As discussed in the introductory notes, these series of lectures focuses both on using central Python packages like tensorflow and scikit-learn as well as writing your own codes for some central ML algorithms. The latter can be written in a language of your choice, be it Python, Julia, R, Rust, C++, Fortran etc. In order to avoid confusion however, in these lectures we will limit our attention to Python, C++ and Fortran.
There are several central software packages for linear algebra and eigenvalue problems. Several of the more popular ones have been wrapped into ofter software packages like those from the widely used text Numerical Recipes. The original source codes in many of the available packages are often taken from the widely used software package LAPACK, which follows two other popular packages developed in the 1970s, namely EISPACK and LINPACK. We describe them shortly here.
$$ \mathbf{A} = \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix}\qquad \mathbf{I} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$
The inverse of a matrix is defined by $$ \mathbf{A}^{-1} \cdot \mathbf{A} = I $$
| Relations | Name | matrix elements |
|---|---|---|
| \( A = A^{T} \) | symmetric | \( a_{ij} = a_{ji} \) |
| \( A = \left (A^{T} \right )^{-1} \) | real orthogonal | \( \sum_k a_{ik} a_{jk} = \sum_k a_{ki} a_{kj} = \delta_{ij} \) |
| \( A = A^{ * } \) | real matrix | \( a_{ij} = a_{ij}^{ * } \) |
| \( A = A^{\dagger} \) | hermitian | \( a_{ij} = a_{ji}^{ * } \) |
| \( A = \left (A^{\dagger} \right )^{-1} \) | unitary | \( \sum_k a_{ik} a_{jk}^{ * } = \sum_k a_{ki}^{ * } a_{kj} = \delta_{ij} \) |
For an \( N\times N \) matrix \( \mathbf{A} \) the following properties are all equivalent
import numpy as np
n = 10
x = np.random.normal(size=n)
print(x)
Here we have defined a vector \( x \) with \( n=10 \) elements with its values given by the Normal distribution \( N(0,1) \). Another alternative is to declare a vector as follows
import numpy as np
x = np.array([1, 2, 3])
print(x)
Here we have defined a vector with three elements, with \( x_0=1 \), \( x_1=2 \) and \( x_2=3 \). Note that both Python and C++ start numbering array elements from \( 0 \) and on. This means that a vector with \( n \) elements has a sequence of entities \( x_0, x_1, x_2, \dots, x_{n-1} \). We could also let (recommended) Numpy to compute the logarithms of a specific array as
import numpy as np
x = np.log(np.array([4, 7, 8]))
print(x)
Here we have used Numpy's unary function \( np.log \). This function is highly tuned to compute array elements since the code is vectorized and does not require looping. We normaly recommend that you use the Numpy intrinsic functions instead of the corresponding log function from Python's math module. The looping is done explicitely by the np.log function. The alternative, and slower way to compute the logarithms of a vector would be to write
import numpy as np
from math import log
x = np.array([4, 7, 8])
for i in range(0, len(x)):
x[i] = log(x[i])
print(x)
We note that our code is much longer already and we need to import the log function from the math module. The attentive reader will also notice that the output is \( [1, 1, 2] \). Python interprets automacally our numbers as integers (like the automatic keyword in C++). To change this we could define our array elements to be double precision numbers as
import numpy as np
x = np.log(np.array([4, 7, 8], dtype = np.float64))
print(x)
or simply write them as double precision numbers (Python uses 64 bits as default for floating point type variables), that is
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
print(x)
To check the number of bytes (remember that one byte contains eight bits for double precision variables), you can use simple use the itemsize functionality (the array \( x \) is actually an object which inherits the functionalities defined in Numpy) as
import numpy as np
x = np.log(np.array([4.0, 7.0, 8.0])
print(x.itemsize)
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
