Practical lesson 2. Linear dependence of vectors and the concept of the basis of the vector system
In this lesson, we deal with the concept of linear dependence of vectors and the concept of the basis of the vector system.
Let's start with this simple task. The slide shows two vector systems. We are going to find out whether they are linearly dependent.
First, we are given a system in which there are two vectors a=(1, 1) and b=(2, 1). If they were linearly dependent, then one vector would be expressed in terms of the second one, which would mean that their lines would be proportional. We don't see any proportionality here, so these vectors are linearly independent. 1 - are linearly independent.
Let us consider another system. There are three vectors given. Let's write them down: a=(1, 1, 2), b=(0, 1, 0), and c=(1, 2, 2). The disproportionality of coordinates does not mean anything here. We can act by definition: we make a linear combination and then analyze when it is equal to the zero vector. However, we can see that the third vector c=(1, 2, 2) can be represented as the sum of the first two vectors. Indeed, if we add up the lines coordinately, the vector c is the sum of the vectors a and b. Since we have a vector that is expressed in terms of other vectors of this system, we conclude that the original system of vectors is linearly dependent. Thus, these three vectors a, b, and c are linearly dependent.
Now let's consider what a basis is. We take the same three vectors. Knowing that they are linearly dependent, we are going to find the basis. The basis is a part of a system that is linearly independent in itself, but any vector of a given system can be expressed in terms of these vectors.
We know that the vector c is expressed through the vectors a and b, which means that the following assumption arises: the vectors a and b might form the basis. Let's put this question. In fact, the answer is positive, since the coordinates of the vectors a and b are not proportional, so these vectors are linearly independent, and the vector c is expressed through a and b. The obvious fact is that when we say: "each vector is expressed in terms of the basis ones" - we understand that the vectors a and b can also be expressed in terms of a and b. What is the vector a? This is 1*a+0*b. The vector b is 0*a+1*b. Thus, by definition, we get that the vectors a and b form the basis.
Now we are going to consider a more complicated problem: to the 3 vectors already considered, we add 2 more: d and e. Let us find the basis for the resulting system of vectors. If we act by definition, we need to find a linearly independent subsystem through which we can express all the vectors of this system. Let's write down what we already have: the vector c is expressed in terms of a and b: c= a+b. The first assumption is that we can express the remaining vectors in terms of a and b. However, this assumption is incorrect, because if we add the vector e with coordinates (0, 0, 2) to the vectors a and b (I will write them under each other (see video)), then the matrix formed by these vectors (see the video) is echelon. We know that the rows of an echelon matrix are linearly independent, so the three vectors (a, b, e) are linearly independent. Therefore, we cannot express the vector e through a and b. Another assumption: a, b, and e might form the basis. We have expressed c in terms of a and b; we can also express the vector c in terms of e by writing the zero coefficient before it: c=a+b+0*e.
Now we need to consider the vector d. If we can express it in terms of the vectors a, b, and e, then we have proved that our triple of vectors forms the basis. The vector d has the coordinates (2, 2, 4). We can see that the vector d is proportional to the vector a, since by multiplying all coordinates of the vector a by 2, we get 2, 2, 4. Thus, the vector d is expressed in terms of a, and so through all the three vectors a, b and e. The vectors b and e are taken with zero coefficients. Taking this fact into account, and the fact that the vectors a, b, and e are linearly independent, we get the answer: the triple of vectors a b and e is the basis of this system of vectors. Note that if there are 3 vectors in this basis, then any basis has 3 vectors. However, this does not mean that any triple of vectors forms the basis. For example, the vectors a, b, and c will not form the basis.
One more task. We have a system of vectors a=(3, 2, 0), b=(2, 1, -1), c=(1, 2, 4) and d=(7, 5, 1). We need to find the rank and basis of this system. After we find the basis, we are going to express the remaining vectors in terms of the basis vectors. To solve the problem, we are going to use the following algorithm: we write our vectors into a matrix and reduce it to a stepwise form (see the video). To make elementary transformations easier I suggest putting the third line in the first place, that is, swapping the first and third lines. Since we have swapped the lines, I am going to mark the corresponding vector next to each line (see the video). Let us subtract the first two lines from the second, the first three from the third line, and the first seven lines from the fourth one (see the video). We get the following equivalent matrix: row 1: 1 2 4; row 2: 0 -3 -9; row 3: 0 -4 -12; row 4: 0 -9 -27. Note that we can reduce all three lines by some numbers. Row 2 is proportional to 3, and to avoid minuses, let us divide it by -3. Row 3 is proportional to 4. We divide it by -4. Finally, We divide row 4 by -9. We get: row 1: 1 2 4, row 2: 0 1 3, row 3: 0 1 3, and row 4: 0 1 3.
