Gain insights into linear algebra essentials for data science, focusing on vectors, matrices, and tensors. Explore practical Python applications, engaging visuals, real datasets, and hands-on projects.

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Linear algebra is a fundamental pillar of data science. In advanced models in data science, like neural networks, the inputs and transformations are based upon vectors, matrices, and tensors which require a reasonable understanding of linear algebra to get the desired results. It is elegant and the most applied mathematics under the umbrella of data science.

This course teaches linear algebra with a focus on data science. This course encompasses several engaging illustrations, including static images and animations. Furthermore, this course presents mathematical modeling through programming in Python. This course contains several executable coding playgrounds on real data sets and a final project with practical applications.

Aside from theoretical implementations, the modern-day world needs its daunting calculations, trajectory mapping, and distance manipulation, all of which linear algebra provides. By the end of this course, you’ll have a working knowledge of all the necessary teachings in linear algebra.

Linear Algebra for Data Science Using Python

## Span
The **span** of a set, $U$, is another set, $V$, consisting of all the linear combinations of vectors in $U$. The set $U$ is referred to as a _**spanning set**_. 

 
### Examples
1. $span\bigg(\bigg\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\bigg\}\bigg)=\R^2$

We can create any vector in $\R^2$, say $\begin{bmatrix}\alpha & \beta\end{bmatrix}^T$, using a linear combination of the spanning set $X=\bigg\{\begin{bmatrix}1 & 0\end{bmatrix}^T, \begin{bmatrix}1 & 0\end{bmatrix}^T\bigg\}$. 
So, $X$ spans $\R^2$. That is: 

$$\alpha\begin{bmatrix}1 \\ 0\end{bmatrix}+\beta\begin{bmatrix}0 \\ 1\end{bmatrix}=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}$$  



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2. $span\bigg(\bigg\{\begin{bmatrix}a\\b\end{bmatrix}\bigg\}\bigg)=$ line in the direction of $\begin{bmatrix}a\\b\end{bmatrix}$
>**Note:** A line consists of points that are also position-vectors.

### Spanning set isn't unique

A spanning set corresponding to a given span isn’t unique.

In a previous example, we showed that
$span\bigg(\bigg\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\bigg\}\bigg)=\R^2$
. . However, there exists infinitely many spanning sets that span $\R^2$. For instance, $span\bigg(\bigg\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}5\\7\end{bmatrix}\bigg\}\bigg)=\R^2$. To prove this, we may show that $\begin{bmatrix}1\\0\end{bmatrix}\in span\bigg(\bigg\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}5\\7\end{bmatrix}\bigg\}\bigg)$ and $\begin{bmatrix}0\\1\end{bmatrix}\in span\bigg(\bigg\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}5\\7\end{bmatrix}\bigg\}\bigg)$. That is:

$$\begin{bmatrix}1\\0\end{bmatrix}=-7\begin{bmatrix}2\\3\end{bmatrix}+3\begin{bmatrix}5\\7\end{bmatrix}$$

$$\begin{bmatrix}1\\0\end{bmatrix}=5\begin{bmatrix}2\\3\end{bmatrix}-2\begin{bmatrix}5\\7\end{bmatrix}$$

>**Note:** A vector, $\bold v$, is in the span of the vectors $\bold{u_1,u_2,...,u_k}$ if $\bold{v}$ is a linear combination of the vectors $\bold{u_1,u_2,...,u_k}$, i.e., $\bold v = \alpha_1\bold{u_1}+\alpha_2\bold{u_2}+\cdots+\alpha_k\bold{u_k}$.

The linear combination test can be viewed as a consistency check of a linear system. In the animation below, the column vectors of the coefficient matrix are linearly independent causing the two by two matrix to be invertible. So, the linear system is consistent for every right-hand-side b.



# Span
The **span** of a set, $U$, is another set, $V$, consisting of all the linear combinations of vectors in $U$. The set $U$ is referred to as a _**spanning set**_. 

 
## Examples
1. $span\bigg(\bigg\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\bigg\}\bigg)=\R^2$

We can create any vector in $\R^2$, say $\begin{bmatrix}\alpha & \beta\end{bmatrix}^T$, using a linear combination of the spanning set $X=\bigg\{\begin{bmatrix}1 & 0\end{bmatrix}^T, \begin{bmatrix}1 & 0\end{bmatrix}^T\bigg\}$. 
So, $X$ spans $\R^2$. That is: 

$$\alpha\begin{bmatrix}1 \\ 0\end{bmatrix}+\beta\begin{bmatrix}0 \\ 1\end{bmatrix}=\begin{bmatrix}\alpha \\ \beta\end{bmatrix}$$  



## Spanning set isn't unique

A spanning set corresponding to a given span isn’t unique.

In a previous example, we showed that
$span\bigg(\bigg\{\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}\bigg\}\bigg)=\R^2$
. . However, there exists infinitely many spanning sets that span $\R^2$. For instance, $span\bigg(\bigg\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}5\\7\end{bmatrix}\bigg\}\bigg)=\R^2$. To prove this, we may show that $\begin{bmatrix}1\\0\end{bmatrix}\in span\bigg(\bigg\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}5\\7\end{bmatrix}\bigg\}\bigg)$ and $\begin{bmatrix}0\\1\end{bmatrix}\in span\bigg(\bigg\{\begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}5\\7\end{bmatrix}\bigg\}\bigg)$. That is:

$$\begin{bmatrix}1\\0\end{bmatrix}=-7\begin{bmatrix}2\\3\end{bmatrix}+3\begin{bmatrix}5\\7\end{bmatrix}$$

$$\begin{bmatrix}1\\0\end{bmatrix}=5\begin{bmatrix}2\\3\end{bmatrix}-2\begin{bmatrix}5\\7\end{bmatrix}$$

>**Note:** A vector, $\bold v$, is in the span of the vectors $\bold{u_1,u_2,...,u_k}$ if $\bold{v}$ is a linear combination of the vectors $\bold{u_1,u_2,...,u_k}$, i.e., $\bold v = \alpha_1\bold{u_1}+\alpha_2\bold{u_2}+\cdots+\alpha_k\bold{u_k}$.

The linear combination test can be viewed as a consistency check of a linear system. In the animation below, the column vectors of the coefficient matrix are linearly independent causing the two by two matrix to be invertible. So, the linear system is consistent for every right-hand-side b.



Learn the concepts of the span, the dimension of a vector space, and the basis of a vector space.

Introduction

Linearity

Matrices

Solving Linear Systems

Singularity

Linear Regression and Least Squares

Vector Space

Vector Spaces of a Matrix

Singular Value Decomposition: SVD

Learning to Find Discriminative Null Space for Face Recognition

Span, Basis, and Dimensions

Span

Examples

Spanning set isn’t unique