print(A)
If we use the shape function we would get \( (3, 3) \) as output, that is verifying that our matrix is a \( 3\times 3 \) matrix. We can slice the matrix and print for example the first column (Python organized matrix elements in a row-major order, see below) as
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
print(A[:,0])
We can continue this was by printing out other columns or rows. The example here prints out the second column
import numpy as np
A = np.log(np.array([ [4.0, 7.0, 8.0], [3.0, 10.0, 11.0], [4.0, 5.0, 7.0] ]))
# print the first column, row-major order and elements start with 0
print(A[1,:])
Numpy contains many other functionalities that allow us to slice, subdivide etc etc arrays. We strongly recommend that you look up the Numpy website for more details. Useful functions when defining a matrix are the np.zeros function which declares a matrix of a given dimension and sets all elements to zero
import numpy as np
n = 10
# define a matrix of dimension 10 x 10 and set all elements to zero
A = np.zeros( (n, n) )
print(A)
or initializing all elements to
import numpy as np
n = 10
# define a matrix of dimension 10 x 10 and set all elements to one
A = np.ones( (n, n) )
print(A)
or as unitarily distributed random numbers (see the material on random number generators in the statistics part)
import numpy as np
n = 10
# define a matrix of dimension 10 x 10 and set all elements to random numbers with x \in [0, 1]
A = np.random.rand(n, n)
print(A)
As we will see throughout these lectures, there are several extremely useful functionalities in Numpy. As an example, consider the discussion of the covariance matrix. Suppose we have defined three vectors \( \hat{x}, \hat{y}, \hat{z} \) with \( n \) elements each. The covariance matrix is defined as $$ \hat{\Sigma} = \begin{bmatrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{zx} & \sigma_{zy} & \sigma_{zz} \end{bmatrix}, $$ where for example $$ \sigma_{xy} =\frac{1}{n} \sum_{i=0}^{n-1}(x_i- \overline{x})(y_i- \overline{y}). $$ The Numpy function np.cov calculates the covariance elements using the factor \( 1/(n-1) \) instead of \( 1/n \) since it assumes we do not have the exact mean values. For a more in-depth discussion of the covariance and covariance matrix and its meaning, we refer you to the lectures on statistics. The following simple function uses the np.vstack function which takes each vector of dimension \( 1\times n \) and produces a $ 3\times n$ matrix \( \hat{W} \) $$ \hat{W} = \begin{bmatrix} x_0 & y_0 & z_0 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ \dots & \dots & \dots \\ x_{n-2} & y_{n-2} & z_{n-2} \\ x_{n-1} & y_{n-1} & z_{n-1} \end{bmatrix}, $$
which in turn is converted into into the \( 3 times 3 \) covariance matrix \( \hat{\Sigma} \) via the Numpy function np.cov(). In our review of statistical functions and quantities we will discuss more about the meaning of the covariance matrix. Here we note that we can calculate the mean value of each set of samples \( \hat{x} \) etc using the Numpy function np.mean(x). We can also extract the eigenvalues of the covariance matrix through the np.linalg.eig() function.
# Importing various packages
import numpy as np
n = 100
x = np.random.normal(size=n)
print(np.mean(x))
y = 4+3*x+np.random.normal(size=n)
print(np.mean(y))
z = x**3+np.random.normal(size=n)
print(np.mean(z))
W = np.vstack((x, y, z))
Sigma = np.cov(W)
print(Sigma)
Eigvals, Eigvecs = np.linalg.eig(Sigma)
print(Eigvals)
import numpy as np
import matplotlib.pyplot as plt
from scipy import sparse
eye = np.eye(4)
print(eye)
sparse_mtx = sparse.csr_matrix(eye)
print(sparse_mtx)
x = np.linspace(-10,10,100)
y = np.sin(x)
plt.plot(x,y,marker='x')
plt.show()
We have an \( N\times N \) matrix A with \( N=100 \) In C/C++ this would be defined as
int N = 100;
double A[100][100];
// initialize all elements to zero
for(i=0 ; i < N ; i++) {
for(j=0 ; j < N ; j++) {
A[i][j] = 0.0;
Note the way the matrix is organized, row-major order.
