Question 1165604: Consider the following functions T : R → R . Show that each is a linear transformation and determine for each the matrix A such that T (x) = A x
a) x1. x1+2x2 +3x3
T x2. = 2x2-3x1+x3
¶ x3
b) x1. 7x1+2x2 +x3
T x2. = 3x1-11x2+2x3
¶ x3
Answer by CPhill(2138)   (Show Source):
You can put this solution on YOUR website! This question asks you to demonstrate that the given transformations $T: \mathbb{R}^3 \to \mathbb{R}^3$ are linear and to find the standard matrix $A$ for each, such that $T(\mathbf{x}) = A\mathbf{x}$.
## General Definition of a Linear Transformation
A transformation $T: V \to W$ is linear if, for all vectors $\mathbf{u}, \mathbf{v}$ in $V$ and all scalars $c$ in $\mathbb{R}$:
1. **Additivity:** $T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$
2. **Homogeneity:** $T(c\mathbf{u}) = cT(\mathbf{u})$
---
## Part (a)
The transformation is defined by:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 + 2x_2 + 3x_3 \\ -3x_1 + 2x_2 + x_3 \\ 0 \end{pmatrix}$$
*(Note: I assumed the third component is $0$ as the input has three components but the output only shows two distinct linear expressions, and a $3 \times 3$ matrix is expected for $T: \mathbb{R}^3 \to \mathbb{R}^3$. I'll proceed by assuming the third component is a constant like 0 or $x_1$ from the format)*
**Based on the layout, it is most likely that the output is meant to be a $3 \times 1$ column vector. Given the formatting ambiguity, I will use the coefficients as presented, and assume the third component is 0 for the linear combination interpretation.**
Let's re-write the output as a $3 \times 1$ vector where the structure dictates the coefficients:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} x_1 + 2x_2 + 3x_3 \\ -3x_1 + 2x_2 + x_3 \\ 0 \end{pmatrix}$$
### A. Finding the Standard Matrix $A$
The standard matrix $A$ is formed by applying $T$ to the standard basis vectors $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$:
$$A = [T(\mathbf{e}_1) \mid T(\mathbf{e}_2) \mid T(\mathbf{e}_3)]$$
1. $$T\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 1(1) + 2(0) + 3(0) \\ -3(1) + 2(0) + 1(0) \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}$$
2. $$T\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1(0) + 2(1) + 3(0) \\ -3(0) + 2(1) + 1(0) \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 2 \\ 0 \end{pmatrix}$$
3. $$T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1(0) + 2(0) + 3(1) \\ -3(0) + 2(0) + 1(1) \\ 0 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ 0 \end{pmatrix}$$
The standard matrix $A$ is:
$$\mathbf{A = \begin{pmatrix} 1 & 2 & 3 \\ -3 & 2 & 1 \\ 0 & 0 & 0 \end{pmatrix}}$$
### B. Proving Linearity
Since the components of $T(\mathbf{x})$ are **linear homogeneous polynomials** (polynomials where every term has degree one) in the variables $x_1, x_2, x_3$ and there is no constant term, $T$ is automatically a linear transformation.
---
## Part (b)
The transformation is defined by:
$$T \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 7x_1 + 2x_2 + x_3 \\ 3x_1 - 11x_2 + 2x_3 \end{pmatrix}$$
*(Note: In this case, $T: \mathbb{R}^3 \to \mathbb{R}^2$, as the output vector has 2 components.)*
### A. Finding the Standard Matrix $A$
The standard matrix $A$ is a $2 \times 3$ matrix formed by applying $T$ to the standard basis vectors $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$:
$$A = [T(\mathbf{e}_1) \mid T(\mathbf{e}_2) \mid T(\mathbf{e}_3)]$$
1. $$T\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 7(1) + 2(0) + 1(0) \\ 3(1) - 11(0) + 2(0) \end{pmatrix} = \begin{pmatrix} 7 \\ 3 \end{pmatrix}$$
2. $$T\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 7(0) + 2(1) + 1(0) \\ 3(0) - 11(1) + 2(0) \end{pmatrix} = \begin{pmatrix} 2 \\ -11 \end{pmatrix}$$
3. $$T\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 7(0) + 2(0) + 1(1) \\ 3(0) - 11(0) + 2(1) \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$
The standard matrix $A$ is:
$$\mathbf{A = \begin{pmatrix} 7 & 2 & 1 \\ 3 & -11 & 2 \end{pmatrix}}$$
### B. Proving Linearity
Since the components of $T(\mathbf{x})$ are again **linear homogeneous polynomials** in $x_1, x_2, x_3$ (no terms with degree other than one and no constant terms), $T$ is a linear transformation.
**Formal Proof for Additivity (Homogeneity is similar):**
Let $\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \\ u_3 \end{pmatrix}$ and $\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix}$.
$$T(\mathbf{u} + \mathbf{v}) = T \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \\ u_3 + v_3 \end{pmatrix}$$
$$T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} 7(u_1 + v_1) + 2(u_2 + v_2) + (u_3 + v_3) \\ 3(u_1 + v_1) - 11(u_2 + v_2) + 2(u_3 + v_3) \end{pmatrix}$$
Rearranging the terms based on $\mathbf{u}$ and $\mathbf{v}$:
$$T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} (7u_1 + 2u_2 + u_3) + (7v_1 + 2v_2 + v_3) \\ (3u_1 - 11u_2 + 2u_3) + (3v_1 - 11v_2 + 2v_3) \end{pmatrix}$$
$$T(\mathbf{u} + \mathbf{v}) = \begin{pmatrix} 7u_1 + 2u_2 + u_3 \\ 3u_1 - 11u_2 + 2u_3 \end{pmatrix} + \begin{pmatrix} 7v_1 + 2v_2 + v_3 \\ 3v_1 - 11v_2 + 2v_3 \end{pmatrix} = T(\mathbf{u}) + T(\mathbf{v})$$
Thus, $T$ is linear.
|
|
|
