Linear Algebra in Python: Matrix Inverses and Least Squares – Real Python (2024)

Getting Started With Linear Algebra in Python

Linear algebra is a branch of mathematics that deals with linear equations and their representations using vectors and matrices. It’s a fundamental subject in several areas of engineering, and it’s a prerequisite to a deeper understanding of machine learning.

To work with linear algebra in Python, you can count on SciPy, which is an open-source Python library used for scientific computing, including several modules for common tasks in science and engineering.

Of course, SciPy includes modules for linear algebra, but that’s not all. It also offers optimization, integration, interpolation, and signal processing capabilities. It’s part of the SciPy stack, which includes several other packages for scientific computing, such as NumPy, Matplotlib, SymPy, IPython, and pandas.

scipy.linalg includes several tools for working with linear algebra problems, including functions for performing matrix calculations, such as determinants, inverses, eigenvalues, eigenvectors, and the singular value decomposition.

In the previous tutorial of this series, you learned how to work with matrices and vectors in Python to model practical problems using linear systems. You solved these problems using scipy.linalg.

In this tutorial, you’re going a step further, using scipy.linalg to study linear systems and build linear models for real-world problems.

In order to use scipy.linalg, you have to install and set up the SciPy library. Besides that, you’re going to use Jupyter Notebook to run the code in an interactive environment. SciPy and Jupyter Notebook are third-party packages that you need to install. For installation, you can use the conda or pip package manager. Revisit Working With Linear Systems in Python With scipy.linalg for installation details.

Next, you’ll go through some fundamental concepts of linear algebra and explore how to use Python to work with these concepts.

Understanding Vectors, Matrices, and the Role of Linear Algebra

A vector is a mathematical entity used to represent physical quantities that have both magnitude and direction. It’s a fundamental tool for solving engineering and machine learning problems. So are matrices, which are used to represent vector transformations, among other applications.

Note: In Python, NumPy is the most used library for working with matrices and vectors. It uses a special type called ndarray to represent them. As an example, imagine that you need to create the following matrix:

With NumPy, you can use np.array() to create it, providing a nested list containing the elements of each row of the matrix:

Python

In [1]: import numpy as npIn [2]: np.array([[1, 2], [3, 4], [5, 6]])Out[2]:array([[1, 2], [3, 4], [5, 6]])

NumPy provides several functions to facilitate working with vector and matrix computations. You can find more information on how to use NumPy to represent vectors and matrices and perform operations with them in the previous tutorial in this series.

A linear system or, more precisely, a system of linear equations, is a set of equations linearly relating to a set of variables. Here’s an example of a linear system relating to the variables x₁ and x₂:

Here you have two equations involving two variables. In order to have a linear system, the values that multiply the variables x₁ and x₂ must be constants, like the ones in this example. It’s common to write linear systems using matrices and vectors. For example, you can write the previous system as the following matrix product:

Comparing the matrix product form with the original system, you can notice the elements of matrix A correspond to the coefficients that multiply x₁ and x₂. Besides that, the values in the right-hand side of the original equations now make up vector b.

Linear algebra is a mathematical discipline that deals with vectors, matrices, and vector spaces and linear transformations more generally. By using linear algebra concepts, it’s possible to build algorithms to perform computations for several applications, including solving linear systems.

When there are just two or three equations and variables, it’s feasible to perform the calculations manually, combine the equations, and find the values for the variables.

However, in real-world applications, the number of equations can be very large, making it infeasible to do calculations manually. That’s precisely when linear algebra concepts and algorithms come handy, allowing you to develop usable applications for engineering and machine learning, for example.

In Working With Linear Systems in Python With scipy.linalg, you’ve seen how to solve linear systems using scipy.linalg.solve(). Now you’re going to learn how to use determinants to study the possible solutions and how to solve problems using the concept of matrix inverses.

Solving Problems Using Matrix Inverses and Determinants

Matrix inverses and determinants are tools that allow you to get some information about the linear system and also to solve it. Before going through the details on how to calculate matrix inverses and determinants using scipy.linalg, take some time to remember how to use these structures.

Using Matrix Inverses to Solve Linear Systems

To understand the idea behind the inverse of a matrix, start by recalling the concept of the multiplicative inverse of a number. When you multiply a number by its inverse, you get 1 as the result. Take 3 as an example. The inverse of 3 is 1/3, and when you multiply these numbers, you get 3 × 1/3 = 1.