We have \( N\times N \) matrices A, B and C and we wish to evaluate \( A=B+C \). $$ \mathbf{A}= \mathbf{B}\pm\mathbf{C} \Longrightarrow a_{ij} = b_{ij}\pm c_{ij}, $$ In C/C++ this would be coded like
for(i=0 ; i < N ; i++) {
for(j=0 ; j < N ; j++) {
a[i][j] = b[i][j]+c[i][j]
We have \( N\times N \) matrices A, B and C and we wish to evaluate \( A=BC \). $$ \mathbf{A}=\mathbf{BC} \Longrightarrow a_{ij} = \sum_{k=1}^{n} b_{ik}c_{kj}, $$ In C/C++ this would be coded like
for(i=0 ; i < N ; i++) {
for(j=0 ; j < N ; j++) {
for(k=0 ; k < N ; k++) {
a[i][j]+=b[i][k]*c[k][j];
At least three possibilities in this course
int N;
double ** A;
A = new double*[N]
for ( i = 0; i < N; i++)
A[i] = new double[N];
Always free space when you don't need an array anymore.
for ( i = 0; i < N; i++)
delete[] A[i];
delete[] A;
#include <iostream>
#include <armadillo>
using namespace std;
using namespace arma;
int main(int argc, char** argv)
{
mat A = randu<mat>(5,5);
mat B = randu<mat>(5,5);
cout << A*B << endl;
return 0;
For people using Ubuntu, Debian, Linux Mint, simply go to the synaptic package manager and install armadillo from there. You may have to install Lapack as well. For Mac and Windows users, follow the instructions from the webpage http://arma.sourceforge.net. To compile, use for example (linux/ubuntu)
c++ -O2 -o program.x program.cpp -larmadillo -llapack -lblas
where the -l option indicates the library you wish to link to.
For OS X users you may have to declare the paths to the include files and the libraries as
c++ -O2 -o program.x program.cpp -L/usr/local/lib -I/usr/local/include -larmadillo -llapack -lblas
#include <iostream>
#include "armadillo"
using namespace arma;
using namespace std;
int main(int argc, char** argv)
{
// directly specify the matrix size (elements are uninitialised)
mat A(2,3);
// .n_rows = number of rows (read only)
// .n_cols = number of columns (read only)
cout << "A.n_rows = " << A.n_rows << endl;
cout << "A.n_cols = " << A.n_cols << endl;
// directly access an element (indexing starts at 0)
A(1,2) = 456.0;
A.print("A:");
// scalars are treated as a 1x1 matrix,
// hence the code below will set A to have a size of 1x1
A = 5.0;
A.print("A:");
// if you want a matrix with all elements set to a particular value
// the .fill() member function can be used
A.set_size(3,3);
A.fill(5.0); A.print("A:");
mat B;
// endr indicates "end of row"
B << 0.555950 << 0.274690 << 0.540605 << 0.798938 << endr
<< 0.108929 << 0.830123 << 0.891726 << 0.895283 << endr
<< 0.948014 << 0.973234 << 0.216504 << 0.883152 << endr
<< 0.023787 << 0.675382 << 0.231751 << 0.450332 << endr;
// print to the cout stream
// with an optional string before the contents of the matrix
B.print("B:");
// the << operator can also be used to print the matrix
// to an arbitrary stream (cout in this case)
cout << "B:" << endl << B << endl;
// save to disk
B.save("B.txt", raw_ascii);
// load from disk
mat C;
C.load("B.txt");
C += 2.0 * B;
C.print("C:");
// submatrix types:
//
// .submat(first_row, first_column, last_row, last_column)
// .row(row_number)
// .col(column_number)
// .cols(first_column, last_column)
// .rows(first_row, last_row)
cout << "C.submat(0,0,3,1) =" << endl;
cout << C.submat(0,0,3,1) << endl;
// generate the identity matrix
mat D = eye<mat>(4,4);
D.submat(0,0,3,1) = C.cols(1,2);
D.print("D:");
// transpose
cout << "trans(B) =" << endl;
cout << trans(B) << endl;
// maximum from each column (traverse along rows)
cout << "max(B) =" << endl;
cout << max(B) << endl;
// maximum from each row (traverse along columns)
cout << "max(B,1) =" << endl;
