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To determine whether the function [tex]\( F \)[/tex] is a linear transformation, we must check whether [tex]\( F \)[/tex] satisfies the two main properties of linearity: preservation of addition and preservation of scalar multiplication.
Given: [tex]\( F: \mathbb{R}^2 \rightarrow \mathbb{R} \)[/tex] defined by [tex]\( F\left(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\right) = |x_1| + |x_2| \)[/tex]:
### Preservation of Addition
A function [tex]\( F \)[/tex] preserves addition if:
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \][/tex]
Let's take two arbitrary vectors [tex]\(\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \)[/tex] and [tex]\(\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \)[/tex] in [tex]\(\mathbb{R}^2\)[/tex].
First, compute [tex]\( F(\mathbf{u} + \mathbf{v}) \)[/tex]:
[tex]\[ \mathbf{u} + \mathbf{v} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix} \][/tex]
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F\left(\begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix}\right) = |u_1 + v_1| + |u_2 + v_2| \][/tex]
Now, compute [tex]\( F(\mathbf{u}) + F(\mathbf{v}) \)[/tex]:
[tex]\[ F(\mathbf{u}) = F\left(\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}\right) = |u_1| + |u_2| \][/tex]
[tex]\[ F(\mathbf{v}) = F\left(\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}\right) = |v_1| + |v_2| \][/tex]
[tex]\[ F(\mathbf{u}) + F(\mathbf{v}) = (|u_1| + |u_2|) + (|v_1| + |v_2|) = |u_1| + |u_2| + |v_1| + |v_2| \][/tex]
For [tex]\( F \)[/tex] to be a linear transformation, the following must hold:
[tex]\[ |u_1 + v_1| + |u_2 + v_2| = |u_1| + |u_2| + |v_1| + |v_2| \][/tex]
However, this is generally false because the absolute values of sums do not necessarily equal the sums of absolute values; for example, if [tex]\( u_1 = 1 \)[/tex] and [tex]\( v_1 = -1\)[/tex]:
[tex]\[ |1 + (-1)| = |0| = 0 \][/tex]
but,
[tex]\[ |1| + |-1| = 1 + 1 = 2 \][/tex]
This example shows that [tex]\( |u_1 + v_1| \neq |u_1| + |v_1| \)[/tex]. Thus, [tex]\( F \)[/tex] does not preserve addition.
### Preservation of Scalar Multiplication
A function [tex]\( F \)[/tex] preserves scalar multiplication if:
[tex]\[ F(c\mathbf{u}) = cF(\mathbf{u}) \][/tex]
For a scalar [tex]\( c \)[/tex] and a vector [tex]\(\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \)[/tex]:
First, compute [tex]\( F(c\mathbf{u}) \)[/tex]:
[tex]\[ c\mathbf{u} = c \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} cu_1 \\ cu_2 \end{bmatrix} \][/tex]
[tex]\[ F(c\mathbf{u}) = F\left(\begin{bmatrix} cu_1 \\ cu_2 \end{bmatrix}\right) = |cu_1| + |cu_2| \][/tex]
Now, compute [tex]\( cF(\mathbf{u}) \)[/tex]:
[tex]\[ cF(\mathbf{u}) = c (|u_1| + |u_2|) = c|u_1| + c|u_2| \][/tex]
For [tex]\( F \)[/tex] to be a linear transformation, the following must hold:
[tex]\[ |cu_1| + |cu_2| = c|u_1| + c|u_2| \][/tex]
This holds when [tex]\( c \)[/tex] is positive. However, if [tex]\( c \)[/tex] is negative, [tex]\( c|u_1| + c|u_2| \)[/tex] will be negative while [tex]\( |c u_1| + |c u_2| \)[/tex] remains positive, which shows that the scalar multiplication property is not generally satisfied for negative scalars.
Given that [tex]\( F \)[/tex] does not preserve both addition and scalar multiplication generally, we can conclude that:
[tex]\[ F \text{ is not a linear transformation.} \][/tex]
Thus, the function [tex]\( F \)[/tex] is not a linear transformation.