With square matrices, you can think of a similar idea. However, instead of 1, you’ll get an identity matrix as the result. An identity matrix has ones in its diagonal and zeros in the elements outside of the diagonal, like the following examples:

The identity matrix has an interesting property: when multiplied by another matrix A of the same dimensions, the obtained result is A. Recall that this is also true for the number 1, when you consider the multiplication of numbers.

This allows you to solve a linear system by following the same steps used to solve an equation. As an example, consider the following linear system, written as a matrix product:

By calling A⁻¹ the inverse of matrix A, you could multiply both sides of the equation by A⁻¹, which would give you the following result:

This way, by using the inverse, A⁻¹, you can obtain the solution x for the system by calculating A⁻¹b.

It’s worth noting that while non-zero numbers always have an inverse, not all matrices have an inverse. When the system has no solution or when it has multiple solutions, the determinant of A will be zero, and the inverse, A⁻¹, won’t exist.

Now you’ll see how to use Python with scipy.linalg to make these calculations.

Calculating Inverses and Determinants With scipy.linalg

You can calculate matrix inverses and determinants using scipy.linalg.inv() and scipy.linalg.det().

For example, consider the meal plan problem that you worked on in the previous tutorial of this series. Recall that the linear system for this problem could be written as a matrix product:

Previously, you used scipy.linalg.solve() to obtain the solution 10, 10, 20, 20, 10 for the variables x₁ to x₅, respectively. But as you’ve just learned, it’s also possible to use the inverse of the coefficients matrix to obtain vector x, which contains the solutions for the problem. You have to calculate x = A⁻¹b, which you can do with the following program:

 1In [1]: import numpy as np 2 ...: from scipy import linalg 3 4In [2]: A = np.array( 5 ...:  [ 6 ...:  [1, 9, 2, 1, 1], 7 ...:  [10, 1, 2, 1, 1], 8 ...:  [1, 0, 5, 1, 1], 9 ...:  [2, 1, 1, 2, 9],10 ...:  [2, 1, 2, 13, 2],11 ...:  ]12 ...: )1314In [3]: b = np.array([170, 180, 140, 180, 350]).reshape((5, 1))1516In [4]: A_inv = linalg.inv(A)1718In [5]: x = A_inv @ b19 ...: x20Out[5]:21array([[10.],22 [10.],23 [20.],24 [20.],25 [10.]])

Here’s a breakdown of what’s happening:

• Lines 1 and 2 import NumPy as np, along with linalg from scipy. These imports allow you to use linalg.inv().

• Lines 4 to 12 create the coefficients matrix as a NumPy array called A.

• Line 14 creates the independent terms vector as a NumPy array called b. To make it a column vector with five elements, you use .reshape((5, 1)).

• Line 16 uses linalg.inv() to obtain the inverse of matrix A.

• Lines 18 and 19 use the @ operator to perform the matrix product in order to solve the linear system characterized by A and b. You store the result in x, which is printed.

  See Also

  5.6: Best Approximation and Least Squares6.5: Orthogonal Least SquaresThe Linear Algebra View of Least-Squares Regression6.5: The Method of Least Squares

You get exactly the same solution as the one provided by scipy.linalg.solve(). Because this system has a unique solution, the determinant of matrix A must be different from zero. You can confirm that it is by calculating it using det() from scipy.linalg:

In [6]: linalg.det(A)Out[6]:45102.0

As expected, the determinant isn’t zero. This indicates that the inverse of A, denoted as A⁻¹ and calculated with inv(A), exists, so the system has a unique solution. A⁻¹ is a square matrix with the same dimensions as A, so the product of A⁻¹ and A results in an identity matrix. In this example, it’s given by the following:

In [7]: A_invOut[7]:array([[-0.01077558, 0.10655847, -0.03565252, -0.0058534 , -0.00372489], [ 0.11287748, -0.00512172, -0.04010909, -0.00658507, -0.0041905 ], [ 0.0052991 , -0.01536517, 0.21300608, -0.01975522, -0.0125715 ], [-0.0064077 , -0.01070906, -0.02325839, -0.01376879, 0.08214713], [-0.00931223, -0.01902355, -0.00611946, 0.1183983 , -0.01556472]])

Now that you know the basics of using matrix inverses and determinants, you’ll see how to use these tools to find the coefficients of polynomials.

Interpolating Polynomials With Linear Systems

You can use linear systems to calculate polynomial coefficients so that these polynomials include some specific points.

For example, consider the second-degree polynomial y = P(x) = a₀ + a₁x + a₂x². Recall that when you plot a second-degree polynomial, you get a parabola, which will be different depending on the coefficients a₀, a₁, and a₂.