Thus, the last two rows are equal to row 2. In other words, after subtracting row 2 from the last two, we get zero rows. We have zero rows that can be crossed out. Let me write them down and cross them out. (See the video). (1 2 4, 0 1 3).
What do we have? Our original matrix has been transformed to a stepwise form. A step matrix has two rows, so its rank is 2. It means that the basis of the original system consists of two vectors. Thus, we need to find two vectors that are linearly independent. We can take any vectors with non-proportional coordinates. We see that there are no vectors with proportional coordinates here. It means any two vectors form the basis. I suggest taking the first vectors that we have written in the matrix b=(2, 1, -1) and c=(1, 2, 4). They form the basis. Thus, (b, c) is the basis.
Now our task is to express vectors a and d in terms of vectors b and c. I suggest doing this in two ways.
The first method is based on a definition. Here we have the vector d, and the vectors b and c. What does the phrase "to express vector d in terms of b and c" mean? It is to represent the vector d as a linear combination of the vectors b and c. That is, d=k1*b+k2*c. We do not know the coefficients k1 and k2. To find them, let's write down the coordinates of the vectors d, b, and c. We have the vector d=(7, 5, 1). In the right part, we need to multiply the vector b by k1, and we get: (2*k1, k1, -k1). We multiply k2 by the vector c and get: (k2, 2*k2, 4*k2). (7, 5, 1)=(2*k1, k1,- k1)+ (k2, 2*k2, 4*k2). Next, we add the first and the second elements in the right part, that is, the two n standing in the right part, and we get the following: (2*k1+k2, k1+2*k2,- k1+4*k2) (see the video). We get these three numbers. Since the left and right parts are equal, then we know what the corresponding coordinates are equal to and we can get the following three equations. Let's write down the resulting system of equations: the first coordinate: 2*k1+k2=7, the second coordinate: k1+2*k2=5, and finally the third coordinate: - k1+4*k2=1. (See the video). We can find the required coefficients from this system of equations.
Let's solve this system. I am not going to use the matrix here, but I will add the last two equations, or rather, add the second equation to the third one. I rewrite the first equation 2*k1+k2=7 and the second one k1+2*k2=5, and I add the second equation to the third one (-k1+k1)+(4*k2+2*k2)=1+5, and I get: 6*k2=6. (See the video). If we don't pay attention to the first equation (for a while), we have a step system without the first equation. We plug in k2=1 into the previous equation: k1+2=5, so k1=3. Now we need to check that the first equation is also true, so that the system is compatible. Let's check by substituting the found coefficients, 2*3+1=7. It all fits together. The system is solved by k1=3, k2=1. Note that we are looking for the vector d in the form of this combination: d=k1*b+k2*c. This means that the vector d has the form: d=3*b+1*c=3*b+c.
Thus, we decompose the vector d according to the basis. Let me remind you that the expansion coefficients are called vector coordinates. In this basis, the vector d has coordinates (3, 1) in the basis (b, c).
Now we have the vector a left. We can also decompose it by the basis according to the definition. However, I want to show you another way to do this, using the calculations that have already been performed. So, let's fix our answer, and remove intermediate calculations. Let's speculate. We want to express the vector a through the vectors b and C. When we transform our matrices, we act with rows, that is, we act with vectors. Where do we end up? Here is the vector a=(3, 2, 0), which is located in the third row. After several transformations, we get a zero row. So we can now do all these transformations with vectors and, in the end, get a zero vector, that is, get a linear combination of the vectors a, b and c; and using it we can easily express the vector a. Let us write down the transformations performed from the second row next to the corresponding row.
I subtract the first two rows, so b is 2*c (see the video). Then, I subtract the first three row from the third one: a-3*c. Since the vector d has already been found, I am not going to convert the fourth row. Then we divide the second row by -3, that is (b-2*c)/-3, and immediately transform (b-2*c)/-3=-1/3*b+2/3*c. This is the second row. Finally, the third row is: (a-3*c)/-4=-1/4*a+3/4*c. Then, we subtract the second row from the third row (see the video). Let's write down the specified transformation corresponding to the third row here, and subtract the second row from the third one. Here we have the second row that we had in the previous step. After this transformation, we get a null row, that is, a null vector: -1/ 4*a+3/4*c-(-1/3*b+2/3*c)=0 (see video). Let's transform the left part, taking into account the properties of operations. Thus, -1/4*a+3/4*c+1/3*b2/3*c=0. Let’s multiply, for example, by 12, so that there are no fractional numbers: -3a+9c+4b-8c=0, give similar ones, and we get c-3a+4b=0. We have the basis (b, c). We need to express the vector a: 3a=4b+c, a=4/3*b+1/3*c. The vector a is decomposed by the basis.