Lecture 2. Homogeneous system of equations
This lecture is devoted to the algorithm for constructing a fundamental system of solutions to a homogeneous system of equations.
You should remember that the set of solutions to an arbitrary homogeneous system of equations forms a vector space. Let’s denote this space with the letter V. Let’s look at the main definitions. First, we know the concept of a basis. A basis on the set of solutions to a homogeneous system is called a fundamental system of solutions. The following theorem holds. We discussed it at the previous lecture. Let’s turn to it. Let there be n variables in our system, if the rank of this homogeneous system is less than the number of variables, this system of linear equations has a fundamental system of solutions, i.e. the basis on the set of solutions, and there are exactly n-r vectors in this basis. If the parameters n and r are equal, in this case, the system has only one trivial solution, the bases cannot be identified.
To construct a fundamental system of solutions, we use the following algorithm. First, we bring the system of equations to a step-by-step form, express the main variables through free variables, give the free variables some value so that they get n-r solutions to this system that will form a basis. In this case, these solutions must be linearly independent. Note that if the parameters r and n are equal, the system has only one trivial solution, the fundamental system of solutions cannot be constructed.
Let’s look at a number of examples. Let such a homogeneous system of linear equations is given. Let’s try to understand whether it is possible to construct a fundamental system for it, and if so, how exactly. We bring it to a step-by-step form. To do this, we will perform standard elementary transformations: subtract the first row from the second and third rows, we can add ½ of the second to the third row and get a step system. However, we can notice that since the last two rows are not proportional, we will not get a zero row after this transformation, i.e. there will be three independent rows in the step system. So, the system rank is 3, i.e. the rank is equal to the number of variables, and, hence, this system has only the trivial solution. There will be no other solutions, so we will not be able to identify the basis here. For our example, a fundamental system cannot be constructed.
Let’s consider the following example. There are 3 equations and 4 unknowns in this system. Note that the fundamental system of solutions definitely exists here because the rank of the system is certainly less than 4, than the number of variables since there are three equations in this system. Let’s define a basis on the set of solutions to this system. We write the coefficient matrix, but we only wrote the main matrix, i.e. it is possible not to write down the column of free terms equal to zero since all the elementary transformations that we perform on the matrix will not change this column, so we simply do not need it (it will only take up extra space). We have written the main matrix, and now we perform the specified transformations on it. We subtract the first row from the third one, subtract the two second rows from the third one, and we get the necessary step system. We fix it and note that there are 3 independent vectors in this system, i.e. the rank of the homogeneous system of equations is 3. This means that since the system has 4 variables, the basis of the solution space consists of exactly 1 vector.
We find this vector. To do this, we write out the resulting step system of equations and express the main variables x1, x2, x4, which corresponds to the leading elements of the matrix through a free variable, in our case, the free variable is x3. We express x4=0, and the variables x2 and x1 are expressed through the third one, i.e. it depends on the value of the variable x3. Since there is only one vector in the basis, we must take any non-trivial solution. To do this, we give the variable x3 any non-trivial value, e.g., 1, and express the remaining variables through this value: x1=-4, x2=-1, x4=0, regardless of x3. The resulting row gives us one basis vector with the coordinates (-4, -1, 1, 0). This vector will form the basis on the set of solutions. So, the fundamental system of solutions is constructed and consists of one vector. Since we know this vector, we can easily express the arbitrary solution. It turns out that any solution of the original system is proportional to the basis vector obtained, i.e. it takes the specified form. This solution can be written in this way if you multiply all the coordinates of the basis vector by the coefficient k (see the video).
Let’s take another example. Let’s take a more complex system of equations and carefully perform elementary transformations. Meanwhile, we repeat the Gauss method. We write a matrix of coefficients. We don’t take into account the zero column of free terms. We perform elementary transformations. After converting three rows, we get the matrix that you see in the video. Check out the transformation completed. Next, we subtract 3 second rows from the third one, and add 2 second rows to the fourth one. To get fewer negative numbers, multiply the second row by -1. As a result, it is the matrix that you see on the screen. This means that the second and third rows will be reset to zero, but the second row after multiplying by -1 will take the following form: 0 1 6 -5. Zero rows can be crossed out, thus, two rows are obtained in the step matrix. This means that the rank of this matrix, and, therefore, of this system of equations is 2, and, hence, the set of solutions to the system of equations has the basis in which there are two vectors.
We find these vectors. Let’s construct a fundamental system of solutions. Here, the variables x1, x2 are main, x3, x4 are free. Again, we express the main variables through free variables, give free variables some values to get a number of solutions to this system, and we need to get two solutions that will form the basis. Since we know that there are two vectors in the basis, we need to choose two linearly independent solutions.
In general, we use the following scheme: we draw the specified table and give some values to free variables in this table so that we get n-r linearly independent rows. The easiest way to do this is as follows: we substitute the so-called unit rows, i.e. one position is 1, and the rest are 0. In our case, we get 2 unit rows: 1 0 and 0 1. For each such a set of values, we calculate the value of the main variables. For the set 1 0, we get x1=8, x2=-6, for the second set of free variables, we get the following values of the main ones: -7 and 5. The first row gives us the first basis vector (8, -6, 1, 0), the second row – the second vector (-7, 5, 0, 1). Given that the unit rows form a step system, the resulting vectors are linearly independent, which means they form the basis.
Let’ draw some conclusion. For the original system of equations, the basis on the set of solutions has two vectors that make up the fundamental system of solutions. In this case, the general solution can be written as a linear combination of basis vectors, i.e. in the form shown on the slide. On the other hand, the same solution can be written differently. The problem is solved, and the fundamental system of solutions is constructed.