Linear Algebra Basics For ML & DL

Introduction for Introduction

We know what vectors and matrices are. Just a set of numbers or objects juxtaposed in a one-dimensional or two-dimensional structures. Now if these numbers are put in three-dimensional or higher-dimensional structures, they are called tensors. We have the following notation:

I think we all are familiar with the transpose and multiplication operations on matrices. Also, the element-wise product operation of matrices which we show with \(\mathbb{A}_1 \odot \mathbb{A}_2\).

Span of Vectors

I assume you know what is linear independence of vectors. Linear combination of the set of vectors \(\{v^{(1)}, v^{(2)}, \dots ,v^{(n)}\}\) is given by \(\sum_{i =1}^n c_i v^{(i)}\), where \(c_i\)s are real coefficients. Set of all points that can be obtained by \(\sum_{i =1}^n c_i v^{(i)}\) (for some \(c_i \in \mathbb{R}, i \in \{1,\dots, n\}\)) is called the span of this set \(\{v^{(1)}, v^{(2)}, \dots ,v^{(n)}\}\). If we add a vector \(v^{(n+1)}\), to the set of vectors which is a linear combinatoin of members of in \(\{v^{(1)}, v^{(2)}, \dots ,v^{(n)}\}\), the span of the new set does not increase. In other words \(\text{span}\{v^{(1)}, v^{(2)}, \dots ,v^{(n)}\} = \text{span}\{v^{(1)}, v^{(2)}, \dots ,v^{(n)}, v^{(n+1)}\}.\)

The set of linear equations \(\mathcal{A}_{m \times n} x_{n\times 1} = b_{m \times 1}\) has a solution if the span of columns of \(\mathcal{A}\), \(\{A^{(1)}, A^{(2)}, \dots, A^{(n)} \}\) includes \(b\). Thus, \(\mathcal{A}\) should have \(m\) linearly independent columns so that equation \(\mathcal{A}x = b\). has a solution for every \(b \in \mathbb{R}^m\).

Good to know: Square matrices with linearly dependent columns are called singular.