Engineering Mathematica · MA25C02-Example-3
F(x,y) = (ax + by, cx + dy) — The General Linear Map
LINEAR ALGEBRA
Given
Find whether \( F: \mathbb{R}^2 \to \mathbb{R}^2 \) defined by \( F(x,y) = (ax+by,\, cx+dy) \), for real constants \( a, b, c, d \), is linear.
Set a, b, c, d — then drag the vector to see F(u) update
◉ drag me
[
1
0.5
0.5
1
]
Input vector u
(3.00, 1.00)
F(u) = (ax+by, cx+dy)
(3.50, 2.50)
Let \( \mathbf{u} = (x_1, y_1) \), \( \mathbf{v} = (x_2, y_2) \) be arbitrary vectors in \( \mathbb{R}^2 \), and let \( k \) be any scalar. (Using \( k \) for the scalar to avoid clashing with the constant \( c \) in \( F \).)
01
Additivity — F(u + v) = F(u) + F(v)
\[
\begin{aligned}
\mathbf{u}+\mathbf{v} &= (x_1+x_2,\, y_1+y_2) \\[0.4em]
F(\mathbf{u}+\mathbf{v}) &= \big(a(x_1+x_2)+b(y_1+y_2),\; c(x_1+x_2)+d(y_1+y_2)\big) \\
&= \big((ax_1+by_1)+(ax_2+by_2),\; (cx_1+dy_1)+(cx_2+dy_2)\big)
\end{aligned}
\]
\[
\begin{aligned}
F(\mathbf{u}) &= (ax_1+by_1,\; cx_1+dy_1), \qquad F(\mathbf{v}) = (ax_2+by_2,\; cx_2+dy_2) \\[0.4em]
F(\mathbf{u})+F(\mathbf{v}) &= \big((ax_1+by_1)+(ax_2+by_2),\; (cx_1+dy_1)+(cx_2+dy_2)\big)
\end{aligned}
\]
Match. \( F(\mathbf{u}+\mathbf{v}) = F(\mathbf{u}) + F(\mathbf{v}) \), so additivity holds.
02
Homogeneity — F(ku) = k·F(u)
\[
\begin{aligned}
k\mathbf{u} &= (kx_1,\, ky_1) \\[0.4em]
F(k\mathbf{u}) &= \big(a(kx_1)+b(ky_1),\; c(kx_1)+d(ky_1)\big) = \big(k(ax_1+by_1),\; k(cx_1+dy_1)\big)
\end{aligned}
\]
\[
k\,F(\mathbf{u}) = k(ax_1+by_1,\; cx_1+dy_1) = \big(k(ax_1+by_1),\; k(cx_1+dy_1)\big)
\]
Match. \( F(k\mathbf{u}) = k\,F(\mathbf{u}) \), so homogeneity holds.
Both properties hold for every a, b, c, d ∈ ℝ — F is a linear transformation.
∎
Why this matters: this is the general form of every linear map from ℝ² to ℝ². The matrix \(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\) acting on \((x,y)^T\) produces exactly this F — so this proof is really showing that every 2×2 matrix transformation is linear.
Watch it worked out Example 3
Engineering Mathematica · Linear Algebra Notes · MA25C01
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