In machine learning and data science, we often use the Eigen theory. It is widely used in the data dimensionality reduction technique; PCA. Also, even Google’s search algorithm ‘PageRank’ is based on this concept. Once you read this article you will be able to understand what exactly the Eigen theory is and how it can be used to solve problems.

在机器学习和数据科学中，我们经常使用本征理论。 它被广泛用于数据降维技术中。 PCA。 同样，甚至谷歌的搜索算法“ PageRank ”也基于此概念。 阅读本文后，您将能够了解本征理论的确切含义以及如何将其用于解决问题。

In linear algebra, we can represent an object with combinations of multiple vectors. When we apply any kind of transformation on an object, we shall observe its impact on all the vectors.

线性代数中，我们可以表示与多个矢量的组合的对象。 当我们对对象应用任何类型的变换时，我们将观察其对所有向量的影响。

When a transformation is applied, some vectors end up staying in the same position. These vectors are called eigenvectors. The factor by which eigenvectors are scaled after the transformation is called eigenvalues.

应用转换后，某些向量最终会停留在相同位置。 这些向量称为特征向量。 变换后缩放特征向量的因素称为特征值。

Now let us understand this with an example of a square shown below(figure1). Although the square is made up of millions of vectors, for our ease of understanding we consider only 3 vectors r, s, v. Vector r, s, v are on plane1, plane3, plane2 respectively.

现在，让我们用下面所示的正方形示例来了解这一点( 图1 )。 尽管正方形由数百万个向量组成，但为了便于理解，我们仅考虑3个向量r，s，v 。 向量r，s，v分别在plane1，plane3，plane2上。

We apply transformation and let A be our transformation matrix.

我们应用变换，使A为变换矩阵。

Let’s have a very quick overview of what exactly the transformation matrix is.

让我们快速了解一下转换矩阵的确切含义。

When vector(s) are represented by a matrix, it must be read column-wise to identify vector(s).

当向量由矩阵表示时，必须逐列读取以识别向量。

Now, let’s see what exactly our transformation matrix A is doing to vector s. The x coordinate of vector s is 1 and the y-coordinate is also 1. (Make sure you are reading matrix A column-wise!)

现在，让我们看看我们的变换矩阵A对向量s到底做了什么。 向量s的x坐标为1 ， y坐标也为1。(确保您按列读取矩阵A ！)

When transformation matrix A is applied, x coordinate of s will move 1 unit in the x-direction and 0 unit in y-direction whereas y cordinate of s will move 0 unit in the x-direction and 2 unit in the y-direction.

当应用变换矩阵A时，s的x坐标将在x方向上移动1个单位，在y方向上移动0个单位，而s的y坐标将在x方向上移动0个单位，在y方向上移动2个单位。

So, we can see that after transformation coordinates of s are (1, 2).

因此，我们可以看到转换后的s坐标 是(1、2)。

The same transformation we apply to all the vector r, s, and v. Once the transformation is applied, our new coordinates are (0, 2), (1, 2), and (1, 0) respectively.

我们适用于所有的矢量R，S，与V相同的转变。 应用转换后，我们的新坐标分别为(0，2)，(1、2)和(1，0)。

As you can see in figure 2, vector r and v are on the same plane (the plane is displayed with a dotted green line; that is plane1 and plane 2) as before whereas vector s is not on the same plane. plane3 is the original plane of vector s but when the transformation is applied it has changed.

从图2中可以看到，向量r和v与以前在同一平面上(该平面显示有一条虚线绿色；即plane1和plane 2 )，而向量s不在同一平面上。 plane3是向量s的原始平面，但是当应用转换时，它已更改。

Please note that when I say on the same plane, it means that those vectors are linearly dependent. Vectors are considered linearly independent if and only if c1 and c2 (equation 0)are equal to zero else it is linearly dependent.

请注意，当我在同一平面上讲话时 ，这意味着这些向量是线性相关的 。 当且仅当c1和c2(等式0)等于零时，向量才被认为是线性独立的，否则它是线性相关的。

For example, we saw vector s is not on the same plane after the transformation. Initially, coordinates of s were (1, 1) later it became (1, 2), let’s call it s’. Look at the formula mentioned below.

例如，我们看到向量s在变换后不在同一平面上。 最初， s的坐标是(1，1)，后来变成(1，2)，我们称它为s' 。 看下面提到的公式。

Therefore, we can conclude that vector r and v are our eigenvectors as they ended up staying in the same position(check figure 2). The length of vector v is the same so the eigenvalue for v is 1 and the length of vector r has been doubled following the transformation so the eigenvalue of r is 2.

因此，我们可以得出结论，向量r和v是我们的特征向量，因为它们最终停留在相同的位置(请参见图2 )。 向量v的长度 是相同的，所以v的特征值 是1并且向量r的长度 变换后已加倍，因此r的特征值是2。

Now, we will calculate all the possible eigenvectors and eigenvalues for the same example.

现在 ，我们将为同一示例计算所有可能的特征向量和特征值。

Our transformation matrix is A. let x be our set of eigenvectors and λ represents eigenvalue(s).