Now, suppose that you’d like to find a specific second-degree polynomial that includes the (x, y) points (1, 5), (2, 13), and (3, 25). How could you calculate a₀, a₁, and a₂, such that P(x) includes these points in its parabola? In other words, you want to find the coefficients of the polynomial in this figure:

For each point that you’d like to include in the parabola, you can use the general expression of the polynomial in order to get a linear equation. For example, taking the second point, (x=2, y=13), and considering that y = a₀ + a₁x + a₂x², you could write the following equation:

This way, for each point (x, y), you’ll get an equation involving a₀, a₁, and a₂. Because you’re considering three different points, you’ll end up with a system of three equations:

To check if this system has a unique solution, you can calculate the determinant of the coefficients matrix and check if it’s not zero. You can do that with the following code:

In [1]: import numpy as np ...: from scipy import linalgIn [2]: A = np.array([[1, 1, 1], [1, 2, 4], [1, 3, 9]])In [3]: linalg.det(A)Out[3]:1.9999999999999996

It’s worth noting that the existence of the solution only depends on A. Because the value of the determinant isn’t zero, you can be sure that there’s a unique solution for the system. You can solve it using the matrix inverse method with the following code:

In [4]: b = np.array([5, 13, 25]).reshape((3, 1))In [5]: a = linalg.inv(A) @ b ...: aOut[5]:array([[1.], [2.], [2.]])

This result tells you that a₀ = 1, a₁ = 2, and a₂ = 2 is a solution for the system. In other words, the polynomial that includes the points (1, 5), (2, 13), and (3, 25) is given by y = P(x) = 1 + 2x + 2x². You can test the solution for each point by inputting x and verifying that P(x) is equal to y.

As an example of a system without any solution, say that you’re trying to interpolate a parabola with the (x, y) points given by (1, 5), (2, 13), and (2, 25). If you look carefully at these numbers, you’ll notice that the second and third points consider x = 2 and different values for y, which makes it impossible to find a function that includes both points.

Following the same steps as before, you’ll arrive at the equations for this system, which are the following:

To confirm that this system doesn’t present a unique solution, you can calculate the determinant of the coefficients matrix with the following code:

In [6]: A = np.array([[1, 1, 1], [1, 2, 4], [1, 2, 4]]) ...: linalg.det(A)Out[6]:0.0

You may notice that the value of the determinant is zero, which means that the system doesn’t have a unique solution. This also means that the inverse of the coefficients matrix doesn’t exist. In other words, the coefficients matrix is singular.

Depending on your computer architecture, you may get a very small number instead of zero. This happens due to the numerical algorithms that det() uses to calculate the determinant. In these algorithms, numeric precision errors make this result not exactly equal to zero.

In general, whenever you come across a tiny number, you can conclude that the system doesn’t have a unique solution.

You can try to solve the linear system using the matrix inverse method with the following code:

In [7]: b = np.array([5, 13, 25]).reshape((3, 1))In [8]: x = linalg.inv(A) @ b---------------------------------------------------------------------------LinAlgError Traceback (most recent call last)<ipython-input-10-e6ee9b06a6fe> in <module>----> 1 x = linalg.inv(A) @ bLinAlgError: singular matrix

Because the system has no solution, you get an exception telling you that the coefficients matrix is singular.

When the system has more than one solution, you’ll come across a similar result. The value of the determinant of the coefficients matrix will be zero or very small, indicating that the coefficients matrix again is singular.

As an example of a system with more than one solution, you can try to interpolate a parabola considering the points (x, y) given by (1, 5), (2, 13), and (2, 13). As you may notice, here you’re considering two points at the same position, which allows an infinite number of solutions for a₀, a₁, and a₂.

Now that you’ve gone through how to work with polynomial interpolation using linear systems, you’ll see another technique that makes an effort to find the coefficients for any set of points.

Minimizing Error With Least Squares

You’ve seen that sometimes you can’t find a polynomial that fits precisely to a set of points. However, usually when you’re trying to interpolate a polynomial, you’re not interested in a precise fit. You’re just looking for a solution that approximates the points, providing the minimum error possible.

This is generally the case when you’re working with real-world data. Usually, it includes some noise caused by errors that occur in the collecting process, like imprecision or malfunction in sensors, and typos when users are inputting data manually.

Using the least squares method, you can find a solution for the interpolation of a polynomial, even when the coefficients matrix is singular. By using this method, you’ll be looking for the coefficients of the polynomial that provides the minimum squared error when comparing the polynomial curve to your data points.

Actually, the least squares method is generally used to fit polynomials to large sets of data points. The idea is to try to design a model that represents some observed behavior.