cout << max(B,1) << endl;
// maximum value in B
cout << "max(max(B)) = " << max(max(B)) << endl;
// sum of each column (traverse along rows)
cout << "sum(B) =" << endl;
cout << sum(B) << endl;
// sum of each row (traverse along columns)
cout << "sum(B,1) =" << endl;
cout << sum(B,1) << endl;
// sum of all elements
cout << "sum(sum(B)) = " << sum(sum(B)) << endl;
cout << "accu(B) = " << accu(B) << endl;
// trace = sum along diagonal
cout << "trace(B) = " << trace(B) << endl;
// random matrix -- values are uniformly distributed in the [0,1] interval
mat E = randu<mat>(4,4);
E.print("E:");
// row vectors are treated like a matrix with one row
rowvec r;
r << 0.59499 << 0.88807 << 0.88532 << 0.19968;
r.print("r:");
// column vectors are treated like a matrix with one column
colvec q;
q << 0.81114 << 0.06256 << 0.95989 << 0.73628;
q.print("q:");
// dot or inner product
cout << "as_scalar(r*q) = " << as_scalar(r*q) << endl;
// outer product
cout << "q*r =" << endl;
cout << q*r << endl;
// sum of three matrices (no temporary matrices are created)
mat F = B + C + D;
F.print("F:");
return 0;
#include <iostream>
#include "armadillo"
using namespace arma;
using namespace std;
int main(int argc, char** argv)
{
cout << "Armadillo version: " << arma_version::as_string() << endl;
mat A;
A << 0.165300 << 0.454037 << 0.995795 << 0.124098 << 0.047084 << endr
<< 0.688782 << 0.036549 << 0.552848 << 0.937664 << 0.866401 << endr
<< 0.348740 << 0.479388 << 0.506228 << 0.145673 << 0.491547 << endr
<< 0.148678 << 0.682258 << 0.571154 << 0.874724 << 0.444632 << endr
<< 0.245726 << 0.595218 << 0.409327 << 0.367827 << 0.385736 << endr;
A.print("A =");
// determinant
cout << "det(A) = " << det(A) << endl;
// inverse
cout << "inv(A) = " << endl << inv(A) << endl;
double k = 1.23;
mat B = randu<mat>(5,5);
mat C = randu<mat>(5,5);
rowvec r = randu<rowvec>(5);
colvec q = randu<colvec>(5);
// examples of some expressions
// for which optimised implementations exist
// optimised implementation of a trinary expression
// that results in a scalar
cout << "as_scalar( r*inv(diagmat(B))*q ) = ";
cout << as_scalar( r*inv(diagmat(B))*q ) << endl;
// example of an expression which is optimised
// as a call to the dgemm() function in BLAS:
cout << "k*trans(B)*C = " << endl << k*trans(B)*C;
return 0;
We start with the linear set of equations $$ \mathbf{A}\mathbf{x} = \mathbf{w}. $$ We assume also that the matrix \( \mathbf{A} \) is non-singular and that the matrix elements along the diagonal satisfy \( a_{ii} \ne 0 \). Simple \( 4\times 4 \) example $$ \begin{bmatrix} a_{11}& a_{12} &a_{13}& a_{14}\\ a_{21}& a_{22} &a_{23}& a_{24}\\ a_{31}& a_{32} &a_{33}& a_{34}\\ a_{41}& a_{42} &a_{43}& a_{44}\\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \\ x_4 \\ \end{bmatrix} =\begin{bmatrix} w_1\\ w_2\\ w_3 \\ w_4\\ \end{bmatrix}. $$
The basic idea of Gaussian elimination is to use the first equation to eliminate the first unknown \( x_1 \) from the remaining \( n-1 \) equations. Then we use the new second equation to eliminate the second unknown \( x_2 \) from the remaining \( n-2 \) equations. With \( n-1 \) such eliminations we obtain a so-called upper triangular set of equations of the form $$ \begin{align} b_{11}x_1 +b_{12}x_2 +b_{13}x_3 + b_{14}x_4=&y_1 \nonumber \\ b_{22}x_2 + b_{23}x_3 + b_{24}x_4=&y_2 \nonumber \\ b_{33}x_3 + b_{34}x_4=&y_3 \nonumber \\ b_{44}x_4=&y_4. \nonumber \label{eq:gaussbacksub} \end{align} $$ We can solve this system of equations recursively starting from \( x_n \) (in our case \( x_4 \)) and proceed with what is called a backward substitution.