我们的变换矩阵是A。 令x为我们的特征向量集， λ代表特征值。

The relation between these 3 terms can be expressed with this equation (equation 1).

这三个项之间的关系可以用该等式( 等式1 )表示。

Here there are 2 possibilities, either x = 0 or (A-λI) = 0. x is a set of eigenvectors that we are supposed to calculate so it can not be zero. Hence, we will consider (A-λI) = 0.

这里有2种可能，要么X = 0或(A-λI)= 0，x是一组特征向量，我们应该计算所以它不能为零。 因此，我们将考虑(A-λI)= 0。

To calculate eigenvectors and eigenvalues follow the steps mentioned below:

要计算特征向量和特征值，请遵循以下步骤：

• With the help of equation 2, calculate all the possible value(s) of λ.

  借助等式2 ，计算所有可能的λ值。

• Substitute the value of λ back to equation 1 and calculate x for each value of λ.

  将λ的值代入等式1，并为每个λ值计算x 。

So, now we got two eigenvalues 1 and 2. For both cases, we will calculate x from equation 1.

因此，现在我们得到两个特征值1和2。对于这两种情况，我们将从等式1计算x 。

For λ2, we calculated that (-x2, 0) = 0. It means that x2 must be zero, however, we do not care what y2 is; y2 can be anything (To verify, replace y2 with any random value, every time you get the same answer ‘0’!). Hence, (0, t) is eigenvectors’ set for eigenvalue 2, where t can be any number. Now check the vector r (Figure 2) from the previous example. For r, λ = 2 and eigenvector is (0, 2) which certainly verifies our above calculations.

对于λ2，我们计算(-x2，0)=0。这意味着x2必须为零，但是，我们不在乎y2是什么； y2可以是任何值(要验证，每次获得相同的答案“ 0”时，请将y2替换为任意随机值！)。 因此，(0，t)是特征值2的特征向量集，其中t可以是任何数字。 现在检查上一个示例中的向量r ( 图2 )。 对于r ， λ = 2，特征向量为(0，2)，这肯定验证了我们的上述计算。

The same explanation can be made for λ = 1 and can be verified by vector v from the previous example. (Figure 2)

可以对λ= 1进行相同的解释，并可以通过前面示例中的向量v进行验证。 ( 图2 )

行使： (Exercise:)

Calculate eigenvectors and eigenvalues for shear operation where the transformation matrix is A.

计算变换矩阵为A的剪切操作的特征向量和特征值。

Answer:

回答：

Set of eigenvectors are: (t, 0), for λ = 1. It has only one eigenvalue and one set of eigenvectors. Answer (t, 0) says that when transformation matrix T is applied, any vector on the x-axis is our eigenvector with 1 as an eigenvalue. In other words, whenever we apply transformation matrix T any vectors on the x-axis end up staying in the same position.

对于λ= 1 ，特征向量集为：(t，0) 。它只有一个特征值和一组特征向量。 答案(t，0)表示，当应用变换矩阵T时，x轴上的任何向量都是我们的特征向量，特征值为1。 换句话说，每当我们应用变换矩阵T时，x轴上的任何向量最终都停留在相同的位置。

Let’s see graphical representation:

让我们看一下图形表示：

As you can see in Figure 4 after the transformation is applied, vector r and s have changed their respective plane; vector r moved from plane 2 to plane 3 and vector s moved from plane 3 to plane 4. In terms of linear algebra, r and r’, s and s’ are linearly independent. However, if you see vector v is on the same plane(Do check figure 4) so it is our eigenvector with eigenvalue of 1. Do check out this animation.

如图4所示 ，应用变换后，向量r和s改变了它们各自的平面。 向量r从平面2移动到平面3 ，向量s从平面3移动到平面4 。 就线性代数而言， r和r' ， s和s'是线性独立的 。 但是，如果您看到向量v在同一平面上(请检查图4)，那么它就是我们的特征向量为1的特征向量。请检查此动画。

真实示例(或幻想世界示例！？！) (Real-World Example (Or Fantasy World Example!?!))

Let’s assume that we are in 2050 and as per Elon Musk’s Mars plan, there are 1 million people on Mars where birth and death rates are 15% and 12% respectively, and 1% of people moving back to the Earth(may be they don’t like Mars!). Also, there are 9 billion people on Earth where birth and death rates are 12% and 10% respectively and 5% of people are moving to Mars every year. Now my question is what would be the population of Earth and Mars in 2100?

