Tuesday, 13 September 2016

If a function $f: A \subset\mathbb{R}\rightarrow\mathbb{R}$ is strictly increasing, then $f$ is injective.

Theorem: Consider a function $f: A \subset\mathbb{R}\rightarrow\mathbb{R}$ that is strictly increasing. Then $f$ is an injective function.

Proof: We know that $f$ is strictly increasing on its domain $A$. Thus, for $x_1 < x_2$, we have $f(x_1) < f(x_2)$. So it's clear that if $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$. Thus $f$ is injective.

Monday, 12 September 2016

If $A$ is a symmetric and regular matrix, then $A^{-1}$ is symmetric.

Theorem: Suppose $A$ is a symmetric and regular matrix, then $A^{-1}$ is symmetric too.

Proof: $A^{-1}$ exists as A is a regular matrix.

So, we have: $AA^{-1} = I_n$ and we have to show that ${(A^{-1})}^{T} = A^{-1}$

Now, transposing everything, we obtain, because ${I_n}^{T} = I_n$ and $(AB)^{T} = B^TA^T$:

$(A^{-1})^TA^T = I_n$, but we also have:  $A^{-1}A = I_n$

So: $(A^{-1})^TA^T = A^{-1}A$

But $A$ is symmetric, thus $A^T = A$

So: $(A^{-1})^TA = A^{-1}A$

And by the theorem: $AB = CB$ and $B$ regular $\Rightarrow A = C$, we get:

$(A^{-1})^T = A^{-1}$

And this is exactly the definition of a symmetric matrix.

Sunday, 11 September 2016

Functions: definition

Definition (function): Given 2 sets X and Y. A function $f: X \rightarrow Y$ is the set of ordered pairs $(x,y)$ such that:

1) $(x,y) \in X \times Y$. This means: $x \in X$ and $y \in Y$. (*)
2) The $x$ in $(x,y)$ occurs only once as first component of the ordered pair $(x,y)$

We call A the domain of f and B the codomain of f (***)
__________________________________________________________________________

Remarks:

1) (*) In fact, this means that a function $f: X \rightarrow Y$ is actually defined as a subset of $X \times Y$. So, we can write that $f = \{(x,f(x))|x \in X\}$
2) (**) Simply stated, this means that we can only have one $y$ for every $x$ in $(x,y)$.
3) (***) When you would compute f(x) for all x in the domain of f, you would get a set of values. This set is called the range of f. Now, it's very important to notice that the range of f is not necessarily equal to the codomain of f! Define, for example, a function g in the following way.

$g: \mathbb{R}\rightarrow \mathbb{R}: x \mapsto f(x) = x^2$.

Then, the codomain of f is the set of the real numbers $\mathbb{R}$ while the range of f is the set of all positive real numbers $\mathbb{R^{+}}$

Examples:

1) Consider the function f:

$f: \mathbb{R}\rightarrow \mathbb{R}: x \mapsto x^2 - x -2$.
- The domain of f is $\mathbb{R}$
- The codomain of f is $\mathbb{R}$
- The range of f is $[\frac{-9}{4},\infty)$

2) The Sign function is defined in the following way:

$Sign(x) := \begin{cases} -1 & \quad x<0\\ 0 & \quad x= 0\\ 1 & \quad x>0 \\ \end{cases}$

Theorem: Let $f: A \to B$ be a function. Then the following statements are equivalent:

(1) $f$ bijective
(2) There is a function $g: B \to A$ such that $g \circ f = 1_A$ and $f \circ g = 1_B$

Proof: $\boxed{(1) \Rightarrow (2)}$

Since $f$ is bijective, for every $b \in B$, there exists a unique $a \in A$ such that $f(a) = b$. Then, define $g: B \to A$ by letting $g(b) := a$. Because $f$ is bijective, $g$ is well defined.

Now, we have that $g \circ f (a) = g(f(a)) = g(b) : a$ and $f \circ g(b) = f(g(b)) = f(a) = b$

$\boxed{(1) \Leftarrow (2)}$