Mathematics - Linear Algebra Functions

 

    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:

  1. 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
  2. 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.
  3. 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!

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