2.6 Linear Maps
2.6.1 Basic Properties and Examples
Definition 2.6.1. A function f : Rn → Rk is called a linear map if the following hold for any x, y ∈Rn and λ∈R:
(i) f(x+y) =f(x) +f(y) (f isadditive),
(ii) f(λx) = λf(x) (f is homogeneous (of degree 1)).
Note that the additions on the left and right hand sides of (i) are different. On the left, we add two vectors inRn and apply the function f for the sum, while on the right we apply the addition in Rk for the images of the two vectors xand y. Also, the scalar multiplication on the left hand side of (ii) is an operation inRn, while on the right hand side it is an operation inRk. so f1 is homogeneous, and hence linear.
2. Let f2 :R2 →R2 be the function which assigns to every plane vector its (orthogonal) projection to the xaxis. It is easy to give a formula for f2:
f2
x y
= x
0
.
The linearity of f2 follows from this formula like in the previous case.
3. Let f3 : R2 → R2 be the rotation about the origin by the angle α, then f3 is a linear map. Indeed, as the sum of two non-zero vectors is the vertex of the (maybe degenerate) parallelogram spanned by them which is different from the origin and the endpoints of the vectors, and the rotation takes the spanned parallelogram into the parallelogram which is spanned by the rotated vectors, we get that the image of the sum is the sum of the images (this holds obviously if at least one of the vectors is the zero vector), hence the additivity off3 follows. Also, the application of a dilation and a rotation after that gives the same result as the application of these transformations in reverse order, so f3 is homogeneous and hence linear. We will give a formula for f3 later.
4. Let f4 :R2 →R2 be the reflection in a line going through the origin. As in the case of the rotation above, it is easy to see that f4 is linear.
5. Let f5 :Rn →Rk be the identically zero map. Since
f5(x) +f5(y) = 0 + 0 = 0 =f5(x+y) and λf5(x) =λ0 = 0 =f5(λx) hold for every x, y ∈Rn and λ∈R, we get that f5 is linear.
6. Let f6 : Rn → Rn be the identity map which maps every vector in Rn to itself. Then f6 is obviously linear.
Here are some basic properties of a linear map:
Proposition 2.6.1. Let f :Rn→Rk be a linear map. Then the following hold:
(i) if 01 is the zero vector in Rn and 02 is the zero vector in Rk, then f(01) = 02,
(ii) f(λ1x1+· · ·+λmxm) =λ1f(x1) +· · ·+λmf(xm) for any vectors x1, . . . , xm ∈Rn and for any scalars λ1, . . . , λm ∈R.
Proof. We havef(01) =f(01+ 01) =f(01) +f(01), becausef is additive. Adding −f(01) = (−1)·f(01) to both sides we get (i). By the repeated application of the additivity and the homogeneity off we get the second statement immediately.
It follows from property (i) above that in the case of the rotationf3 in the examples above it is important that we rotate about the origin. Rotation about any other point of the plane does not fix the origin and hence it is not linear. Similarly, if we reflect in a line which does not go through the origin, then this transformation is not linear.
We assign two important sets to every linear map:
Definition 2.6.2. Assume that f : Rn →Rk is a linear map. The kernel of f is the set of vectors in the domain Rn of f that are mapped to the zero vector 0∈Rk of the range of f.
That is,
kerf ={x∈Rn : f(x) = 0}.
The image of f is the set of the vectors in the range of the map f. That is, Imf ={y∈Rk : ∃x∈Rn, f(x) =y}={f(x) : x∈Rn}.
Examples. Let f1, . . . , f6 denote the same maps as in the examples above.
1. Every vector of R2 is an image of f1 since x
y
=f1
0
−x y
,
hence Imf1 =R2. The vector(x, y, z)T is in kerf1 if and only if x−y= 0
x+z = 0.
The solutions of this system are the vectors of the form (α, α,−α)T for some α∈R, so kerf1 = span{(1,1−1)T}.
2. In the case of the projection to thex axis every value of f2 is (obviously) on thex axis and we get every point of this axis as an image, so Imf2 is the x axis. The projection yields the zero vector if and only if the x coordinate of the projected vector is zero, hence kerf2 is the y axis.
3. Rotations and Reflections are a bijections of the plane, so Imf3 = Imf4 = R2. The origin is mapped to itself in both cases so kerf3 = kerf4 ={0}.
4. In the case of f5 every vector is mapped to 0, so Imf5 ={0} and kerf5 =Rn. For the identity map f6 we obviously have Imf6 =Rn and kerf ={0}.
The image and the kernel have special structure in the previous examples, namely, they are subspaces. As the following theorem shows, this is true in general:
Theorem 2.6.2. If f :Rn→Rk is a linear map, then kerf is a subspace of Rn and Imf is a subspace of Rk.
Proof. The kernel of f is non-empty, since 0∈Rn is in it by part (i) of the previous propo-sition. If x, y ∈kerf, then by the additivity of f we have
f(x+y) =f(x) +f(y) = 0 + 0 = 0,
sox+y∈kerf. Similarly, if λ∈R then by the homogeneity of f we have f(λx) =λf(x) =λ0 = 0,
soλx ∈kerf. This means thatkerf is a non-empty set ofRn which is closed under addition and scalar multiplication, so it is a subspace of Rn.
Now we prove the statement for the image of f. We have f(0) = 0 ∈ Imf, so it is non-empty. If x, y ∈Imf, then there is a u∈Rn for which f(u) =x, and similarly, there is av ∈Rn for which f(v) = y. Then by the additivity off we have
x+y=f(u) +f(v) =f(u+v),
sox+y∈Imf. Similarly, if λ∈R, then by the homogeneity of f we have λx=λf(u) = f(λu),
henceλx ∈Imf. That is, Imf is non-empty and closed under addition and scalar multipli-cation, so it is a subspace of Rk.
As we have seen above, a linear map is not always injective, i.e. the image of different vectors can be the same. It turns out that there is a connection between the injectivity of a linear map and its kernel:
Theorem 2.6.3. Assume thatf :Rn →Rkis a linear map andx, y ∈Rn. Thenf(x) = f(y) holds if and only if x−y ∈kerf. Consequently, f is injective if and only if kerf ={0}.
Proof. Iff(x) = f(y) holds, then by the linearity of f this is equivalent to 0 =f(x)−f(y) =f(x−y),
sof(x) = f(y)holds if and only if x−y∈kerf.
If f is not injective, then there are different vectors u, v ∈ Rn for which f(u) = f(v), and hence 0 6= u−v ∈ kerf. On the other hand, if kerf 6= {0}, then there is a vector 06=u∈kerf, and then f(u) = f(0) = 0, so f is not injective.