Memory-Baguette

Body isn't all me.

View the Project on GitHub LLan-ovoy/Memory-Baguette

Deep Learning

This is the reading summary of Deep Learning by Ian Goodfellow, Yoshua Bengio and Aaron Courville.

Introduction

Why deep learning?
What is deep learning?
The history of deep learning

Review of Linear Algebra

I will only listed some concepts here unless those I am not familiar with, denoted in italic.

Mathematiclal objects

operation

A system of linear equations

A is a metric, x is a vector and b is an vector, \(Ax = b\) In which, \(A_{1,:} x =b1\) more detailed, \(A_{1,1}x_1 + A_{1,2}x_2 + ... + A_{1,n}x_n = b1\)

Indentity and Inverse Matrices

So we can solve the linear system by \(Ax = b \\ A^{-1}Ax = A^{-1}b \\ I_n x = A^{-1}b \\ x = A^{-1}b\) This process depends on it being possible to find $A^{-1}$.

However, a digital computer can only find the closed form of $A^{-1}$ with limited precision, and this is the reason why this solution of such linear system is more of a theoretical tool while not actually be used in practice.

Algorithms that make use of the value of b can usually obtain more accurate estimates of $x$.

Linear dependent and Span

It is possibly that $Ax = b$ have no solution or infinite solutions but impossible to have more than one but less than infinite solutions for a particular $b$, because if both $x$ and $y$ are solutions of this quation, then $z$ in the following equation should also be the solution of $Ax = b$ . \(z = \alpha x + (1-\alpha)y\) To Analyze how many solutions it has, think of columns of A as specifying different directions we can trvel in from the origin $[0,0,…,0]^T$ to $b$. So each element of $x$ specifying how far to move in each direction of column $i$, and this is a linear combination: \(Ax = \sum_i x_i A_{:,i}\) This is also a span of a set of vectors, which is just columns of matrix $A$. It is the set of all points obtainable by linear combination of the original vectors.

Example, consider a $3\times2$ matrix $A$ with a $b\in \mathbb{R}^3$, since the column space of $A$ is 2-D while $b$ is 3-D, only when $b$ lies on that plane, $Ax = b$ has a solution. In this case, it is not for every $b \in \mathbb{R}^3$.