Hello it's drifter1 again. Today we will get into some **Linear Function Examples** finishing off our **Linear Algebra **series for now. You should check out the posts about Linear Functions and the Linear Function Matrix first if you haven't already. So, without further do let's get into it!

## Example 1:

Which of the following functions/morphisms are linear?

f: R^2->R^3, f(x, y) = (x + 1, 3y, y - x)

g: R^3->R, g(x, y, z) = (2x - 3y + 4z)

h: R^2->R^2, h(x, y) = (2x + y, x + 2y)

If you remember from our theory post, **a function f: V->W is linear only when**:

* f(ku + lv) = k*f(u) + l*f(v)*, where u, v are vectors of V and k, l are real numbers

So, we will test out all the functions to see if they fulfill this requirement.

**f:**

Suppose two vectors u=(x1, y1) and v=(x2, y2)

We have that:

f(ku + lv) = f(k(x1, y1) + l(x2, y2)) = f((kx1, ky1) + (lx2, ly2)) = f(kx1 + lx2, ky1 + ly2) =>

f(ku + lv) = (kx1 + lx2 + 1, 3(ky1 + ly2), ky1 + ly2 - (kx1 + lx2)).

After this line we see that we can break up the second and third dimension, but the first dimension contains this "+1" that makes it impossible to create f(ku) + f(lv), because not both can "go back".

So, we will end up with either:

f(ku + lv) = f(ku) + (lx2, 3ly2, ly2 - lx2)

or

f(ku + lv) = (kx1, 3ky1, ky1 - kx1) + f(lv)

This means that** f is not a linear function**!

**g:**

Suppose two vectors u=(x1, y1, z1) and v=(x2, y2, z2)

We have that:

g(ku + lv) = g(k(x1, y1, z1) + l(x2, y2, x2)) = g((kx1, ky1, kz1) + (lx2, ly2, lz2)) =>

g(ku + lv) = g(kx1 + lx2, ky1 + ly2, kz1 + lz2) = (2(kx1 +lx2) -3(ky1 + ly2) + 4(kz1 + lz2)).

This time we can go back and create the g(ku) and g(lv) terms and so:

g(ku + lv) = (2kx1 -3ky1 + 4kz1) + (2lx2 - 3ly2 + 4lz2) = g(ku) + g(lv)

This means that **g is a linear function**!

**h:**

Suppose two vectors u=(x1, y1) and v=(x2, y2)

We have that:

h(ku + lv) = h(k(x1, y1) + l(x2, y2)) = h((kx1, ky1) + (lx2, ly2)) = h(kx1 + lx2, ky1 + ly2) =>

h(ku + lv) = (2(kx1 + lx2) + ky1 + ly2, kx1 + lx2 + 2(ky1 + ly2)).

We again can go back and create the h(ku) and h(lv) terms and so:

h(ku + lv) = (2kx1 +ky1, kx1 + 2ky1) + (2lx2 + ly2, lx2 + 2ly2) = h(ku) + h(lv)

This means that **h is also a linear function**!

So, from these examples we can see that a function is linear when it contains only linear combinations of the differenent dimension values. This means that when there is some kind of constant added (like +1 in function f) this constant will be passed only to one of the two terms/vectors and so we can't get the linear result.

## Example 2:

Suppose the function f: R^3->R^3, f(x, y, z) = (ax - y - 2z, x + ay + 3z, x + 3y + az).

You can show that it is a linear function with ease and we also already showed some good examples with how you proof it and so I would not get into how you proof it.

So, let's find out for **which integer values of a** (a in Z) the function/morphism **f is an monomorphism**.

In the Linear Functions post I said that f is a monomorphism when the **nullspace Kerf = {0}**.

The nullspace contains the solutions for the homogeneous system that this function's matrix represents and so we will have to find the values of a for which the system has no solutions. Because the matrix is a 3x3 square matrix this is pretty easy, cause we simply have to see if the system has one solution that is the all-zero solution and means that the determinant is non-zero.

The matrix A is:

a -1 -2

1 a 3

1 3 a

So, using the rule of Sarrus the determinant is(we have talked about it in Determinants):

det(A) = a^3 - 6a - 9 = (a - 3)(a^2 + 3a + 3).

And so we have that det(A) = 0 when:

a = 3

a1,2 = [-3 +-(9 - 12)] / 2 = [-3 +-3]/2 => a = 0 and a = -3

That way we proofed that **f is a monomorphism when** **a = (Z - {0, 3, -3}).**

## Example 3:

Suppose the function f: R^4 -> R^4, f(x1, x2, x3, x4) = (x2, x3, x4, 0).

a) find the fuction matrix that corresponds to the standard basis of R^4

b) find the rank(A) of A and a basis for the nullspace N(A)

a)

The first part is pretty simple.

We know that the standard basis is {e1, e2, e3, e4} and so:

f(e1) = (0, 0, 0, 0) = 0*e1 + 0*e2 + 0*e3 + 0*e4

f(e2) = (1, 0, 0, 0) = 1*e1 + 0*e2 + 0*e3 + 0*e4

f(e3) = (0, 1, 0, 0) = 0*e1 + 1*e2 + 0*e3 + 0*e4

f(e4) = (0, 0, 1, 0) = 0*e1 + 0*e2 + 1*e3 + 0*e4

So, the **function matrix A of f** is:

**0 0 0 0**

**1 0 0 0**

**0 1 0 0**

**0 0 1 0**

b)

rank(A) equals to the number of non-zero rows and so rank(A) = 3.

We know that dimV = dimKerf + dimImf, where dimImf = rank(A) and dimV = dimR^4 = 4.

This means that 4 = dimKerf + 3 => dimKerf = 1.

We also know that N(A) = Kerf and so the nullspace basis contains one vector.

We can already see from here that the linear system A represents has infinite solutions, cause rank(A) < variables. So, finding the "form" of those solutions we will find a basis of N(A).

From the matrix A and solving the homogeneous system AX = 0 we have that:

x1 = x2 = x3 = 0

x4 is arbitrary

So, the solutions are of the form:

[0, 0, 0, x4], x4 arbitrary

or

[0, 0, 0, 1]*x4

And so** N(A) = span{[0 0 0 1]}** and **[0 0 0 1] is a basis of N(A)**.

This example hopefully made you understand better what we mean with Nullspace. You can see that the result of function f will never have a value different to 0 in the 4th dimension value, but N(A) has 0's in the other dimension values and a value different to 0 in the 4th dimension. So, N(A) + f(R^4) = R^4 and you can proof it by yourself if you want, but it's actually logically if you think about what I told you right now, cause function f gives us all vectors that have values different to 0 in the first 3 dimension and N(A) in the 4th dimension and so both combined together give us the whole vector space R^4.

And this is actually it and I hope you enjoyed it!

From next time we will get into Mathematical Analysis (limits, derivatives, integrals etc.) for our Mathematics posts and I will also start getting into Programming again, cause math really took us away of programming.

Until next time...Bye!