Given: [tex]\( F: \mathbb{R}^2 \rightarrow \mathbb{R} \)[/tex] defined by [tex]\( F\left(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\right) = |x_1| + |x_2| \)[/tex]:
### Preservation of Addition
A function [tex]\( F \)[/tex] preserves addition if:
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \][/tex]
Let's take two arbitrary vectors [tex]\(\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \)[/tex] and [tex]\(\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} \)[/tex] in [tex]\(\mathbb{R}^2\)[/tex].
First, compute [tex]\( F(\mathbf{u} + \mathbf{v}) \)[/tex]:
[tex]\[ \mathbf{u} + \mathbf{v} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} + \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix} \][/tex]
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F\left(\begin{bmatrix} u_1 + v_1 \\ u_2 + v_2 \end{bmatrix}\right) = |u_1 + v_1| + |u_2 + v_2| \][/tex]
Now, compute [tex]\( F(\mathbf{u}) + F(\mathbf{v}) \)[/tex]:
[tex]\[ F(\mathbf{u}) = F\left(\begin{bmatrix} u_1 \\ u_2 \end{bmatrix}\right) = |u_1| + |u_2| \][/tex]
[tex]\[ F(\mathbf{v}) = F\left(\begin{bmatrix} v_1 \\ v_2 \end{bmatrix}\right) = |v_1| + |v_2| \][/tex]
[tex]\[ F(\mathbf{u}) + F(\mathbf{v}) = (|u_1| + |u_2|) + (|v_1| + |v_2|) = |u_1| + |u_2| + |v_1| + |v_2| \][/tex]
For [tex]\( F \)[/tex] to be a linear transformation, the following must hold:
[tex]\[ |u_1 + v_1| + |u_2 + v_2| = |u_1| + |u_2| + |v_1| + |v_2| \][/tex]
However, this is generally false because the absolute values of sums do not necessarily equal the sums of absolute values; for example, if [tex]\( u_1 = 1 \)[/tex] and [tex]\( v_1 = -1\)[/tex]:
[tex]\[ |1 + (-1)| = |0| = 0 \][/tex]
but,
[tex]\[ |1| + |-1| = 1 + 1 = 2 \][/tex]
This example shows that [tex]\( |u_1 + v_1| \neq |u_1| + |v_1| \)[/tex]. Thus, [tex]\( F \)[/tex] does not preserve addition.
### Preservation of Scalar Multiplication
A function [tex]\( F \)[/tex] preserves scalar multiplication if:
[tex]\[ F(c\mathbf{u}) = cF(\mathbf{u}) \][/tex]
For a scalar [tex]\( c \)[/tex] and a vector [tex]\(\mathbf{u} = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \)[/tex]:
First, compute [tex]\( F(c\mathbf{u}) \)[/tex]:
[tex]\[ c\mathbf{u} = c \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} cu_1 \\ cu_2 \end{bmatrix} \][/tex]
[tex]\[ F(c\mathbf{u}) = F\left(\begin{bmatrix} cu_1 \\ cu_2 \end{bmatrix}\right) = |cu_1| + |cu_2| \][/tex]
Now, compute [tex]\( cF(\mathbf{u}) \)[/tex]:
[tex]\[ cF(\mathbf{u}) = c (|u_1| + |u_2|) = c|u_1| + c|u_2| \][/tex]
For [tex]\( F \)[/tex] to be a linear transformation, the following must hold:
[tex]\[ |cu_1| + |cu_2| = c|u_1| + c|u_2| \][/tex]
This holds when [tex]\( c \)[/tex] is positive. However, if [tex]\( c \)[/tex] is negative, [tex]\( c|u_1| + c|u_2| \)[/tex] will be negative while [tex]\( |c u_1| + |c u_2| \)[/tex] remains positive, which shows that the scalar multiplication property is not generally satisfied for negative scalars.
Given that [tex]\( F \)[/tex] does not preserve both addition and scalar multiplication generally, we can conclude that:
[tex]\[ F \text{ is not a linear transformation.} \][/tex]
Thus, the function [tex]\( F \)[/tex] is not a linear transformation.
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