In this video, we will learn that if a set is given to you, how can you say whether it constitutes a basis of a vector space or not? Further, we will see that how can you find a dimension of a vector space. What do we mean by basis for a real vector space? Let V be a vector space over R, over R means real vector space. Then a subset B of V is set to be a basis of V if these two conditions hold. What are these two conditions? The first condition is that B consists of linearly independent vectors. The second one is B generates V, that means the span of B must be equal to V. See this example. This example of the set this contains three vectors. This is first vector, second vector, and third vector, and these three vectors basically, this vector is subset of R^3. Now, I'll make claim that it's a basis of R^3. To show that it's a basis of R^3, I have to show two things. Number 1, this set is linearly independent. Number 2, the span of this set generates real time data. First of all, how can I show that this set is LI (linearly independent)? To show that this set is linearly independent, this set contains linear independent vectors. For that, I have to take a linear combination of these three vectors, put it equal to 0, 0, 0. If all three vectors comes out to be zero, this means the set is linearly independent. Take linear combination of these three vectors, 1, 0, 1, Beta times 0, 2, 1, plus Gamma times minus 1, 1, 0, put it equal to 0, 0, 0. What I obtain from here, it is Alpha minus Gamma equal to 0. Then from the second equation, it is 2 Beta plus Gamma equal to 0. The third equation is Alpha plus Beta equal to 0. For one linearly independent Alpha, Beta, Gamma all must be zero. That is the only single solution of this system of equation that is the previous solution which is Alpha equal to 0, Beta equal to 0, and Gamma equal to 0. That means this sum it should be, it is 1, 0, 1, it is 0, 2, 1, it is minus 1, 1, 0. This is Alpha, Beta, and Gamma and it is equal to 0, 0, 0. For non-trivial solution of this homogeneous system of linear equations, the determinant of this matrix must be zero. If you take the determinant of this matrix, then the determinant come out to be this one, it is 2 minus 1 is minus 1 then it is in a plus 0, which is of course 0 minus 1. This times 0 minus 2, which comes out to be minus 1 plus 2, which is 1 which is not equal to 0. This implies Alpha equal to Beta equal to Gamma equal to 0. This means a set is linearly independent. Or you can directly solve the system of linear equations and you can see it comes out to be Alpha equal to 0, Beta equal to 0, and Gamma equal to 0. The second thing is that we have to show that this is the span of this generates data vector space . Data standard space. Now next to show that span of this set is equal to R^3. That means if you take this set, I suppose S, then the span of S is equal to R^3. How to show this? That means that we have to show that you take any element a, b, c, and R^3 and we have to show that a, b, c can be expressed as linear combination of these three elements. In order to show this, take any element a, b, c in R^3, write as a linear combination of these three vectors, 1, 0, 1, Beta times 0, 2, 1, and Gamma times minus 1, 1, 0. In order to show that any a, b, c in our three can be expressed as linear combination of these three vectors, we have to express these Alpha, Beta, Gamma in terms of a, b, c. Then only we can see. How to express? See, you can write the equations as a is equal to the first component here is Alpha minus Gamma. B is equal to what? B is equal to 2 Beta plus Gamma. C is equals to Alpha plus Beta. Now you can take the first two equations. The first two equations give a plus b is equal to Alpha plus 2 Beta. The third equation is what? C is equals to Alpha plus Beta. Now subtract these two. This will give Beta equal to a plus b minus c. We have expressed Beta in terms of a, b, and c. Now from the third equation, Alpha is what? Alpha is c minus Beta. This implies this is 2c minus a minus b. From the first equation, Gamma is what? Gamma is alpha minus a. Alpha is what? Alpha is 2c minus a minus b minus a, so, this is 2c minus 2a minus b. This Alpha, Beta, and Gamma all are expressed in terms of a, b, c. Hence, whatever a, b, c I will pick here that the corresponding pillars of these three vectors we can easily determine using these expressions. Hence, we can say that whatever vector in R^3 we will take, we can always express the vector in terms of linear combination of these three vectors. Hence, we can say that the span of this vector generates in data. Proving these two properties, we can say that this set is a basis of what? Now consider another example of simple example of matrices of 2 cross 2. Now if you take these four matrices, 1, 0, 0, 0 and similarly others, this constitute a basis of what? Why consider basis of R^3? This you can easily show. First of all, this set is linearly independent. You did Alpha times first element, Beta times the second element, Gamma times the third element, and Delta time the fourth element. Put it equal to 0, 0, zero matrix. You can easily see that Alpha equal to, Beta equal to, Gamma equal to, Delta equal to 0. This means the set is linearly independent. Now the next property to show is the span of this set generates all matrices. That we can easily show, you can take a general matrix a, b, c, d in M. This a, b, c, d can be expressed as a times first element. We can easily see b times the second element, c times the third element, and d times the last element. We can see that the span of this generates in data hence we can see that this is the basis of this vector space M consisting of all 2 plus 2 data. Now, number of elements in the basis of a vector space is called is dimension. For example, we have just now we have considered this set. This set is a basis of R^3 and number of element is what? Three. The dimension of R^3 is three. The important point here is that basis of a vector space is not unique. But dimension of a vector space is always unique. One can easily see here, say you take vector spaces, real vector spaces R^2. You're going to easily see that this 1, 0, 0, 1 is a basis of R^2, because if this set is LI and the spans generates in data vectors space. Similarly, you can see that 1, 0, 1, 1, this set is also linearly independent and spans and generates data R^2. It is also a basis of R^2. Similarly, I can construct many examples which are the basis of R^2. But dimension is always two, dimension of this is two, dimensional of this is two, and this is also two. The point is that if you take a vector space, like you can find number of basis of vector space, but the dimension will be same. In this video, we have learned that what do you mean by basis of a vector space? We have also seen that how can we find out dimension of a given vector space? One important property of the basis and dimension is, that basis of a vector space is not unique. However, dimension of vectors is always unique.