Note: If a linear system has a unique solution, then the least squares solution will be equal to that unique solution.

For example, you could design a model to try to predict car prices. For that, you could collect some real-world data, including the car price and some other features like the mileage, the year, and the type of car. With this data, you can design a polynomial that models the price as a function of the other features and use least squares to find the optimal coefficients of this model.

Soon, you’re going to work on a model to address this problem. But first, you’re going to see how to use scipy.linalg to build models using least squares.

Building Least Squares Models Using scipy.linalg

To solve least squares problems, scipy.linalg provides a function called lstsq(). To see how it works, consider the previous example, in which you tried to fit a parabola to the points (x, y) given by (1, 5), (2, 13), and (2, 25). Remember that this system has no solution, since there are two points with the same value for x.

Just like you did before, using the model y = a₀ + a₁x + a₂x², you arrive at the following linear system:

Using the least squares method, you can find a solution for the coefficients a₀, a₁, and a₂ that provides a parabola that minimizes the squared difference between the curve and the data points. For that, you can use the following code:

Python

 1In [1]: import numpy as np 2 ...: from scipy import linalg 3 4In [2]: A = np.array([[1, 1, 1], [1, 2, 4], [1, 2, 4]]) 5 ...: b = np.array([5, 13, 25]).reshape((3, 1)) 6 7In [3]: p, *_ = linalg.lstsq(A, b) 8 ...: p 9Out[3]:10array([[-0.42857143],11 [ 1.14285714],12 [ 4.28571429]])

In this program, you’ve set up the following:

See Also

Unit II: Least Squares, Determinants and Eigenvalues | Linear Algebra | Mathematics | MIT OpenCourseWare

• Lines 1 to 2: You import numpy as np and linalg from scipy in order to use linalg.lstsq().

• Lines 4 to 5: You create the coefficients matrix A using a NumPy array called A and the vector with the independent terms b using a NumPy array called b.

• Line 7: You calculate the least squares solution for the problem using linalg.lstsq(), which takes the coefficients matrix and the vector with the independent terms as input.

lstsq() provides several pieces of information about the system, including the residues, rank, and singular values of the coefficients matrix. In this case, you’re interested only in the coefficients of the polynomial to solve the problem according to the least squares criteria, which are stored in p.

As you can see, even considering a linear system that has no exact solution, lstsq() provides the coefficients that minimize the squared errors. With the following code, you can visualize the solution provided by plotting the parabola and the data points:

Python

 1In [4]: import matplotlib.pyplot as plt 2 3In [5]: x = np.linspace(0, 3, 1000) 4 ...: y = p[0] + p[1] * x + p[2] * x ** 2 5 6In [6]: plt.plot(x, y) 7 ...: plt.plot(1, 5, "ro") 8 ...: plt.plot(2, 13, "ro") 9 ...: plt.plot(2, 25, "ro")

This program uses matplotlib to plot the results:

• Line 1: You import matplotlib.pyplot as plt, which is typical.

• Lines 3 to 4: You create a NumPy array named x, with values ranging from 0 to 3, containing 1000 points. You also create a NumPy array named y with the corresponding values of the model.

• Line 6: You plot the curve for the parabola obtained with the model given by the points in the arrays x and y.

• Lines 7 to 9: In red ("ro"), you plot the three points used to build the model.

The output should be the following figure:

Notice how the curve provided by the model tries to approximate the points as well as possible.

Besides lstsq(), there are other ways to calculate least squares solutions using SciPy. One of the alternatives is using a pseudoinverse, which you’ll explore next.

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 1In [1]: import numpy as np 2 ...: from scipy import linalg 3 4In [2]: A = np.array([[1, 1, 1], [1, 2, 4], [1, 2, 4]]) 5 ...: b = np.array([5, 13, 25]).reshape((3, 1)) 6 7In [3]: A_pinv = linalg.pinv(A) 8 9In [4]: p2 = A_pinv @ b10 ...: p211Out[4]:12array([[-0.42857143],13 [ 1.14285714],14 [ 4.28571429]])

In [5]: A_pinvOut[5]:array([[ 1. , -0.14285714, -0.14285714], [ 0.5 , -0.03571429, -0.03571429], [-0.5 , 0.17857143, 0.17857143]])

Example: Predicting Car Prices With Least Squares

In this example, you’re going to build a model using least squares to predict the price of used cars using the data from the Used Cars Dataset. This dataset is a huge collection with 957 MB of vehicle listings from craigslist.org, including very different types of vehicles.

