✅ T(x,y)=(x+2y, 3x−y) is Linear ❌ T(x,y)=xy is NOT! | Linear Algebra Proof| Linear Transformation | Example 2

Engineering Mathematica · MA25C01

Linear vs Not Linear: T(x,y)=(x+2y, 3x−y) vs T(x,y)=xy

Example 2
LINEAR ALGEBRA

Given

Show that \( T(x,y) = (x+2y,\, 3x-y) \) is linear for \( T: \mathbb{R}^2 \to \mathbb{R}^2 \), while \( T(x,y) = xy \) is not linear for \( T: \mathbb{R}^2 \to \mathbb{R}^2 \).

Drag each vector — watch one transform predictably, the other break ◉ drag to test

LINEAR T(x,y) = (x+2y, 3x−y)

u

(2.00, 1.00)

T(u)

(4.00, 5.00)

2·T(u) vs T(2u)

(8.00, 10.00) = (8.00, 10.00) ✓

NOT LINEAR T(x,y) = xy

u

(2.00, 1.00)

T(u)

2.00

2·T(u) vs T(2u)

4.00 ≠ 8.00 ✗

Part 1 — Proving T(x,y) = (x+2y, 3x−y) IS linear

Let \( \mathbf{u} = (x_1, y_1) \) and \( \mathbf{v} = (x_2, y_2) \) be vectors in \( \mathbb{R}^2 \), and let \( c \) be any scalar.

01

Additivity — T(u + v) = T(u) + T(v)

\[ \begin{aligned} \mathbf{u}+\mathbf{v} &= (x_1+x_2,\, y_1+y_2) \\[0.4em] T(\mathbf{u}+\mathbf{v}) &= \big((x_1+x_2)+2(y_1+y_2),\; 3(x_1+x_2)-(y_1+y_2)\big) \\ &= \big((x_1+2y_1)+(x_2+2y_2),\; (3x_1-y_1)+(3x_2-y_2)\big) \end{aligned} \]

\[ \begin{aligned} T(\mathbf{u}) &= (x_1+2y_1,\; 3x_1-y_1), \qquad T(\mathbf{v}) = (x_2+2y_2,\; 3x_2-y_2) \\[0.4em] T(\mathbf{u})+T(\mathbf{v}) &= \big((x_1+2y_1)+(x_2+2y_2),\; (3x_1-y_1)+(3x_2-y_2)\big) \end{aligned} \]

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

02

Homogeneity — T(cu) = c·T(u)

\[ \begin{aligned} c\mathbf{u} &= (cx_1,\, cy_1) \\[0.4em] T(c\mathbf{u}) &= \big(cx_1+2(cy_1),\; 3(cx_1)-cy_1\big) = \big(c(x_1+2y_1),\; c(3x_1-y_1)\big) \end{aligned} \]

\[ c\,T(\mathbf{u}) = c(x_1+2y_1,\; 3x_1-y_1) = \big(c(x_1+2y_1),\; c(3x_1-y_1)\big) \]

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

Both properties are satisfied — T(x,y) = (x+2y, 3x−y) is a linear transformation. ∎

Part 2 — Proving T(x,y) = xy is NOT linear

One failed check is enough to disprove linearity. Let \( \mathbf{u} = (x_1, y_1) \) and let \( c \) be any scalar.

✗

Homogeneity — T(cu) = c·T(u)

\[ \begin{aligned} c\mathbf{u} &= (cx_1,\, cy_1) \\[0.4em] T(c\mathbf{u}) &= (cx_1)(cy_1) = c^2(x_1 y_1) \end{aligned} \]

\[ c\,T(\mathbf{u}) = c(x_1 y_1) \]

No match. \( c^2(x_1y_1) \neq c(x_1y_1) \) in general (they differ whenever \( c \neq 0,1 \)), so homogeneity fails.

Homogeneity fails — T(x,y) = xy is not a linear transformation. ∎

Watch it worked out Example 2

Engineering Mathematica · Linear Algebra Notes · MA25C01

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