General linear equations
Linear mapping = linear transformation = linear function
Definition. Given vector spaces V1 and V2, a
mapping L : V1 → V2 is linear if
L(x + y) = L(x) + L(y),
L(rx) = rL(x)
for any x, y ∈ V1 and r ∈ R.
Basic properties of linear mappings:
• L(r1v1 + · · · + rkvk) = r1L(v1) + · · · + rkL(vk)
for all k ≥ 1, v1, . . . , vk ∈ V1, and r1, . . . , rk ∈ R.
• L(01) = 02, where 01 and 02 are zero vectors in
V1 and V2, respectively.
• L(−v) = −L(v) for any v ∈ V1.
Range and kernel
Let V,W be vector spaces and L : V → W be a
Definition. The range (or image) of L is the set
of all vectors w ∈ W such that w = L(v) for some
v ∈ V. The range of L is denoted L(V).
The kernel of L, denoted ker L, is the set of all
vectors v ∈ V such that L(v) = 0.
Theorem (i) The range of L is a subspace of W.
(ii) The kernel of L is a subspace of V.
General linear equations
Definition. A linear equation is an equation of the form
L(x) = b,
where L : V → W is a linear mapping, b is a given vector
from W, and x is an unknown vector from V.
The range of L is the set of all vectors b ∈ W such that the
equation L(x) = b has a solution.
The kernel of L is the solution set of the homogeneous linear
equation L(x) = 0.
Theorem If the linear equation L(x) = b is solvable then the
general solution is
x0 + t1v1 + · · · + tkvk ,
where x0 is a particular solution, v1, . . . , vk is a basis for the
kernel of L, and t1, . . . , tk are arbitrary scalars.
Linear equation: L(x) = b, where
Linear operator L :
Linear equation: Lu = b, where b(x) = e2x .
It can be shown that the range of L is the entire
space C(R) while the kernel of L is spanned by the
functions sin x and cos x.
By linearity, is a particular solution.
Thus the general solution is
Any m×n matrix A gives rise to a transformation
L : Rn → Rm given by L(x) = Ax, where x ∈ Rn
and L(x) ∈ Rm are regarded as column vectors.
This transformation is linear.
Indeed, L(x + y) = A(x + y) = Ax + Ay = L(x) + L(y),
L(rx) = A(rx) = r (Ax) = rL(x).
Let e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) be the
standard basis for R3. We have that L(e1) = (1, 3, 0),
L(e2) = (0, 4, 5), L(e3) = (2, 7, 8). Thus L(e1), L(e2), L(e3)
are columns of the matrix.
Problem. Find a linear mapping L : R3 → R2
such that L(e1) = (1, 1), L(e2) = (0,−2),
L(e2) = (3, 0), where e1, e2, e3 is the standard
basis for R3.
Columns of the matrix are vectors L(e1), L(e2), L(e3).
Theorem Suppose L : Rn → Rm is a linear map. Then
there exists an m×n matrix A such that L(x) = Ax for all
x ∈ Rn. Columns of A are vectors L(e1), L(e2), . . . , L(en),
where e1, e2, . . . , en is the standard basis for Rn.
Linear transformations of R2
Any linear mapping f : R2 → R2 is represented as
multiplication of a 2-dimensional column vector by a
2×2 matrix: f (x) = Ax or
Linear transformations corresponding to particular
matrices can have various geometric properties.
|Rotation by 90o|
|Rotation by 45o|
the vertical axis
the line x − y = 0
|Vertical projection on
the horizontal axis
on the line x + y = 0