An injective function is one which is one-to-one: that it maps distinct inputs to distinct outputs.

constituents

• A function \(T: V \to W\)

requirements

\(T\) is injective if \(Tu = Tv\) implies \(u=v\).

additional information

injectivity implies that null space is \(\{0\}\)

Proof: let \(T \in \mathcal{L}(V,W)\); \(T\) is injective IFF \(null\ T = \{0\}\).

given injectivity

Suppose \(T\) is injective.

Now, we know that \(0\), because it indeed gets mapped by \(T\) to \(0\), is in the null space of \(T\).

constituents

• \(V\) a vector space
• \((u,v)\), an ordered pair of vectors in \(V\) (its not commutative!)

requirements

We define \(\langle u, v \rangle \in \mathbb{F}\) as the inner product of \((u,v)\) in that order!. It carries the following properties:

1. positivity: \(\langle v, v\rangle \geq 0, \forall v \in V\)
2. definiteness: \(\langle v, v\rangle = 0\) IFF \(v = 0\)
3. additivity in the first slot: \(\langle u+v, w\rangle = \langle u, w \rangle + \langle v, w \rangle\)
4. homogeneity in the first slot: \(\langle \lambda u, v \rangle = \lambda \langle u, v \rangle\)
5. conjugate symmetry: \(\langle u,v \rangle = \overline{\langle v,u \rangle}\)

additional information

Inner Product Space

An Inner Product Space is a vector space with a well-defined inner product. For instance, \(\mathbb{F}^{n}\) has the canonical inner product named Euclidean Inner Product (see below, a.k.a. dot product for reals). The existence of such a well-defined inner product makes \(\mathbb{F}^{n}\) an Inner Product Space.

an integer (\(\mathbb{Z}\)) is the natural numbers, zero, and negative numbers: …,-4,-3,-2,-1,0,1,2,2,3

representing integers

• what are the limitations of computational arithmetic
• how to perform efficient arithmetic
• how to encode data more compactly and efficiently

See also computer number system

The integrating factor \(\rho(x)\) is a value that helps undo the product rule. For which:

\begin{equation} log(\rho(x)) = \int P(x)dx \end{equation}

for some function \(P(x)\).

Separating the \(\rho(x)\) out, we have therefore:

\begin{equation} e^{\int P dx} = \rho(x) \end{equation}

Why is this helpful and undoes the product rule? This is because of a very interesting property of how \(\rho(x)\) behaves.