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

## Definition
The column space of a matrix, 
$A$, often denoted by $C(A)$, is a vector space spanned by the column vectors of $A$.



>The column-space of an $m \times n$ matrix $A$ is a subspace of $\R^m$. The dimensions of the column space are the number of linearly independent columns, that is, the rank of the matrix $A$.





### Examples

1. The column space of $\begin{bmatrix}1 &0\\0&1\end{bmatrix}$ is $\R^2$.


2. The column space of $\begin{bmatrix}1 &2\\2&4\end{bmatrix}$ is a one-dimensional subspace of $\R^2$. This is because there's a single linearly independent vector as $2\bold{c_1} =\bold{c_2}$. Any of these columns can be a basis for the subspace.



3. The column space of $\begin{bmatrix}0 &0\\0&0\end{bmatrix}$ is a zero-dimensional subspace of $\R^2$.

4. The column space of an $n \times n$ invertible matrix $A$ is $\R^n$. This is because all the columns of such a matrix are linearly independent. That is, $rank(A)=n$.



>The columns of every $n \times n$ invertible matrix form a basis of $R^n$.



## Basis of column space
A **basis** of the column space of a matrix, $A$, is a set of linearly independent columns of $A$. One algorithm for computing the basis is as follows:

1. Take the transpose of the matrix to convert the columns into rows.

2. Compute  the $rref$ (reduced row echelon form) of the transposed matrix.
3. The set of non-zero rows in $rref$ is a linearly-independent set. These form the basis of the column space.


>Note that the basis found by the algorithm above may not contain the original column vectors, but their linear combinations as elementary row operations have transformed the matrix into $rref$.

Let's try it out by using the following code: 


# Definition
The column space of a matrix, 
$A$, often denoted by $C(A)$, is a vector space spanned by the column vectors of $A$.



>The column-space of an $m \times n$ matrix $A$ is a subspace of $\R^m$. The dimensions of the column space are the number of linearly independent columns, that is, the rank of the matrix $A$.





## Examples

1. The column space of $\begin{bmatrix}1 &0\\0&1\end{bmatrix}$ is $\R^2$.


2. The column space of $\begin{bmatrix}1 &2\\2&4\end{bmatrix}$ is a one-dimensional subspace of $\R^2$. This is because there's a single linearly independent vector as $2\bold{c_1} =\bold{c_2}$. Any of these columns can be a basis for the subspace.



3. The column space of $\begin{bmatrix}0 &0\\0&0\end{bmatrix}$ is a zero-dimensional subspace of $\R^2$.

4. The column space of an $n \times n$ invertible matrix $A$ is $\R^n$. This is because all the columns of such a matrix are linearly independent. That is, $rank(A)=n$.



>The columns of every $n \times n$ invertible matrix form a basis of $R^n$.



# Basis of column space
A **basis** of the column space of a matrix, $A$, is a set of linearly independent columns of $A$. One algorithm for computing the basis is as follows:

1. Take the transpose of the matrix to convert the columns into rows.

2. Compute  the $rref$ (reduced row echelon form) of the transposed matrix.
3. The set of non-zero rows in $rref$ is a linearly-independent set. These form the basis of the column space.


>Note that the basis found by the algorithm above may not contain the original column vectors, but their linear combinations as elementary row operations have transformed the matrix into $rref$.

Let's try it out by using the following code: 


Learn about  a matrix’s column space and its relationship with the solution of a linear system.


Column Space

Learn about  a matrix’s column space and its relationship with the solution of a linear system.

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

Column Space

Definition

Examples