Linear Transformation

3. Linear transformation

The following example should convince you that the expression Ax is linear:

A(x + y) = Ax + Ay, A(cx) = cAx.

A more general proof can be found here.

Example Let

 A = [ a b ] c d

The corresponding linear transformation is

 T[ x1 ] = [ ax1 + bx2 ] : R2 → R2 x2 cx1 + dx2

We have

 T[ x1 + y1 ] = [ a(x1 + y1) + b(x2 + y2) ] = [ ax1 + bx2 ] + [ ay1 + by2 ] = T[ x1 ] + T[ y1 ] x2 + y2 c(x1 + y1) + d(x2 + y2) cx1 + dx2 cy1 + dy2 x2 y2

and (we use k instead of c as the scalar to avoid confusion in notations)

 T[ kx1 ] = [ akx1 + bkx2 ] = [ k(ax1 + bx2) ] = k[ ay1 + by2 ] = kT[ x1 ] kx2 ckx1 + dkx2 k(cx1 + dx2) cy1 + dy2 x2

You are strongly encouraged to verify the linearity for matrices of other sizes.

The discussion suggests us to introduce the following concept.

A transformation T: RnRm is a linear transformation if it satisfies

T(u + v) = T(u) + T(v), T(cu) = cT(u).

By taking c = 0 in the definition, we have

T(0) = 0.

Moreover, by combining the two properties in the definition, we see that linear transformations preserve linear combinations.

T(c1u1 + c2u2 + ... + ckuk) = c1T(u1) + c2T(u2) + ... + ckT(uk).

Example Suppose T: R2R2 is a linear transformation. Suppose we also know (recall e1 = (1, 0), e2 = (0, 1))

T(e1) = (1, 2), T(e2) = (3, 4)

Then by

(5, -3) = 5 (1, 0) - 3(0,1) = 5e1 - 3e2,

(-2, 4) = -2e1 + 4e2,

we have

T(5, -3) = 5T(e1) - 3T(e2) = 5(1, 2) - 3(3, 4) = (-4, -2),

T(-2, 4) = -2T(e1) + 4T(e2) = -2(1, 2) + 4(3, 4) = (10, 12).

In general, by

(x1, x2) = x1e1 + x2e2,

we have

T(x1, x2) = x1T(e1) + x2T(e2) = x1(1, 2) + x2(3, 4) = (x1 + 3x2, 2x1 + 4x2).

We remark that T is a matrix transformation given by

 [ 1 3 ] = [T(e1) T(e2)] 2 4

Example The transformation

S(x1, x2, x3) = (2x1 + 3x2 - 7x3 + 1, 10x1 - 4x2 - 8x3, - 4x1 + x2 - 5)

is not a linear transformation because

S(0) = S(0, 0, 0) = (1, 0, -5) ≠ 0

The transformation

T(x, y) = (x + y, xy)

does satisfy T(0) = 0. However, we have

T(1, 0) + T(0, 1) = (1 + 0, 1×0) + (0 + 1, 0×1) = (1, 0) + (1, 0) = (2, 0).

This is different from

T((1, 0) + (0, 1)) = T(1, 1) = (1 + 1, 1×1) = (2, 1).

Therefore T is not linear.

Since the transformations are not linear, they are not matrix transformations (see this example).

The linearity of Ax helps us to construct solutions of systems of linear equations. The key idea is that, if x and x' are respectively solutions of Ax = b and Ax = b', then cx + c'x' is a solution of Ax = cb + c'b'.

Example The vectors x = (1, 1, 2) and x' = (0, 1, -1) are the respective solutions of the following systems.

 x1 - x2 + x3 = 2 3x1 + x2 - x3 = 2 2x1 + 2x2 + x3 = 6
 x1 - x2 + x3 = -2 3x1 + x2 - x3 = 2 2x1 + 2x2 + x3 = 1

Note that the two systems have the same left side. The following systems

 x1 - x2 + x3 = 0 3x1 + x2 - x3 = 4 2x1 + 2x2 + x3 = 7
 x1 - x2 + x3 = 1 3x1 + x2 - x3 = 1 2x1 + 2x2 + x3 = 3
 x1 - x2 + x3 = 8 3x1 + x2 - x3 = -4 2x1 + 2x2 + x3 = 3

still have the same left side but different right sides. By

(0, 4, 7) = (2, 2, 6) + (-2, 2 1),
(1, 1, 3) = 1/2 (2, 2, 6),
(8, -4, 3) = (2, 2, 6) - 3(-2, 2 1),

we see that

x + x' = (1, 1, 2) + (0, 1, -1) = (1, 2, 1),
1/2 x = 1/2 (1, 1, 2) = (1/2, 1/2, 1),
x - 3x' = (1, 1, 2) - 3(0, 1, -1) = (1, -2, 5),

are the respective solutions of the three new systems.