In this video, we will learn what vector subspaces are. For that we will illustrate a few examples based on this. So, what vector subspaces are? If V is a real vector space, then a subset W of V is said to be a vector subspace of V, if W itself is a vector space, respect to the operations star and dot in V. The star is sometimes also called vector addition and this dot is also called scalar multiplication. If you take any subset of this V, and if this W constitute itself a vector space under these operations star and dot, then we say that this W is nothing but a subspace of V. How can we show that a given subset of V is a subspace of V? Let us illustrate by this definition. See if a non-empty subset W of a vector space V over real numbers is a subspace of V if and only if the following conditions are satisfied. What are these two conditions? First of all, for every u, v, in W, you take any two elements in W, then applying the operation star on these two element, this must be in W. That means this operation star must be a closed operation. The second property is that for every u in W, and for every Alpha in real numbers, then Alpha dot u must be W. If we prove these two properties for any subset W of V, then it is sufficient to show that it's a subspace of the given vector space V. Let us take one simple example of R^2. How the operations are defined? Operations R^2 are defined as usual addition and usual scalar multiplication. That means I defined operation star as usual addition, x plus x^0, y plus y^0. How dot is defined? Dot is defined in this way, Alpha x, Alpha y. Now if you take this subset, this is your w. If you take this subset here, then this subset of R^2 is a subspace of R^2. How can we prove that is a subspace? We have to show these two properties. First of all, it must be closed under star and if you take any find real numbers and any u in W, then Alpha dot u must be in W. You take any x, y, you take u equal to x_1, y_1, and v equal to x_2, y_2. Now you take u star v. U star v will be what? x_1, y_1, u star v x_2, y_2. This will be equal to x_1 plus x_2, y_1 plus y_2 by this definition. Where u and v are? U and v are in W, and we have to show that u star v is also in W. If u and v are in W that means it satisfies the properties of W. This means, this implies that x_1 plus y_1 must be 0 and x_2 plus y_2 must be 0. In order to show that this is in W, we have to show that this element plus this element must be zero which is true because x_1 plus x_2 plus y_1 plus y_2 if you take, then x_1 plus y_1 is zero because of this condition and x_2 plus y_2 is zero because of this condition. This is zero that means this element belongs to W. We have seen that this is closed under this operation. Now for dot, if you want to show, we have to show that if you take any element Alpha in R and any element u in W, then Alpha dot u must be in W, this we have to show. If you take u as x, y, this implies x plus y equal to 0 because this belongs to W. You multiply both sides by Alpha for any Alpha in R, so it is Alpha x plus Alpha y equal to 0. This implies Alpha x, Alpha y is also in W. That means Alpha dot u is in W. We have shown that the two given property are satisfied for this subset W of V, hence, we can say that it's a subspace of V. Now similarly, if you take this example, vector space consisting of all two close to real matrices under usual addition and usually scalar multiplication. Then if we define this subset a, 0, 0, 0 which belongs to M, where a is a real number, is a subspace of M. How we can show it? We can take two matrices, a and b belongs to this set. Suppose the set is, I'm saying suppose this is S, so belongs to S, then this implies the sum Alpha, 0, 0, 0 is equals to A. Suppose Beta, 0, 0, 0 equal to suppose B, then if you take Alpha star B, which is nothing but usual addition of matrices, it is equals to what? Alpha plus Beta, 0, 0, 0, which also belongs to M. Similarly, if you take Alpha dot with A, which is usually scalar multiplication, suppose it is a, 0, 0, 0, which is equals to Alpha a, 0, 0, 0, it also belongs to M. We have seen that it is satisfied the two given properties of addition and scalar multiplication. Hence, we can see that it constitute a subspace of M. Now see this subset W, which is consisting of all those matrices in M whose determinant is zero. This is also a subset of M, but it is not constitute a subspace of M. Why it is not constituents of subspace? In order to show that it is not constituting subspace, we have to give a counter example saying that it is not closed either respect to that addition or scalar multiplication. Take one simple example. Suppose you take matrix A as you take two elements, A and B in W. That means determinant of A is zero, determinant of B is zero. Suppose you take A as 1, 0, 0, 0, which belongs to W because the determinant is zero. You take another matrix B, which is in W, say 0, 0, 0, 1. The determinant is also zero so it belongs to W. Now if you take vector addition which is usual addition in this case. This is 1, 0, 0, 1 and its determinant is what? Determinant of A plus B is what? Is one, so which is not zero. This means this does not belongs to W. However, see A and B belongs to W, but when you apply the operation star on this A and B, it is not in W. That means this operation star is not closed here. This implies that this will not constitute a subspace of the given vector space M. In this video, we have seen that if a subset of a vector space is given to you and you want to show that it's a subspace of the given vector space, then it is sufficient to put the two properties. The first property is that you have to show that it is close respect to operation star and the second one is that Alpha dot u should be in that set for every Alpha in R and for every u in the W.
