数据科学与机器学习的线性代数 (LINEAR ALGEBRA FOR DATA SCIENCE AND MACHINE LEARNING)

We are going to treat two of the most used calculations for matrices, multiplications, and inversion, let’s start with multiplication and how to do it in distinct ways.

我们将处理两个最常用的矩阵，乘法和求逆计算，让我们从乘法以及如何以不同的方式开始。

矩阵乘法 (Matrix Multiplication)

To be able to multiplicate matrices, their sizes have to be compatible, the number of rows of the first matrix has to match the number of columns of the second matrix.

为了能够相乘矩阵，它们的大小必须兼容，第一矩阵的行数必须与第二矩阵的列数匹配。

We will run all the examples on the same two 2 by 2 matrixes:

我们将在相同的两个2 x 2矩阵上运行所有示例：

点积 (Dot product)

The first way to multiplicate them is by using the dot product, that is, multiplicate every row per every column and the index that matches between the two vectors is the position of the result, let’s calculate C = A B.

将它们相乘的第一种方法是使用点积，即将每列的每一行相乘，并且两个向量之间匹配的索引是结果的位置，让我们计算C = AB 。

Let’s explain how to calculate c11 and c12,

让我们解释一下如何计算c11和c12 ，

After calculating all the dot products, using the next mathematical expression, we get the C matrix.

在计算所有点积之后，使用下一个数学表达式，我们得到C矩阵。

向量乘法 (Vector multiplication)

As we’ve seen in the dot product, we are multiplying rows per columns, the dot product can be upgraded by using vector products, that have fastest computing times, so let’s think about matrix multiplications as vectors, where:

正如我们在点积中所看到的，我们在每列中增加行，可以使用具有最快计算时间的矢量积来升级点积，因此让我们考虑将矩阵乘法作为矢量，其中：

块乘法 (Block multiplication)

This last strategy for multiplication is not intuitive, but it ends up doing the same multiplications and getting the same results, the strategy here is to divide the matrixes into compatible submatrices for the multiplication, for example, if you have two 10 by 10 matrix, you can divide it as 4 5 by 5 matrices and do the following:

最后一种乘法策略不直观，但最终会进行相同的乘法并获得相同的结果，此处的策略是将矩阵划分为兼容的子矩阵进行乘法运算，例如，如果您有两个10 x 10矩阵，您可以将其除以4 5除以5矩阵，然后执行以下操作：

矩阵求逆 (Matrix inversion)

To be reversible, a matrix has to have the same number of rows and columns and there should be no linear combination in their rows or columns.

为了可逆，矩阵必须具有相同数量的行和列，并且其行或列中不应有线性组合。

To invert a matrix we use the following condition, which says that a matrix by the inverse of it we get the identity matrix. The identity matrix is the one that’s composed of 1 at the diagonal.

为了使矩阵求逆，我们使用以下条件，即通过逆矩阵可以得到单位矩阵。 单位矩阵是由对角线1组成的矩阵。

The easiest way to check if a matrix has no linear combinations and is invertible is calculate her determinant, if it’s 0, it’s not invertible.

检查矩阵是否没有线性组合并且是可逆的，最简单的方法是计算其行列式，如果为0，则表示不可逆。

To calculate the inverse, the first idea becomes use the property:

要计算逆，首先要使用属性：

So, we can obtain the inverse solving the equation system that we get when we multiplicate the matrices:

因此，当矩阵相乘时，我们可以获得方程组的逆求解：

Instead of using this method, we can get the inverse using a better strategy, following the next steps:

代替使用此方法，我们可以按照以下步骤使用更好的策略获得逆函数：

• Create an augmented matrix adding the identity matrix a the right of A:创建一个增强矩阵，在A的右边添加单位矩阵：

• Use linear combinations to get the identity matrix at the left and the resulting right matrix will be the inverse of A:使用线性组合在左侧获得恒等矩阵，而所得的右矩阵将为A的逆：

So, the inverse is:

因此，相反是：

摘要 (Summary)

We learned how to multiplicate and invert matrices. This is the real basis of lineal algebra methods used in data science, on top of that, we will build the methods that all deep learning models use.

我们学习了如何对矩阵进行乘法和求逆。 这是数据科学中线性代数方法的真正基础，最重要的是，我们将构建所​​有深度学习模型都使用的方法。

This is the fifteenth post of my particular #100daysofML, I will be publishing the advances of this challenge at GitHub, Twitter, and Medium (Adrià Serra).

这是我特别＃第十五后100daysofML，我会发布在GitHub上，Twitter和中型企业(这一挑战的进步阿德里亚塞拉 )。

https://twitter.com/CrunchyML

https://twitter.com/CrunchyML

https://github.com/CrunchyPistacho/100DaysOfML

https://github.com/CrunchyPistacho/100DaysOfML