00:00 Introduction

00:08 Example with a 3x4 matrix

00:32 Definition of the kernel

01:53 Row operations don’t change the kernel

03:10 Creating zeros in second column

03:54 Row echelon form

04:45 Stating equation with variables

05:36 Backwards substitution

06:50 Solution set

07:45 Rewriting solution set as a span

08:30 Basis of the kernel

1 00:00:00,000 –> 00:00:02,000 Hello and welcome.

2 00:00:02,000 –> 00:00:06,740 Today I show you how to calculate the kernel of a given matrix.

3 00:00:06,740 –> 00:00:13,447 And this will be an easy example of a 3x4 matrix.

4 00:00:13,447 –> 00:00:18,000 Well, this is the matrix, we will consider here.

5 00:00:18,000 –> 00:00:24,708 And we will consider it over the real numbers, and as you can see, it has 3 rows.

6 00:00:24,708 –> 00:00:32,102 So 3x4 and 4 is the number of the columns here.

7 00:00:32,102 –> 00:00:36,248 However, first of all, you have to know what the kernel is.

8 00:00:36,248 –> 00:00:45,467 So the kernel of a matrix A is, by definition, the set of all vectors that are sent to 0 by this matrix.

9 00:00:45,467 –> 00:00:55,815 So this means that we have here all vectors x and they come from R^4, since we have 4 columns here.

10 00:00:55,815 –> 00:01:04,131 So we need a vector with 4 rows and then we can give Ax a meaning. So Ax.

11 00:01:04,131 –> 00:01:11,548 So this will be a vector in R^3 and this goes to the zero vector in R^3.

12 00:01:11,548 –> 00:01:22,149 So Ax = 0. Well, this subspace of R^4 is exactly the kernel of A.

13 00:01:22,149 –> 00:01:28,289 And as you can see, this is just a system of linear equations.

14 00:01:28,289 –> 00:01:36,934 So we calculate the kernel by solving the system of linear equations given by the matrix and the right-hand side: 0.

15 00:01:36,934 –> 00:01:44,867 Therefore, this means that we can row operations on this system of linear equations, here.

16 00:01:44,867 –> 00:01:48,867 And they don’t change the system, so the solution set!

17 00:01:48,867 –> 00:01:58,867 We want the solution set here for the kernel. So we can just do row operations on this matrix inside the kernel here.

18 00:01:58,867 –> 00:02:02,867 And we don’t change the kernel! So this is what we do now.

19 00:02:02,867 –> 00:02:08,867 We want to change the second and the third row.

20 00:02:08,867 –> 00:02:14,104 In such a way that we can create zeros here in the first column.

21 00:02:14,104 –> 00:02:25,067 So, meaning, we have to add, on the second row, a multiple of the first row. And this is just 3.

22 00:02:25,067 –> 00:02:32,067 So, -3 times the first row gives us what we want here.

23 00:02:32,067 –> 00:02:41,934 So 0 here; there we have -2; and -2; and 1.

24 00:02:41,934 –> 00:02:51,400 Then, we do the same for the third row. So third row minus 4 times the first row.

25 00:02:51,400 –> 00:03:06,000 And then we have the zero as wanted, and -8 (so -4 -4); and -8 as well; and +4 here.

26 00:03:06,000 –> 00:03:13,000 Okay, now we have the zeros in the first column and we want to do the same in the second column as well.

27 00:03:13,000 –> 00:03:15,134 So we want to create a zero here.

28 00:03:15,134 –> 00:03:29,267 As you can see, this is very easy now. So the new third row here gets just 4 times the new second row here.

29 00:03:29,267 –> 00:03:34,334 And just with minus again.

30 00:03:34,334 –> 00:03:41,808 And then we get: 0, 0, 0, 0. So very boring here.

31 00:03:41,808 –> 00:03:50,067 But, of course, this was obvious since just the third row here is just a multiple of the second row.

32 00:03:50,067 –> 00:03:57,734 And now we are finished since we arrived at the so-called row echelon form here.

33 00:03:57,734 –> 00:04:07,467 Or in other words, just a triangle form here since all elements below this diagonal here are zero.

