{array}\right]\] Thus, \(T\left[\begin{array}{r}-1 \ 3\end{array}\right] = \left[\begin{array}{r}1 \ -1\end{array}\right]\).

Problem Statement for Part d

Given a linear transformation \(T: \mathbb{R}^2 \rightarrow \mathbb{R}^2\) with \(T\left[\begin{array}{c}1 \ -1\end{array}\right] = \left[\begin{array}{c}0 \ 1\end{array}\right]\) and \(T\left[\begin{array}{c}1 \ 1\end{array}\right] = \left[\begin{array}{c}1 \ 0\end{array}\right]\), find \(T\left[\begin{array}{r}1 \ -7\end{array}\right]\).

Express Vector in Terms of Known Vectors for Part d

Write \(\left[\begin{array}{r}1 \ -7\end{array}\right]\) as a linear combination of \(\left[\begin{array}{c}1 \ -1\end{array}\right]\) and \(\left[\begin{array}{c}1 \ 1\end{array}\right]\): \[\left[\begin{array}{r}1 \ -7\end{array}\right] = 3\left[\begin{array}{c}1 \ -1\end{array}\right] - 4\left[\begin{array}{c}1 \ 1\end{array}\right]\] Apply \(T\) using linearity: \[T\left[3 \left[\begin{array}{c}1 \ -1\end{array}\right] - 4 \left[\begin{array}{c}1 \ 1\end{array}\right]\right] = 3T\left[\begin{array}{c}1 \ -1\end{array}\right] - 4T\left[\begin{array}{c}1 \ 1\end{array}\right]\] Substitute the known values: \[3 \left[\begin{array}{c}0 \ 1\end{array}\right] - 4 \left[\begin{array}{c}1 \ 0\end{array}\right] = \left[\begin{array}{r}-4 \ 3\end{array}\right]\] Thus, \(T\left[\begin{array}{r}1 \ -7\end{array}\right] = \left[\begin{array}{r}-4 \ 3\end{array}\right]\).

Problem Statement for Part e

Given a linear transformation \(T: \mathbf{P}_2 \rightarrow \mathbf{P}_2\) with \(T(x+1) = x\), \(T(x-1) = 1\), and \(T(x^2) = 0\), find \(T(2 + 3x - x^2)\).

Express Polynomial in Terms of Known Basis for Part e

Express \(2 + 3x - x^2\) as a sum of the known polynomials: \[2 + 3x - x^2 = 2(x+1) + (3-2)(x-1) - x^2 \] Apply \(T\) using linearity: \[T(2 + 3x - x^2) = 2T(x+1) + (3-2)T(x-1) - T(x^2)\] Substitute the transformations: \[2(x) + (1)(1) - 0 = 2x + 1\] Thus, \(T(2 + 3x - x^2) = 2x + 1\).

Problem Statement for Part f

Given a linear transformation \(T: \mathbf{P}_2 \rightarrow \mathbb{R}\) with \(T(x+2) = 1\), \(T(1) = 5\), and \(T(x^2 + x) = 0\), find \(T(2 - x + 3x^2)\).

Express Polynomial in Terms of Known Basis for Part f

Express \(2 - x + 3x^2\) in terms of known values: \[2 - x + 3x^2 = 2(1) - (x+2) + 3(x^2 + x)\] Apply \(T\) using linearity: \[T(2 - x + 3x^2) = 2T(1) - T(x+2) + 3T(x^2 + x)\] Substitute the transformations: \[2(5) - 1 + 3(0) = 10 - 1 = 9\] Thus, \(T(2 - x + 3x^2) = 9\).

Key concepts

Vector Spaces

Vector spaces are fundamental structures in linear algebra. A vector space consists of a set of vectors, along with two operations: vector addition and scalar multiplication. These operations must satisfy specific properties:

• Closure under addition: Adding any two vectors results in another vector within the same space.
• Closure under scalar multiplication: Multiplying a vector by a scalar results in another vector within the same space.
• Existence of zero vector: A unique vector, usually denoted as \(\mathbf{0}\), acts as an additive identity.
• Existence of additive inverses: For every vector \(\mathbf{v}\), there exists a vector \(-\mathbf{v}\) such that \(\mathbf{v} + (-\mathbf{v}) = \mathbf{0}\).
• Associative and commutative properties: Vector addition and scalar multiplication need to follow associative and commutative rules.

Understanding these properties helps you understand how linear transformations operate within vector spaces and ensures all modifications remain consistent with the structure of the space.

Polynomial Transformation

Polynomial transformations involve applying a linear transformation to polynomial expressions. Consider a transformation \(T: \mathbf{P}_2 \rightarrow \mathbf{P}_2\), where \(\mathbf{P}_2\) represents the vector space of polynomials of degree at most 2. In this context, each polynomial is transformed while maintaining the polynomial structure.

• Basis for polynomials: Common basis elements for \(\mathbf{P}_2\) include \(1, x, x^2\).
• Transformation properties: Applying transformations to these basis elements through known relations simplifies more complex expressions. For example, if \(T(x)\) or \(T(1)\) is known, it aids in computing transformations for larger polynomials.
• Linearity applications: Transform a polynomial like \(2 + 3x - x^2\) by expressing it as a combination of basis elements such as \(2(1) + 3x - x^2\), then applying \(T\) to each.

By understanding the polynomial basis and transformation rules, complex polynomial transformations become straightforward and manageable.

Linearity of Transformations

The linearity property of transformations is vital in simplifying calculations. A transformation \(T\) is linear if it satisfies two essential rules for any vectors \(\mathbf{u}, \mathbf{v}\) and scalars \(a, b\):

• Additivity: \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\).
• Homogeneity of degree 1: \(T(a\mathbf{u}) = aT(\mathbf{u})\).

These rules allow you to break down transformations into simpler parts, transform each segment separately, and recombine them for the final result.
For example, if \(T(\mathbf{v}_1)\) and \(T(\mathbf{v}_2)\) are known, you can find \(T(3\mathbf{v}_1 - 5\mathbf{v}_2)\) using linear combinations: \[ T(3\mathbf{v}_1 - 5\mathbf{v}_2) = 3T(\mathbf{v}_1) - 5T(\mathbf{v}_2) \] This method highlights the power of linearity through the combination of smaller parts into comprehensible solutions. Embracing these properties enables you to tackle more complex problems with confidence.