Our actual \( 4\times 4 \) example reads after the first operation $$ \begin{bmatrix} a_{11}& a_{12} &a_{13}& a_{14}\\ 0& (a_{22}-\frac{a_{21}a_{12}}{a_{11}}) &(a_{23}-\frac{a_{21}a_{13}}{a_{11}}) & (a_{24}-\frac{a_{21}a_{14}}{a_{11}})\\ 0& (a_{32}-\frac{a_{31}a_{12}}{a_{11}})& (a_{33}-\frac{a_{31}a_{13}}{a_{11}})& (a_{34}-\frac{a_{31}a_{14}}{a_{11}})\\ 0&(a_{42}-\frac{a_{41}a_{12}}{a_{11}}) &(a_{43}-\frac{a_{41}a_{13}}{a_{11}}) & (a_{44}-\frac{a_{41}a_{14}}{a_{11}}) \\ \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ x_3 \\ x_4 \\ \end{bmatrix} =\begin{bmatrix} y_1\\ w_2^{(2)}\\ w_3^{(2)} \\ w_4^{(2)}\\ \end{bmatrix}, $$ or $$ \begin{align} b_{11}x_1 +b_{12}x_2 +b_{13}x_3 + b_{14}x_4=&y_1 \nonumber \\ a^{(2)}_{22}x_2 + a^{(2)}_{23}x_3 + a^{(2)}_{24}x_4=&w^{(2)}_2 \nonumber \\ a^{(2)}_{32}x_2 + a^{(2)}_{33}x_3 + a^{(2)}_{34}x_4=&w^{(2)}_3 \nonumber \\ a^{(2)}_{42}x_2 + a^{(2)}_{43}x_3 + a^{(2)}_{44}x_4=&w^{(2)}_4, \nonumber \\ \label{_auto2} \end{align} $$
The new coefficients are $$ \begin{equation} b_{1k} = a_{1k}^{(1)} \quad k=1,\dots,n, \label{_auto3} \end{equation} $$ where each \( a_{1k}^{(1)} \) is equal to the original \( a_{1k} \) element. The other coefficients are $$ \begin{equation} a_{jk}^{(2)} = a_{jk}^{(1)}-\frac{a_{j1}^{(1)}a_{1k}^{(1)}}{a_{11}^{(1)}} \quad j,k=2,\dots,n, \label{_auto4} \end{equation} $$ with a new right-hand side given by $$ \begin{equation} y_{1}=w_1^{(1)}, \quad w_j^{(2)} =w_j^{(1)}-\frac{a_{j1}^{(1)}w_1^{(1)}}{a_{11}^{(1)}} \quad j=2,\dots,n. \label{_auto5} \end{equation} $$ We have also set \( w_1^{(1)}=w_1 \), the original vector element. We see that the system of unknowns \( x_1,\dots,x_n \) is transformed into an \( (n-1)\times (n-1) \) problem.