34 00:04:07,467 –> 00:04:11,467 This is what we want. Then the system is solved.

35 00:04:11,467 –> 00:04:23,067 Now, we call these variables here, given by these non-zero elements in each column, the bound variables (= leading variables).

36 00:04:23,067 –> 00:04:31,000 And the other ones here, so the 2 here, are the free variables.

37 00:04:31,000 –> 00:04:35,067 And now we can just write it down.

38 00:04:35,067 –> 00:04:45,267 The second row here is an equation. So please keep in mind that we just have the equation with right-hand side zero.

39 00:04:45,267 –> 00:05:00,267 That means that we have, here for the second row, we have -2 times x_2 minus 2 x_3 plus x_4 is equal to zero.

40 00:05:00,267 –> 00:05:05,000 So, now we already set here the free variables.

41 00:05:05,000 –> 00:05:12,534 So, we can just choose here another letter, maybe an alpha and here a beta.

42 00:05:12,534 –> 00:05:20,934 And then, we can just write the bound variable in terms of the free variables.

43 00:05:20,934 –> 00:05:28,800 So here x_2 (so this implies) x_2 is equal to minus alpha.

44 00:05:28,800 –> 00:05:30,480 (so we just divide by 2)

45 00:05:30,480 –> 00:05:34,800 Plus one half beta.

46 00:05:34,800 –> 00:05:51,734 And the equation in the first row gives us x_1 plus x_2 plus 2 times x_3 plus x_4 equals to zero.

47 00:05:51,734 –> 00:06:02,334 But we already know that we can write (maybe I put this here into orange)

48 00:06:02,334 –> 00:06:09,000 We can write x_2 as minus alpha plus one half beta.

49 00:06:09,000 –> 00:06:17,000 We know that this here is just 2 times alpha and this is just beta.

50 00:06:17,000 –> 00:06:32,134 So we can write x_1 also in terms of alpha and beta. And this is just minus alpha (so 2 alpha minus one)

51 00:06:32,134 –> 00:06:39,267 And minus three half beta.

52 00:06:39,267 –> 00:06:50,467 Of course, this is the solution set of our system of linear equations given by Ax = 0.

53 00:06:50,467 –> 00:06:55,267 So now we can just write the solution set, that is the kernel.

54 00:06:55,267 –> 00:07:06,000 So, set brackets here and then we have the vector. So it’s minus alpha minus three half beta.

55 00:07:06,000 –> 00:07:12,200 This comes from the x_1 component. And then we have the x_2 component.

56 00:07:12,200 –> 00:07:24,667 Minus alpha plus one half beta. And x_3 and x_4 were the free variables, so just alpha and beta here.

57 00:07:24,667 –> 00:07:33,067 And most importantly, don’t forget to write “alpha, beta are coming from R”. So they are elements in R.

58 00:07:33,067 –> 00:07:35,667 And this is the whole solution set.

59 00:07:35,667 –> 00:07:41,667 So an arbitrary vector given in this form where alpha and beta are coming from R.

60 00:07:41,667 –> 00:07:50,734 So, of course, we can split this up into two vectors. So we use the alpha coefficient.

61 00:07:50,734 –> 00:07:57,467 Then we have the vector: -1, -1, 1, and zero.

62 00:07:57,467 –> 00:08:09,067 Plus beta and the vector with coefficients in front of beta, so: minus three half, plus one half, 0 and 1.

63 00:08:09,067 –> 00:08:17,000 And the same again: we have to mention that alpha and beta are coming from the real numbers.

64 00:08:17,000 –> 00:08:23,934 So obviously, this is the same set. So we just wrote the vector in another way.

65 00:08:23,934 –> 00:08:29,534 And this is the whole kernel. And by our calculation, we know

66 00:08:29,534 –> 00:08:39,134 that these two vectors here are linearly independent and therefore form a basis of the kernel.

67 00:08:39,134 –> 00:08:44,867 Of this kernel of A.

68 00:08:44,867 –> 00:08:54,800 That means: for this example, the kernel is a two-dimensional subset of R^4.

69 00:08:54,800 –> 00:09:02,198 And this is all for today. Thank you for watching!