In this video, we will learn some more concepts on real vector spaces. This, we will try to understand by demonstrating more examples. Now, see this example. You consider set R2, what is R2? R2 is basically all x comma y, such that x, y belongs to R. So, what does it mean? It means that, it is x-axis and y-axis, so it is basically all the points which lie in this x, y plane. Say you take 1.2, 3, then 2, 3 also belongs to R2. Similarly, you might be having R3. R3 is what? R3 is all x, y, z, so that x, y, z belongs to R. Basically, it's a three-dimensional geometry. Now we'll define the two operation star and dot like this. The star means the usual component-wise addition. The first component, we'll add the first component here. The second component, we'll add to the second component here. How dot is defined? Dot is defined in this way, this Alpha dot x, y is nothing but alpha x alpha y. So, our claim is that this R2 is a real vector space. Now how can we show that it's a real vector space? In order to prove it we have to prove all the properties. The first property respect to the star, so respect to the star the first property is closure. Closure means that it's a binary operation. Of course, it's a binary operation. You can see from the definition itself that this element is in R2 and you have planned the operation is star. The resulting element is also in R2. So that means star satisfy binary operation. The second property is commutative. The star is commutative, so you take one element, say x, y, star with x naught, y naught. By this definition it is equal to x plus x naught, y plus y naught, which is same as x naught plus x because real numbers satisfy commutative property, y naught plus y and which is equal to x naught, y naught, star with x, y. So, this means this star satisfy commutative property. In the same lines, we can also show that this star satisfy associative property. The associative property means you take three elements, say x, y, x naught, y naught, x^1, y^1. Then this is same as x, y star with x naught, y naught, star with x^1, y^1. This we can easily show in the same lines. Next, is existence of identity element. Now, identity means that you apply the operation, this star, with any element of the set R2, it must be equal to itself. Suppose you take e as x naught, y naught, and you take any element u as x, y. Now, you take e star u, this must be u for every u. So, what does this imply? This implies that the x naught, y naught is star with x, y must be equal to x, y. This implies x naught plus x, y naught plus y, should be equal to x, Y, which is true only when x naught, y naught both are zero. So, what would be e then? e would be 0, 0 which is in R2. Of course, if you add 0, 0 with any element x, y then it will be itself, so that means identity element exist. Now, further you have to show the existence of inverse element. Now, if you see any element say u as x, y and suppose v as x^1, y^1, then u with star v must be e. So this implies, if you apply the operation star here, this is x plus x^1, y plus y^1 must be equal to 0, 0 and this implies that x^1 is minus x and y^1 is minus y. That means v will be nothing but minus u which is the inverse element of this u respect to this operation star. So, we have shown that all the property respect to the star are satisfying. Now, what about the properties respect to dot? So, first of all, respect to dot, we have to show that it satisfies closure property. I mean, closure property, see it's very easy to show that you take any real number that define in R and any x, y in R2, you apply the operation star, the resultant element is also in R2. So that means this dot satisfy closure property, this is the first thing. Now the other properties, See, for example, you take alpha dot u star v. It must be equals to alpha naught u dot u star with alpha dot v. So this you have to show. To show it, take the left-hand side, so alpha dot, what is u? Suppose you take u as x, y and suppose you take v as z, w. If we apply the operation as star here, it is x plus z, y plus w. When you take this operation dot here, is a usual multiplication component-wise, it is x plus z, it is Alpha times y plus w, which is equal to what? Which is Alpha x plus Alpha z, Alpha y plus Alpha w. Which is equals to Alpha x, Alpha y. We star with Alpha z, Alpha w. By the definition of star. This is Alpha naught x y, star with Alpha naught zw. This is Alpha u, the star with Alpha v. We have shown that this property holds. Similarly, if you want to see that L1 naught u is equal to u. This is easy to verify because 1 naught x y is by the definition is x y, which is equal to u for every u. The other property, suppose we want to show that Alpha plus Beta naught with u is Alpha naught u star with Beta naught u. Either you can solve left-hand side separately, right-hand side separately and you can show that left-hand side is equal to right-hand side. Or you can simplify the left-hand side and try to obtain the right-hand side from that expression itself. See what is left-hand side, it is Alpha plus Beta dot u, which is Alpha plus Beta x, Alpha plus Beta y. If u is x y, which is Alpha x, Alpha y, star with Beta x Beta y. It is Alpha naught x y dot x y star with Beta dot x y, which is Alpha dot u star Beta dot. We have shown this property. The last property can also be shown using this, which is Alpha naught Beta naught u is same as Alpha Beta naught u. You take this side, which is Beta naught u is Beta x Beta y. If you take u equal to x y, which is same as Alpha Beta x, Alpha Beta y. It is same as Alpha Beta naught with x y by the definition of naught and the Alpha Beta naught u. This is equal to this. Hence, we have seen that under these operations, star and dot, these are to satisfy all the properties of vector space, hence it constitutes a vector space. If you see this example, you have defined star operation the same way as we did in the earlier example, but the dot operation is defined in this way. Alpha x, y. This is not constituting a vector space. Why this is not concentrating vector space? Because you can easily see that if you take Alpha as this and Beta as this, this is Alpha, this is Beta. Dot with 1, 2. This left-hand side is. See, 2 plus 3 is 5. This is a usual addition, dot with 1,2. How is this dot is defined? This dot is defined in this way. Alpha is only multiplying with the first component, but not to the second component. This is equal to 5,2. Now see the right-hand side. See if it's a vector space, it must be equal to this for every x, y naught 2. You see the right-hand side. The right-hand side is basically 2.1,2. What is this value is 2, 2, star with 3, 2. Because this 3 will not multiply with 2 as per the definition of the dot. This is now a definition of a star, if you apply this is 5,4, which is not equal to this. This is not equal to this. This is 5,2, this is 5,4. Since one property is not satisfying, hence we can say that it will not constitute a vector space. Take another example. Take the collection of all two cross two real matrices and take usual addition and scalar multiplication. We noted how we can add two matrices. We know that if you multiply a matrix with a scalar say, Alpha, how we can multiply. Under usual addition and usual multiplication, this will constitute a vector space. That one can easily prove it using the definition of vector space. In this video, we have seen that if a star operation or dot operation is given to you and you want to show that it's a vector space or not, then how can you show? If it is not a vector space, then you have to give a counterexample to show that it's not a vector space.