This step is called forward substitution. Proceeding with these substitutions, we obtain the general expressions for the new coefficients $$ \begin{equation} a_{jk}^{(m+1)} = a_{jk}^{(m)}-\frac{a_{jm}^{(m)}a_{mk}^{(m)}}{a_{mm}^{(m)}} \quad j,k=m+1,\dots,n, \label{_auto6} \end{equation} $$ with \( m=1,\dots,n-1 \) and a right-hand side given by $$ \begin{equation} w_j^{(m+1)} =w_j^{(m)}-\frac{a_{jm}^{(m)}w_m^{(m)}}{a_{mm}^{(m)}}\quad j=m+1,\dots,n. \label{_auto7} \end{equation} $$ This set of \( n-1 \) elimations leads us to an equations which is solved by back substitution. If the arithmetics is exact and the matrix \( \mathbf{A} \) is not singular, then the computed answer will be exact.
Even though the matrix elements along the diagonal are not zero, numerically small numbers may appear and subsequent divisions may lead to large numbers, which, if added to a small number may yield losses of precision. Suppose for example that our first division in \( (a_{22}-a_{21}a_{12}/a_{11}) \) results in \( -10^{-7} \) and that \( a_{22} \) is one. one. We are then adding \( 10^7+1 \). With single precision this results in \( 10^7 \).
The LU decomposition method means that we can rewrite this matrix as the product of two matrices \( \mathbf{L} \) and \( \mathbf{U} \) where $$ \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14} \\ a_{21} & a_{22} & a_{23} & a_{24} \\ a_{31} & a_{32} & a_{33} & a_{34} \\ a_{41} & a_{42} & a_{43} & a_{44} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ l_{21} & 1 & 0 & 0 \\ l_{31} & l_{32} & 1 & 0 \\ l_{41} & l_{42} & l_{43} & 1 \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} & u_{13} & u_{14} \\ 0 & u_{22} & u_{23} & u_{24} \\ 0 & 0 & u_{33} & u_{34} \\ 0 & 0 & 0 & u_{44} \end{bmatrix}. $$
LU decomposition forms the backbone of other algorithms in linear algebra, such as the solution of linear equations given by $$ \begin{align} a_{11}x_1 +a_{12}x_2 +a_{13}x_3 + a_{14}x_4=&w_1 \nonumber \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + a_{24}x_4=&w_2 \nonumber \\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 + a_{34}x_4=&w_3 \nonumber \\ a_{41}x_1 + a_{42}x_2 + a_{43}x_3 + a_{44}x_4=&w_4. \nonumber \end{align} $$ The above set of equations is conveniently solved by using LU decomposition as an intermediate step.
The matrix \( \mathbf{A}\in \mathbb{R}^{n\times n} \) has an LU factorization if the determinant is different from zero. If the LU factorization exists and \( \mathbf{A} \) is non-singular, then the LU factorization is unique and the determinant is given by $$ det\{\mathbf{A}\}=det\{\mathbf{LU}\}= det\{\mathbf{L}\}det\{\mathbf{U}\}=u_{11}u_{22}\dots u_{nn}. $$
There are at least three main advantages with LU decomposition compared with standard Gaussian elimination:
With the LU decomposition it is rather simple to solve a system of linear equations $$ \begin{align} a_{11}x_1 +a_{12}x_2 +a_{13}x_3 + a_{14}x_4=&w_1 \nonumber \\ a_{21}x_1 + a_{22}x_2 + a_{23}x_3 + a_{24}x_4=&w_2 \nonumber \\ a_{31}x_1 + a_{32}x_2 + a_{33}x_3 + a_{34}x_4=&w_3 \nonumber \\ a_{41}x_1 + a_{42}x_2 + a_{43}x_3 + a_{44}x_4=&w_4. \nonumber \end{align} $$
This can be written in matrix form as $$ \mathbf{Ax}=\mathbf{w}. $$
where \( \mathbf{A} \) and \( \mathbf{w} \) are known and we have to solve for \( \mathbf{x} \). Using the LU dcomposition we write $$ \mathbf{A} \mathbf{x} \equiv \mathbf{L} \mathbf{U} \mathbf{x} =\mathbf{w}. $$
The previous equation can be calculated in two steps $$ \mathbf{L} \mathbf{y} = \mathbf{w};\qquad \mathbf{Ux}=\mathbf{y}. $$
To show that this is correct we use to the LU decomposition to rewrite our system of linear equations as $$ \mathbf{LUx}=\mathbf{w}, $$ and since the determinant of \( \mathbf{L} \) is equal to 1 (by construction since the diagonals of \( \mathbf{L} \) equal 1) we can use the inverse of \( \mathbf{L} \) to obtain $$ \mathbf{Ux}=\mathbf{L^{-1}w}=\mathbf{y}, $$ which yields the intermediate step $$ \mathbf{L^{-1}w}=\mathbf{y} $$ and as soon as we have \( \mathbf{y} \) we can obtain \( \mathbf{x} \) through \( \mathbf{Ux}=\mathbf{y} \).
