<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/a6608cea-8e7c-4b49-acc7-ac1b22a9914c/Field.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/a6608cea-8e7c-4b49-acc7-ac1b22a9914c/Field.png" width="40px" /> A field $\mathbb F$ is a set with two operations defined on it: addition and multiplication

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💃Example: The set of real numbers $\mathbb F= \mathbb R$ is a field, as is the set of complex numbers $\mathbb F= \mathbb C$

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/66937b1e-9251-4e43-ab2e-f9fbd2b56226/vector_space.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/66937b1e-9251-4e43-ab2e-f9fbd2b56226/vector_space.png" width="40px" /> Vector space: Let $\mathbb F$ be a field. A vector space $V$ over $\mathbb F$ is a set of objects ( 💃Example:$\vec u, \vec v, \vec w$ ) that is satisfying the following properties:

  1. đŸĸ The set is closed under addition: such that if $\vec u, \vec v \in V$ then $\vec w=\vec u + \vec v \in V$

  2. đŸĸ Addition is commutative

    $$ a+b=b+a $$

  3. đŸĸ Addition is associative

    $$ (a+b)+c=a+(b+c) $$

  4. 🏎ī¸ The set is closed under multiplication by a scalar: $\vec u \in V$ then $\lambda \vec u\in V$ for $\lambda \in \mathbb F$

  5. 🏎ī¸ Multiplication is associative

    $$ \lambda (\mu \vec u)=(\lambda \mu) \vec u \quad \text{for} \quad \lambda, \mu\in \mathbb F $$

  6. 🏎ī¸ Multiplication is scalar distributive

    $$ (\lambda+\mu)\hat u=\lambda \vec u+\mu \vec u \quad \text{for} \quad \lambda, \mu \in \mathbb F $$

  7. 🏎ī¸ Multiplication is vector distributive

    $$ \lambda(\vec u+ \vec v)=\lambda \vec u+ \lambda \vec v $$

  8. 🏎ī¸ Multiplication by unity leaves a vector unchanged $1\times \vec u =\vec u$

  9. 🏴‍☠ī¸ There exists a null vector $0$ defined such that $\vec u + 0 =\vec u$

  10. ☚ī¸ All vectors have a corresponding negative vector $-\vec u$ such that $\vec u+(-\vec u)=0$ </aside>

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/8100b855-d65c-4093-b97d-684751b56ef2/linear_independence.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/8100b855-d65c-4093-b97d-684751b56ef2/linear_independence.png" width="40px" /> Linear independence: A set of vectors $\{ \vec u_i \;\text{for} \; i=1,2,\ldots,n \}$ is linearly independent if the equation

$$ \sum^n_{i=1}\lambda _i \vec u_j=0 $$

has only one solution: $\lambda _i=0 \;\forall i$. Otherwise, they are said to be linearly dependent.

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💃 Example: The following vectors $\vec u= ( 1,\, 1 )$, $\vec w=(1,\,-1)$ are linearly independent, to show this we calculate $\alpha \vec u + \beta \vec w=0$

$$ \begin{aligned} \alpha + \beta&=0 \\ \alpha - \beta &=0 \end{aligned} $$

we see that the solutions are $\alpha ,\beta =0$ thus linearly independent

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/594990cb-2ec3-4b29-9ca6-d763d7a83bc5/dimensionality.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/594990cb-2ec3-4b29-9ca6-d763d7a83bc5/dimensionality.png" width="40px" /> Dimensionality: take a set of vectors $\{\vec u_i \;\text{for} \; i=1,2,\ldots,n\}$, a vector space $V$ has dimension $N$ if it can accommodate no more than $N$ linearly independent vectors $\vec u_i$

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🗒ī¸ Note: vectors space are often denotated $V^N(\mathbb R)$ or $V^N(\mathbb C)$

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/608e3001-d852-4873-bac0-41546fd32e3b/span.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/608e3001-d852-4873-bac0-41546fd32e3b/span.png" width="40px" /> Span: the span of a set of vectors $\{\vec u_i\,\text{for}\,i=1,2,\ldots,n\}$ is the set of all vectors that can be written as a linear combination of the $\vec u_i$ vectors

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💃 Example: any arbitrary vector in $\R^2$ is spanned by $\{\hat u,\hat v\}$ because it can be decomposed into $\hat u$ and $\hat v$ in the 2D plane

🧔‍♀ī¸ Theorem: In an $N$-dimensional vector space $V^N$ any vector $\vec u$ can be written as a linear combination of linearly independent basis vectors

đŸ’Ģ Proof: This follows from the definition of linearly independence and dimensionality: Since there are no more than $N$ linearly independent vectors, therefore there must be a relation of the form

$$ \sum^N_{i=1}\lambda_i \hat e_i+\lambda_0 \vec u=0 $$

where $\vec u \in V^N$ is an arbitrary vector, and not all $\lambda_i$ are zero. In particular the definition of the linear dependence requires $\lambda_0\ne 0$. We therefore have

$$ \vec u=-\frac{1}{\lambda_0}\sum^N_{i=1}\lambda_i \hat e_i=\sum^N_{i=1}u_i\hat e_i $$

where $u_i=-\lambda_i/\lambda_0$

<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/3e91b7a7-3b22-4ea7-a1a2-ca2a60478655/basis.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/3e91b7a7-3b22-4ea7-a1a2-ca2a60478655/basis.png" width="40px" /> Basis: Any set of linearly independent vectors in $V^N$ is called a basis, and they are said to span $V^N$ or synonymously, they are complete. This allow us to write any vector $\vec v\in V^N$ as,

$$ \vec v= \sum^N_{i=1}v_i \hat e_i $$

where the set $\{ \hat e_j\}^N_{j=1}$ is a complete basis

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💃 Example: If $\{ e_i\}$ is a basis of $V^N$ prove that for any $\vec u \in V^N$ the coefficients $u_i$ in the expansion $\vec u=\sum^N_{i=1} u_i \hat e_i$ are unique

🧔‍♀ī¸ Theorem: Any set of $M$ linearly independent vectors $\{ \hat e_i\}^M_{i=1}$ in $V^N$ span a subspace $V^M$ of $V^N$

It must satisfy the vector space properties

  1. It must contain the zero vector $\vec 0$
  2. It must be closed under addition and scalar multiplication

💃 Example: Consider the vector space $\R^3$, this is the set of numbers $(x,y,z)$ where $x,y,z\in \R$. A subspace of $\R^3$ is the set of vectors $(x,y,0)$ where $x,y\in \R$, which define $xy-$plane in $\R^3$.

We can test the conditions