📐 Prove T(x,y,z)=(z, x+y) is Linear | ℝ³ → ℝ² Transformation | Linear Transformation | Example 4

Engineering Mathematica · MA25C01 Example 4

T(x,y,z) = (z, x+y) — Linear Map ℝ³ → ℝ²

Example 4
LINEAR ALGEBRA

Given

Show that the transformation \( T: \mathbb{R}^3 \to \mathbb{R}^2 \) defined by \( T(x,y,z) = (z,\, x+y) \) is linear.

A/M-24-R-21

Move x, y, z — watch T(x,y,z) land in ℝ² ◉ 3 inputs → 2 outputs

x2

y1

z3

Input (x, y, z) ∈ ℝ³

(2.00, 1.00, 3.00)

T(x,y,z) = (z, x+y) ∈ ℝ²

(3.00, 3.00)

Let \( \mathbf{u} = (x_1, y_1, z_1) \) and \( \mathbf{v} = (x_2, y_2, z_2) \) be vectors in \( \mathbb{R}^3 \), and let \( c \) be any scalar.

01

Additivity — T(u + v) = T(u) + T(v)

\[ \begin{aligned} \mathbf{u}+\mathbf{v} &= (x_1+x_2,\, y_1+y_2,\, z_1+z_2) \\[0.4em] T(\mathbf{u}+\mathbf{v}) &= \big(z_1+z_2,\; (x_1+x_2)+(y_1+y_2)\big) \\ &= \big(z_1+z_2,\; (x_1+y_1)+(x_2+y_2)\big) \end{aligned} \]

\[ \begin{aligned} T(\mathbf{u}) &= (z_1,\; x_1+y_1), \qquad T(\mathbf{v}) = (z_2,\; x_2+y_2) \\[0.4em] T(\mathbf{u})+T(\mathbf{v}) &= \big(z_1+z_2,\; (x_1+y_1)+(x_2+y_2)\big) \end{aligned} \]

Match. \( T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \), so additivity holds.

02

Homogeneity — T(cu) = c·T(u)

\[ \begin{aligned} c\mathbf{u} &= (cx_1,\, cy_1,\, cz_1) \\[0.4em] T(c\mathbf{u}) &= \big(cz_1,\; cx_1+cy_1\big) = \big(cz_1,\; c(x_1+y_1)\big) \end{aligned} \]

\[ c\,T(\mathbf{u}) = c(z_1,\; x_1+y_1) = \big(cz_1,\; c(x_1+y_1)\big) \]

Match. \( T(c\mathbf{u}) = c\,T(\mathbf{u}) \), so homogeneity holds.

Both properties are satisfied — T(x,y,z) = (z, x+y) is a linear transformation. ∎

Why this works: each output coordinate — z and x+y — is just a sum of the inputs with coefficient 1, no squares, products, or added constants. That pattern is what makes a map from ℝⁿ to ℝᵐ linear on sight.

Watch it worked out Example 4

Engineering Mathematica · Linear Algebra Notes · MA25C01

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