Practical lesson 1. Arithmetic vector space
Our practical lesson is devoted to the n-dimensional arithmetic space. We are going to solve a number of tasks to understand the concepts introduced at the lecture.
We can perform addition and multiplication operations with vectors. Here is the task. We have an equation in which the vectors a, b, and c are known, and the vector x is to be found and the specified equation should be correct.
In order to transform the equation, you need to use a number of properties that are valid in vector space. Namely, you need to open the brackets and multiply the vector by a scalar. Let's do it. We transform the left part of our equation: 2a-6x, open the second bracket, add the scalar 3, -3b + 3x =-x. Our task: we need to express the vector x. To do this, we present the similars on the left side, the properties allow us to do this. In other words, we take x out of the bracket and calculate 3-6. We get - 3x + 2 a -3 b = c-x. Then, we add the vector x to both parts. We can even do something else: we can add the 3x vector so that it disappears on the left, and subtract the c vector from both parts. In this case, we get 2s -3 b-c, and on the right we have 3x-x=2x. The next step is to express the vector x. We divide all the coefficients by 2. a – 3/2 b -½. Thus, the vector x is expressed through the vectors a, b, and c using the operations. In other words, we represent the vector x as a linear combination of the vectors a, b, and c.
Now we need to find this vector, taking into account a, b, and c. We substitute: x= (3, -1, 2) – (3/2, 3, -3/2) – (3/2, -1, -1/2). Then we calculate. 3/2 plus 3/2 is 3, from 3 we subtract 3, we get 0. The second coordinate is -3 and, finally, the third coordinate is 4. We have the final answer: the vector x is found, it is a triple (0, -3,4).
Let us consider another problem. Using the operations of adding vectors and multiplying them by a number, we need to get a zero vector having three other vectors, and we have to do it using different methods. First, we can get the zero vector using zero coefficients. Let me remind you that we denote the zero vector with a bold zero, so as not to write down all its coordinates. Obviously, if I multiply each of the vectors by 0, I get a zero vector. It is always true. Now we are going to get the zero vector using some other ways, so that the coefficients are not zero. To do this, we write down the equation that we want to have in the end. That is, we can multiply each vector by a scalar, and then we can add these vectors together. Our task is to choose the numbers k1, k2 and k3, so that in the end we get a zero vector, that is, we need three zeros. We decipher this equation by substituting the vectors a, b and c. We multiply k1 by the vector (1, -1, 1), k2 is multiplied by the vector (1, 2, 2), k3 is multiplied by the vector (2, 1, 3). This results in a zero vector with all coordinates equal to zero. Let us we again recall how operations are performed. Multiplying by a number means multiplying each coordinate (k1, - k1, k1) + (k2, 2k2, 2k2) + (2k3, k3, 3k). Again, everything equals to 0. Then we add it. We add coordinately. What is the first coordinate? k1+k2 + 2k3. What is it equal to? Zero. The situation is the same with the second coordinate. We get it. What is it equal to? It is equal to zero. The same is true for the third coordinate. In other words, we get a system of three equations. Let's write down this system. Let's write down the first coordinate k1+k2 + 2k3, it should turn to 0. The second coordinate k1+2k2 + k32 is also 0. Finally, the third coordinate is k1+2k2 + 3k3. This is also 0. Thus, the problem is reduced to the study of a system of linear equations with three unknowns, and all the free terms are equal to 0. This system is called a homogeneous system. Our task is to solve it. We already know how to solve such systems. To do this, we can use, for example, the Gauss method, which is based on the elementary transformations. Let's write a matrix of coefficients, and the last column is zero. Let's carry out the transformations: we add the first line to the second one, and subtract the first one from the third one. You must understand why we are doing this. I'm not talking about this in detail.
We get the second line 0 3 3 0, the third line changes, and we get 0 1 1 0. Note that the last column does not change because it contains all zeros. What do we see? The last two lines are proportional, that is, we can subtract three thirds from the second line, or subtract 1/3 of the second line from the third one. Then, we get a null row, and we cross it out. After that, we get a step matrix. I am not going to write the zero row, but I will write a step matrix immediately.
Let's go to the system. At the same time, I suggest dividing the second line by 3. This is the system we have. We can consider the variable k3 free. Let's express k2= - k3. From the first level, we express k1 = - k3. The system has infinitely many solutions. Here is one of them. Let the variable k3 have the value 1. Then k2 and k1 will be equal to -1. This means that a triple (-1, -1, 1) is the solution of this system. In other words, we have found a set of required coefficients. Thus, the combination -a - b + c gives a zero vector. It is clear that instead of k3, we could have set a different value, for example, -2. Then k2 and k1 would be equal to 2. We would get another decomposition of the zero vector: 2a + 2b-2c=0. Note that writing 0 in this equation, we understand it as a 0 vector, because another vector is on the left. This is the example of decomposition. It is clear that one can get infinitely many more decompositions in the same way. The required problem is solved. Note that these 3 vectors are somehow connected to each other. If we add the vectors a and b, that is, add their coordinates, we get the vector c. If we had understood this initially, we would not have had to solve our system. When we express one vector through another, as here, c = a + b, then we can get a linear combination from this equation, that is, we can express the zero vector. To do this, we transfer all the vectors to one part; we have the vectors a and b with the coefficients -1, that is - a - b + c, and the zero vector is on the right. That is, we have the same thing that we have received in the process of solving the task.
Another task. You need to get the vector with coordinates (5, λ, 7) using operations from the vectors with coordinates (1, 0, 2), (-1, 0, 1). You need to understand at what value of λ this is possible, and what coefficients in the decomposition will be? Let's write down the required equations. We want to get the vector (5, λ, 7) from the specified vectors. We follow the same way. We can multiply this vector by some numbers: (1, 0, 2) by some number k1, and another vector (-1, 0, 1) by some number k2. Then we add these vectors. We should get the required vector. Let's carry out the transformations in the right part. I described everything in detail in the previous task. Now I suggest reducing the transformations a little. That is, I immediately multiply the first coordinate by k1, do the same with the second summand and add them. Then I get the following three coordinates: (see the video). Pay your attention to the second coordinate. It's zero. Finally, the third coordinate is 2k1 + k2. Nothing has changed on the left: (5, λ, 7). What do we have? Using the operations, we can obtain the desired vector, which is written on the left, only when the second coordinate is equal to zero. No matter what numbers k1, k2 are, we have the zero. Thus, if λ is not 0, decomposition is impossible. λ=0 is the obligatory condition.
However, we still need to analyze whether we can find k1 and k2 so that the equation is correct. Let's analyze it. What do we know? The second coordinate is zero, the first coordinate is k1 - k2 = 5. 2 k1 + k2 =7. Again, we get a system of linear equations. We solve it. I don’t go to the matrix here. The system is simple. I carry out the simplest conversion: I add the second line to the first one, thereby getting rid of the k2 variable. We add up, and get 3k1 = 12. l rewrite the second equation. I get a system equivalent to this one. We find k1, which is 4, and express k2 by substituting it into the second equation. We have 2 by 4, it is 8, plus k2 is 7, hence k2 is -1. It turns out that the coefficients exist. For λ=0, the required representation is possible. Thus, the vector (5, 0, 7) can be expressed in terms of these vectors with the following coefficients: before the first vector, the coefficient is 4 (4(1, 0, 2)), the second vector has the coefficient -1. Thus, -1(1, 0, 1). This decomposition is the answer.