When working with real data, it’s often necessary to perform some steps of filtering and cleaning in order to use the data to build a model. In this case, it’s necessary to narrow down the types of vehicles that you’ll include, in order to get better results with your model.

Since your main focus here is on using least squares to build the model, you’ll start with a cleaned dataset, which is a small subset from the original one. Before you start working on the code, get the cleaned data CSV file by clicking the link below and navigating to vehicles_cleaned.csv:

In the downloadable materials, you can also check out the Jupyter Notebook to learn more about data preparation.

(linalg) $ conda install pandas

In [1]: import pandas as pd ...: cars_data = pd.read_csv("vehicles_cleaned.csv")

In [2]: cars_data.columnsOut[2]:Index(['price', 'year', 'condition', 'cylinders', 'fuel', 'odometer', 'transmission', 'size', 'type'], dtype='object')

In [3]: cars_data.iloc[0]Out[3]:price 7000year 2011condition goodcylinders 4 cylindersfuel gasodometer 76202transmission automaticsize compacttype sedanName: 0, dtype: object

In [4]: cars_data_dummies = pd.get_dummies( ...:  cars_data, ...:  columns=[ ...:  "condition", ...:  "cylinders", ...:  "fuel", ...:  "transmission", ...:  "size", ...:  "type", ...:  ], ...:  drop_first=True, ...: )

In [5]: cars_data_dummies.columnsOut[5]:Index(['price', 'year', 'odometer', 'condition_fair', 'condition_good', 'condition_like new', 'condition_new', 'condition_salvage', 'cylinders_6 cylinders', 'fuel_gas', 'transmission_manual', 'size_full-size', 'size_mid-size', 'size_sub-compact', 'type_hatchback', 'type_sedan', 'type_wagon'], dtype='object')

In [6]: cars_data_dummies["intercept"] = 1

In [7]: A = cars_data_dummies.drop(columns=["price"]).to_numpy() ...: b = cars_data_dummies.loc[:, "price"].to_numpy()

In [8]: from scipy import linalgIn [9]: p, *_ = linalg.lstsq(A, b) ...: pOut[9]:array([ 8.47362988e+02, -3.53913729e-02, -3.47144752e+03, -1.66981155e+03, -1.80240398e+02, -7.15885691e+03, -6.36540791e+03, 3.76583261e+03, -1.84837210e+03, 1.31935783e+03, 6.60484388e+02, 6.38913933e+02, 1.54163679e+02, -1.76423109e+03, -1.99439766e+03, 6.97365788e+02, -1.68998811e+06])

In [10]: p2 = linalg.pinv(A) @ b ...: p2Out[10]:array([ 8.47362988e+02, -3.53913729e-02, -3.47144752e+03, -1.66981155e+03, -1.80240398e+02, -7.15885691e+03, -6.36540791e+03, 3.76583261e+03, -1.84837210e+03, 1.31935783e+03, 6.60484388e+02, 6.38913933e+02, 1.54163679e+02, -1.76423109e+03, -1.99439766e+03, 6.97365788e+02, -1.68998811e+06])

In [11]: cars_data_dummies.drop(columns=["price"]).columnsOut[11]:Index(['year', 'odometer', 'condition_fair', 'condition_good', 'condition_like new', 'condition_new', 'condition_salvage', 'cylinders_6 cylinders', 'fuel_gas', 'transmission_manual', 'size_full-size', 'size_mid-size', 'size_sub-compact', 'type_hatchback', 'type_sedan', 'type_wagon', 'intercept'], dtype='object')

In [12]: import numpy as np ...: car = np.array( ...:  [2010, 50000, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1] ...: )

In [13]: predicted_price = p @ car ...: predicted_priceOut[13]:6159.510724281656

Conclusion

Congratulations! You’ve learned how to use some linear algebra concepts with Python to solve problems involving linear models. You’ve discovered that vectors and matrices are useful for representing data and that, by using linear systems, you can model practical problems and solve them in an efficient manner.

In this tutorial, you’ve learned how to:

• Study linear systems using determinants and solve problems using matrix inverses
• Interpolate polynomials to fit a set of points using linear systems
• Use Python to solve linear regression problems
• Use linear regression to predict prices based on historical data

Linear algebra is a very broad topic. For more information on some other linear algebra applications, check out the following resources:

• Working With Linear Systems in Python With scipy.linalg
• Scientific Python: Using SciPy for Optimization
• Hands-On Linear Programming: Optimization With Python
• NumPy, SciPy, and pandas: Correlation With Python

Keep studying, and feel free to leave any questions or comments below!