假设我们到2050年，并且按照埃隆·马斯克(Elon Musk)的火星计划 ， 火星上有100万人的出生率和死亡率分别为15％和12％，并且有1％的人返回地球(也许他们没有不喜欢火星！)。 此外，地球上每年有90亿人的出生率和死亡率分别为12％和10％，每年有5％的人移居火星。 现在我的问题是2100年地球和火星的人口将是多少？

Firstly, we need to put the data into a proper format so we will calculate the transformation matrix as mentioned below:

首先，我们需要将数据放入适当的格式，因此我们将如下所述计算转换矩阵：

Where dE/dT means a change in Earth’s population per year and dM/dT means a change in Mars’s population per year.

dE / dT表示每年地球人口的变化，而dM / dT表示每年火星人口的变化。

Transformation matrix T represents vector E(Earth’s) and M(Mars’s). (Read column-wise)

变换矩阵T表示向量E(地球)和M(火星)。 (逐栏阅读)

Today(I mean in 2050!), the population of Earth and Mars is 9 billion and 1 million respectively. So vector P would be (9B, 1M). To calculate the population in 2100 we need to perform T⁵⁰ x P.

今天(我的意思是在2050年！)，地球和火星的人口分别为90亿和100万。 因此向量P为(9B，1M)。 为了计算2100年的人口，我们需要执行T⁵⁰xP。

So in 2100, the population of Earth and Mars would be 5.6 billion and 26.4 billion respectively. Luckily in this problem, we dealt with just 2x2 matrix, however, in the real-world matrix can be so huge that let our systems run out space and time, if still, we use this method!

因此，到2100年，地球和火星的人口分别为56亿和264亿。 幸运的是，在此问题中，我们只处理了2x2矩阵，但是，在现实世界中，矩阵可能是如此之大，以至于我们的系统用尽了空间和时间，即使如此，我们仍然使用这种方法！

Now just imagine what if we had a diagonal matrix? Then finding nth power of the matrix would be so easy, right?

现在，假设我们有一个对角矩阵怎么办？ 那么找到矩阵的n次方将非常容易，对吧？

Here eigen theory comes into the picture. To find the nth power of matrix T we will use eigenvectors and eigenvalues. This method called Diagonalization of the matrix. Steps are mentioned below:

在这里， 本征理论就出现了。 为了找到矩阵T的n次幂，我们将使用特征向量和特征值。 这种方法称为矩阵的对角线化。 步骤如下：

• Change the basis vectors of matrix T and let eigenvectors be a new basis vector or, in another word, moving vectors from the standard coordinate system (1,0), (0,1) to (x1, y1), (x2, y2). Call this matrix D.

  更改矩阵T的基本向量，并将特征向量设为新的基本向量，或者换句话说，将向量从标准坐标系(1,0)，(0,1)移至( x1，y1)，(x2，y2 )。 将此矩阵称为D。

• Let C be a diagonal matrix which is nothing but a matrix of eigenvectors.

  令C为对角矩阵，它仅是特征向量矩阵。

• Calculate the nth power of matrix C (in our case n is 50).

  计算矩阵C的n次幂(在我们的情况下，n为50)。

• Again change the basis and bring back vectors to the standard coordinate system/ or whatever coordinate system was used before.再次更改基础，然后将矢量恢复到标准坐标系/或以前使用的任何坐标系。

We have already prepared our transformation matrix T. Now we will calculate its eigenvalues and eigenvectors using the same method mentioned above. (However, calculation of eigenvectors is not mentioned here but you can directly calculate from here or review calculation 1.2 again). Calculations for matrix diagonalization is mentioned below:

我们已经准备好了变换矩阵T。现在，我们将使用上述相同的方法来计算其特征值和特征向量。 (但是，此处未提及特征向量的计算，但是您可以从此处直接进行计算，或者再次查看计算1.2 )。 矩阵对角化的计算方法如下：

Using this diagonalization, we have drastically reduced the number of computations and it just became so easy. Now just multiply T⁵⁰ with vector P (which denotes the populations). Results are here:

使用这种对角线化，我们大大减少了计算量，并且变得非常容易。 现在，只需将T⁵⁰乘以矢量P(表示总体)即可。 结果在这里：

That is it! Now we know what would be the population of Earth and Mars in 2100. Numbers are just fascinating, right! Also, there might be some minor errors in the answers from both the methods due to round off so just ignore that.

这就对了！ 现在我们知道2100年地球和火星的人口是多少。数字令人着迷，对！ 此外，由于四舍五入，两种方法的答案中可能都存在一些小错误，因此请忽略该错误。

Also, look at figure 5 to see how the population expands every year. (X-axis is for Earth’s population and Y-axis is for Mars’s population).

另外，请参见图5以查看人口每年的增长情况。 (X轴代表地球的人口，Y轴代表火星的人口)。

如何用Python计算？ (How to calculate in Python?)

In python, using numpy library we can calculate eigenvalues and eigenvectors.

在python中，使用numpy库，我们可以计算特征值和特征向量。

import numpy.linalg as la       #numpy’s linear algebra libraryimport numpy as npT = np.array([[0.97, 0.01]      #defining the transformatin matrix              [0.05, 1.02]])#calculating eigenvalues and eigenvectors of matrix TeigenValues, eigenVectors = la.eig(T) print(eigenValues, eigenVectors)

In case of any doubts or queries feel free to contact me at 7mayurpshah@gmail.com

如有任何疑问或疑问，请随时通过7mayurpshah@gmail.com与我联系。

Peace!

和平！