For our four-dimentional example this takes the form $$ \begin{align} y_1=&w_1 \nonumber\\ l_{21}y_1 + y_2=&w_2\nonumber \\ l_{31}y_1 + l_{32}y_2 + y_3 =&w_3\nonumber \\ l_{41}y_1 + l_{42}y_2 + l_{43}y_3 + y_4=&w_4. \nonumber \end{align} $$
and $$ \begin{align} u_{11}x_1 +u_{12}x_2 +u_{13}x_3 + u_{14}x_4=&y_1 \nonumber\\ u_{22}x_2 + u_{23}x_3 + u_{24}x_4=&y_2\nonumber \\ u_{33}x_3 + u_{34}x_4=&y_3\nonumber \\ u_{44}x_4=&y_4 \nonumber \end{align} $$
This example shows the basis for the algorithm needed to solve the set of \( n \) linear equations.
The algorithm goes as follows
ludcmp(double a, int n, int indx, double &d). This functions returns the LU decomposed matrix \( \bf A \), its determinant and the vector indx which keeps track of the number of interchanges of rows. If the determinant is zero, the solution is malconditioned.lubksb(double a, int n, int indx, double w) which uses the LU decomposed matrix \( \bf A \) and the vector \( \bf w \) and returns \( \bf x \) in the same place as \( \bf w \). Upon exit the original content in \( \bf w \) is destroyed. If you wish to keep this information, you should make a backup of it in your calling function.If the inverse exists then $$ \mathbf{A}^{-1}\mathbf{A}=\mathbf{I}, $$ the identity matrix. With an LU decomposed matrix we can rewrite the last equation as $$ \mathbf{LU}\mathbf{A}^{-1}=\mathbf{I}. $$
If we assume that the first column (that is column 1) of the inverse matrix can be written as a vector with unknown entries $$ \mathbf{A}_1^{-1}= \begin{bmatrix} a_{11}^{-1} \\ a_{21}^{-1} \\ \dots \\ a_{n1}^{-1} \\ \end{bmatrix}, $$ then we have a linear set of equations $$ \mathbf{LU}\begin{bmatrix} a_{11}^{-1} \\ a_{21}^{-1} \\ \dots \\ a_{n1}^{-1} \\ \end{bmatrix} =\begin{bmatrix} 1 \\ 0 \\ \dots \\ 0 \\ \end{bmatrix}. $$
In a similar way we can compute the unknow entries of the second column, $$ \mathbf{LU}\begin{bmatrix} a_{12}^{-1} \\ a_{22}^{-1} \\ \dots \\ a_{n2}^{-1} \\ \end{bmatrix}=\begin{bmatrix} 0 \\ 1 \\ \dots \\ 0 \\ \end{bmatrix}, $$ and continue till we have solved all \( n \) sets of linear equations.
#include <iostream>
#include "armadillo"
using namespace arma;
using namespace std;
int main()
{
mat A = randu<mat>(5,5);
vec b = randu<vec>(5);
A.print("A =");
b.print("b=");
// solve Ax = b
vec x = solve(A,b);
// print x
x.print("x=");
// find LU decomp of A, if needed, P is the permutation matrix
mat L, U;
lu(L,U,A);
// print l
L.print(" L= ");
// print U
U.print(" U= ");
//Check that A = LU
(A-L*U).print("Test of LU decomposition");
return 0;
}