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Sagot :
To determine whether the function [tex]\( F: \mathbb{R}^2 \rightarrow \mathbb{R}^3 \)[/tex] defined by
[tex]\[ F\left(\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}\right) = \begin{pmatrix} x_1 - x_2 \\ -x_1 + x_2 \\ x_2 \end{pmatrix} \][/tex]
is a linear transformation, we need to check two properties:
1. Additivity: [tex]\( F(\mathbf{u} + \mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \)[/tex]
2. Homogeneity (Scalar multiplication): [tex]\( F(c \mathbf{v}) = c F(\mathbf{v}) \)[/tex]
Let's denote vectors [tex]\(\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}\)[/tex] and [tex]\(\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\)[/tex], and a scalar [tex]\( c \)[/tex].
### Step 1: Check Additivity
First, calculate [tex]\( \mathbf{u} + \mathbf{v} \)[/tex]:
[tex]\[ \mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix} \][/tex]
Now apply [tex]\( F \)[/tex] to [tex]\( \mathbf{u} + \mathbf{v} \)[/tex]:
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F\left(\begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix}\right) = \begin{pmatrix} (u_1 + v_1) - (u_2 + v_2) \\ -(u_1 + v_1) + (u_2 + v_2) \\ u_2 + v_2 \end{pmatrix} = \begin{pmatrix} u_1 - u_2 + v_1 - v_2 \\ -u_1 + u_2 - v_1 + v_2 \\ u_2 + v_2 \end{pmatrix} \][/tex]
Now calculate [tex]\( F(\mathbf{u}) + F(\mathbf{v}) \)[/tex]:
[tex]\[ F(\mathbf{u}) = \begin{pmatrix} u_1 - u_2 \\ -u_1 + u_2 \\ u_2 \end{pmatrix} \][/tex]
[tex]\[ F(\mathbf{v}) = \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} \][/tex]
[tex]\[ F(\mathbf{u}) + F(\mathbf{v}) = \begin{pmatrix} u_1 - u_2 \\ -u_1 + u_2 \\ u_2 \end{pmatrix} + \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 - u_2 + v_1 - v_2 \\ -u_1 + u_2 - v_1 + v_2 \\ u_2 + v_2 \end{pmatrix} \][/tex]
Clearly,
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \][/tex]
### Step 2: Check Homogeneity
Calculate [tex]\( c\mathbf{v} \)[/tex]:
[tex]\[ c\mathbf{v} = c \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} cv_1 \\ cv_2 \end{pmatrix} \][/tex]
Apply [tex]\( F \)[/tex] to [tex]\( c\mathbf{v} \)[/tex]:
[tex]\[ F(c \mathbf{v}) = F\left(\begin{pmatrix} cv_1 \\ cv_2 \end{pmatrix}\right) = \begin{pmatrix} cv_1 - cv_2 \\ -cv_1 + cv_2 \\ cv_2 \end{pmatrix} = c \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} \][/tex]
Now calculate [tex]\( c F(\mathbf{v}) \)[/tex]:
[tex]\[ c F(\mathbf{v}) = c \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} = \begin{pmatrix} cv_1 - cv_2 \\ -cv_1 + cv_2 \\ cv_2 \end{pmatrix} \][/tex]
Clearly,
[tex]\[ F(c \mathbf{v}) = c F(\mathbf{v}) \][/tex]
### Conclusion
Since both properties (additivity and homogeneity) are satisfied, the function [tex]\( F \)[/tex] is a linear transformation.
[tex]\[ F\left(\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}\right) = \begin{pmatrix} x_1 - x_2 \\ -x_1 + x_2 \\ x_2 \end{pmatrix} \][/tex]
is a linear transformation, we need to check two properties:
1. Additivity: [tex]\( F(\mathbf{u} + \mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \)[/tex]
2. Homogeneity (Scalar multiplication): [tex]\( F(c \mathbf{v}) = c F(\mathbf{v}) \)[/tex]
Let's denote vectors [tex]\(\mathbf{u} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix}\)[/tex] and [tex]\(\mathbf{v} = \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}\)[/tex], and a scalar [tex]\( c \)[/tex].
### Step 1: Check Additivity
First, calculate [tex]\( \mathbf{u} + \mathbf{v} \)[/tex]:
[tex]\[ \mathbf{u} + \mathbf{v} = \begin{pmatrix} u_1 \\ u_2 \end{pmatrix} + \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix} \][/tex]
Now apply [tex]\( F \)[/tex] to [tex]\( \mathbf{u} + \mathbf{v} \)[/tex]:
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F\left(\begin{pmatrix} u_1 + v_1 \\ u_2 + v_2 \end{pmatrix}\right) = \begin{pmatrix} (u_1 + v_1) - (u_2 + v_2) \\ -(u_1 + v_1) + (u_2 + v_2) \\ u_2 + v_2 \end{pmatrix} = \begin{pmatrix} u_1 - u_2 + v_1 - v_2 \\ -u_1 + u_2 - v_1 + v_2 \\ u_2 + v_2 \end{pmatrix} \][/tex]
Now calculate [tex]\( F(\mathbf{u}) + F(\mathbf{v}) \)[/tex]:
[tex]\[ F(\mathbf{u}) = \begin{pmatrix} u_1 - u_2 \\ -u_1 + u_2 \\ u_2 \end{pmatrix} \][/tex]
[tex]\[ F(\mathbf{v}) = \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} \][/tex]
[tex]\[ F(\mathbf{u}) + F(\mathbf{v}) = \begin{pmatrix} u_1 - u_2 \\ -u_1 + u_2 \\ u_2 \end{pmatrix} + \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} = \begin{pmatrix} u_1 - u_2 + v_1 - v_2 \\ -u_1 + u_2 - v_1 + v_2 \\ u_2 + v_2 \end{pmatrix} \][/tex]
Clearly,
[tex]\[ F(\mathbf{u} + \mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \][/tex]
### Step 2: Check Homogeneity
Calculate [tex]\( c\mathbf{v} \)[/tex]:
[tex]\[ c\mathbf{v} = c \begin{pmatrix} v_1 \\ v_2 \end{pmatrix} = \begin{pmatrix} cv_1 \\ cv_2 \end{pmatrix} \][/tex]
Apply [tex]\( F \)[/tex] to [tex]\( c\mathbf{v} \)[/tex]:
[tex]\[ F(c \mathbf{v}) = F\left(\begin{pmatrix} cv_1 \\ cv_2 \end{pmatrix}\right) = \begin{pmatrix} cv_1 - cv_2 \\ -cv_1 + cv_2 \\ cv_2 \end{pmatrix} = c \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} \][/tex]
Now calculate [tex]\( c F(\mathbf{v}) \)[/tex]:
[tex]\[ c F(\mathbf{v}) = c \begin{pmatrix} v_1 - v_2 \\ -v_1 + v_2 \\ v_2 \end{pmatrix} = \begin{pmatrix} cv_1 - cv_2 \\ -cv_1 + cv_2 \\ cv_2 \end{pmatrix} \][/tex]
Clearly,
[tex]\[ F(c \mathbf{v}) = c F(\mathbf{v}) \][/tex]
### Conclusion
Since both properties (additivity and homogeneity) are satisfied, the function [tex]\( F \)[/tex] is a linear transformation.
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