Today we continue with **Linear Algebra** getting into **Linear Functions**. Linear Vector Spaces and much more we covered in our series until now can be useful, so you should check them out to get a better understanding. I will talk about some more things in this topice next time to split it up. So, without further do let's get started!

## Linear Functions:

Suppose **two linear vector spaces V and W**. A **function f: V -> W** (from V to W) **is linear** **when**:

- f(u + v) = f(u) + f(v), for every u, v in V
- f(k*v) = kf(v), for every v in V

We sometimes also call a linear function a **morphism or homomorphism** of V. A **function from a vector space to itself** (f: V -> V) is called a **endomorphism**. The "+" operation in the first bullet in the left side was inside of vector space V and in the right side in the vector space W. The same is true in the "*" operation.

So, we can generally say that** a function f: V -> W is linear when f(ku + lv) = kf(u) + lf(v)**, where u, v in V and k, l are real numbers, combining both bullets into one.

### "Properties":

Suppose **f: V -> W a linear morphism** between the vector spaces V, W then:

- f(0) = 0, where the zeros represent the 0 in V and W respectively
- f(-v) = -f(v), v in V

When** V, U and W are vector spaces** and **f: V -> U and g: U -> W are linear morphisms** then the **composition **of g and f**, gof: V-> W is **also** a linear morphism**. I will use them in random alternate order, cause I don't have a preference.

### Examples:

- With V, W being vector spaces we have that f: V -> W, where f(v) = 0 (
**zero morphism**) and Iv: V -> V, where Iv(v) = v (**identity morphism**) are both linear functions - f: R -> R with
**f(x) = ax**, a in R is a linear function. The function f(x) = ax + b is not a linear function tho. - f: R^3 -> R^2, with f(x, y, z) = (x + y, x - z) is also a linear function.

Let's prove the last one.

For (x1, y1, z1), (x2, y2, z2) of R^3 we have:

*f( (x1, y1, z1) + (x2, y2, z2) ) = f( (x1 + x2, y1 + y2, z1 + z2) ) = f( x1 + x2 + y1 + y2, x1 + x2 - z1 - z2 )*

For (x, y , z) in R^3 and k real we have:

*f( k*(x, y, z) ) = f( (kx, ky, kz) ) = f( (kx + ky, kx - kz) ) = k*f(x + y, x - z) = k*f(x, y, z)*

[It may seem a little wierd but we simply follow our function that says that (x, y, z) = (x + y, x - z)]

### Hom-set:

With V and W being vector spaces **Hom(V, W) is the set of all linear morphisms from V to W**.

For all f, g in Hom(V, W) we define an **addition and scalar multiplication**:

- (f + g)(x) = f(x) + g(x), x in V
- (kf)(x) = kf(x), x in V, k in R

As we already said in the beginning the linear function from V to itself is called an **Endomorphism **and we can defineit using Hom like that: **End(V) = Hom(V, V)**.

Kernel and Image:

Suppose f: V -> W is a linear morphism between the vector spaces V and W.

The** kernel of the function f **is a **subset of V** and is defined as:

Ker f = { x in V : f(x) = 0 }

The **image of the function f** is a **subset of W **and is defined as:

f(V) = Im f = { w in W : w = f(v), v in V }

We can prove using what we already know that the **kernel and image are not only subsets but also subspaces**! I will leave it to you if you want to make a quick refreshment of your knowledge :D

**Example:**

Let's get the same function as before that was: f(x, y, z) = (x + y, x - z) where f: R^3 -> R^2.

Now suppose we have v = (x, y, z) that is in the Kerf.

Using the definition of the kernel we have that:

f(x, y, z) = 0 => (x + y, x - z) = (0, 0) => x + y = 0 and x - z = 0 => x = -y and x = z.

So, the random element (x, y, z) of the Kerf is in the form:

(x, -x, x) = x*(1, -1, 1)

That way Kerf = span{ (1, -1, 1) } and dimKerf = 1. (see how everything gets connected?)

Let's now suppose a w in Imf = f(V) then from the defintion of the image we know that there is a v = (x, y, z) in R^3 so that:

w = f(x, y, z) = (x + y, x - z) =>

w = (x, x) + (y, 0) - (0, z) = x*(1, 1) + y*(1, 0) + z*(0, 1) [where the minus can be removed]

So, that way Imf = span{ (1, 1), (1, 0), (0, 1) }.

But, (1, 1) = (1, 0) + (0, 1) and so is linear dependent and that way we have that:

Imf = span{ (1, 0), (0, 1) } and dimImf = 2

You can see that dimKerf + dimImf = 1 + 2 = 3 = dimR^3 that is not random and we will explain it now!

### Monomorphism, Epimorphism and Isomorphism:

Suppose V, W two vector spaces and f: V -> W a linear function.

The function f is called a **monomorphism when f is mono**.

The function f is called a **epimorphism when f is epi**.

The function f is called a **isomorphism when f is mono and epi**.

**Mono **means that for every f(v1) = f(v2) => v1 = v2. It also means that f is invertible.

**Epi** means that f(V) = W and so f "is on top" of W

To check mono, epi and iso-morphism we use the following **Rules**:

- A
**linear function f: V -> W**is a**monomorphism when Kerf = {0}**and vise versa - - || - is a
**epimorphism when Imf = W**and vise versa - -|| - is a
**isomorphism when Kerf = {0} and Imf = W**and vise versa

**Example:**

Suppose f: R -> R^2 where f(x) = (x, 0) for every x in R

For a x in Kerf we have that:

f(x) = 0 => (x, 0) = (0, 0) => x = 0

That way Kerf = {0} and f is a monomorphism

### Monomorphism Rule:

Suppose V, W are two vector spaces and f: V -> W is a monomorphism. When **v1, v2, ..., vn of V **are **linear independent **then** f(v1), f(v2), ..., f(vn) of W** are also **linear independent**.

Also span{ v1, v2, ..., vn } = span{ f(v1), f(v2), ..., f(vn) } ( = means isomorphic )

So, that way when V and W are finite-dimension with dimV <= dimW we know that the subspace E of V goes to the vector space of V using a monomorphism function f and so W = f(E).

### Dimension Rules:

Suppose V and W are finite vector spaces and f: V -> W is a linear function then:

**dimV = dimKerf + dimImf**(as we already found out in an example previously)- When
**dimV = dimW then V = W**(isomorphic) and vise versa

The last one can be proven by saying that when V = W the function f is an isomorphism.

Because f is an isomorphism (mono-part) dimKerf = 0 => dimV = dimImf,

Also (epi-part), Imf = W => dimImf = dimW

And so dimV = dimW

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

Next time in Linear Algebra we will continue with Linear Functions getting into the function matrix and some special cases and maybe some more examples.

Bye!