To show that a mapping \( T: \mathbb{R}^3 \to \mathbb{R}^2 \) defined by \( T(x, y, z) = (2x + y + z, 3x + 2y + 4z) \) is linear, we need to verify two properties:
1. Additivity: ( T({u} +{v}) = T(\mathbf{u}) + T(\mathbf{v}) \)
2. Homogeneity of degree 1: \( T(a \mathbf{u}) = a T(\mathbf{u}) \)
Where \( \mathbf{u} \) and \( \mathbf{v} \) are arbitrary vectors in \( \mathbb{R}^3 \) and \( a \) is a scalar.
Let's proceed to verify these properties:
1. **Additivity:**
Consider two vectors \( \mathbf{u} = (x_1, y_1, z_1) \) and \( \mathbf{v} = (x_2, y_2, z_2) \) in \( \mathbb{R}^3 \). We need to show that:
\[ T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \]
Let's calculate each side separately:
Left-hand side:
\[ T(\mathbf{u} + \mathbf{v}) = T(x_1 + x_2, y_1 + y_2, z_1 + z_2) \]
\[ = (2(x_1 + x_2) + (y_1 + y_2) + (z_1 + z_2), 3(x_1 + x_2) + 2(y_1 + y_2) + 4(z_1 + z_2)) \]
\[ = (2x_1 + y_1 + z_1 + 2x_2 + y_2 + z_2, 3x_1 + 2y_1 + 4z_1 + 3x_2 + 2y_2 + 4z_2) \]
Right-hand side:
\[ T(\mathbf{u}) + T(\mathbf{v}) = (2x_1 + y_1 + z_1, 3x_1 + 2y_1 + 4z_1) + (2x_2 + y_2 + z_2, 3x_2 + 2y_2 + 4z_2) \]
\[ = (2x_1 + y_1 + z_1 + 2x_2 + y_2 + z_2, 3x_1 + 2y_1 + 4z_1 + 3x_2 + 2y_2 + 4z_2) \]
As we can see, the left-hand side is equal to the right-hand side, which means that the additivity property holds.
2. **Homogeneity of Degree 1:**
Now, let's verify the homogeneity property for a scalar \( a \) and a vector \( \mathbf{u} = (x, y, z) \):
\[ T(a \mathbf{u}) = T(ax, ay, az) \]
\[ = (2(ax) + (ay) + (az), 3(ax) + 2(ay) + 4(az)) \]
\[ = (a(2x + y + z), a(3x + 2y + 4z)) \]
\[ = a(2x + y + z, 3x + 2y + 4z) \]
\[ = a T(x, y, z) \]
The homogeneity property is satisfied.
Since both additivity and homogeneity of degree 1 properties hold for the mapping \( T \), we can conclude that the mapping \( T \) is linear from \( \mathbb{R}^3 \) to \( \mathbb{R}^2 \).
