00:01
In this problem, it is given that v is the set of all x comma y, such that x and y are in the set of real numbers.
00:10
So, v is nothing but r2.
00:13
In the set v, the addition is denoted as u plus v and it is defined as u1 plus v1 comma u2 plus v2 plus 1, where u is equal to u1 u 2 and v is equal to v 1 comma v 2 and the scale of multiplication k times u is defined as k times u .m.
00:45
K times u 2 in part a we are given with u is equal to 0 .4 and v is equal to 1.
00:55
Minus 3 and the value of k is given as 2 so we are as to find the value of u plus v and k times u if we substitute u1 is equal to 0 u2 is equal to 4 v1 is equal to 1 and v2 is equal to minus 3 in this expression we will get the value of u plus v so it will be equal to 0 plus 1 which is nothing but 2 comma 4 minus 3 plus 1 which is nothing but 2 so this is the value of u plus v similar similarly, we can compute the value of k times u.
01:36
K is 2 and u is given as 0 comma 4 and the scalar multiplication is defined like this.
01:44
So we have to multiply each of u1 and u2 with k.
01:48
So it will be equal to 2 times 0 which is nothing but 0 times 2 times 4 which is 8.
01:55
So these are the answers for the first part of the problem.
01:59
Let's move on to part b.
02:01
In part b, we are as to prove.
02:04
That the vector 0 .0 is not the 0 element of the given vector space.
02:12
Suppose 0 .0 is the 0 element of the given vector space v, then for any element you belongs to the set v, the vector addition, u plus 0 .0 will give the vector u itself.
02:30
If we consider u as u1 comma u2, this expression will give us u1 comma u2 plus 0 comma 0.
02:45
If we use the definition of vector addition what is given to us, then what we can say is sum of these two vectors are equal to u1 plus 0 plus 1 which is nothing but u1 plus 1 comma u2 plus 1.
03:01
It is not equal to u1 comma u2 we are not getting u1 comma u2 when we add the zero element so 0 comma 0 cannot be the 0 vector of the given vector field of the given vector space so this is the answer for the second part of the problem now let's move on to part c in part c we are as to prove that minus 1 comma 1 is the 0 element of the given vector space so we need to prove that for any u belongs to v, u plus minus 1 comma minus 1 gives the 0 vector, gives the u vector itself.
03:46
Let's assume u is equal to u1 comma u2.
03:51
If we add this u with the vector minus 1 comma minus 1 we should get the vector u1 comma u2 itself.
04:00
Let's check this.
04:01
If we use the given definition of vector addition, on the left hand side of this expression, we will get u1 minus 1 plus 1 comma u2 minus 1 plus 1 plus 1.
04:15
Minus 1 plus 1 is equal to 0.
04:18
So it will be equal to u1 comma u2.
04:22
Observe that when we add u with minus 1 comma minus 1 we get u itself...