00:01
In this question, we have given that w is contained in r square and where x is x1 and x2.
00:12
In the first part they are asking w where w is the x, it consists of all the vectors such that x1 is equal to 2 of x2.
00:21
We do find out if it is a subspace or not.
00:24
For this, consider a vector x1, y1, belonging to w with the condition given that x1 is 2 of y1.
00:31
And another factor x2 y2 which is belonging to w with the condition given that x2 is 2 of y2 by adding them x1 y1 plus x2 which implies is equal to x1 plus x2 and y1 plus y2 x1 plus x2 in order to prove that this element belongs to w x1 plus x2 should equal to twice of y1 plus y2 x1 is 2y1 and it is to y2 clearly this property holds so now let's find out the second property to be of the subspace consider any element x1 comma y1 belonging to x1 .m .y1 belonging to x1 belongs to w with the condition given that x1 is 2 of y1 clearly for any alpha belonging to r this implies that alpha x1 y1 will belongs to w because in that case x1 or y1 to be of 2 which implies alpha x1 over alpha y1 is 2.
01:56
Hence w in this case w is a subspace.
02:03
Now in the second part we have given w2b consists of all those vectors such that x1 minus x2 is equals to 2.
02:13
Clearly additive identity does not belongs to w that is 0 0 does not belong to w which implies that because 0 minus 0 is always equal to 0 it is not equal to 2 it implies that w is not a subspace in the next part in the next part w is of the definition that is w consists of all the vectors x such that x1 is equal to x2 or either x1 is equal to minus of x2.
02:57
Consider x1 x2 belongs to w v and first element and x3 x4 w belongs to w with the condition that x1 is equal to x2 and x3 is equal to x4.
03:10
When we commute them under usual addition, that is x1 comma x2 plus x3 comma x4, it is x1 plus x3, x2 plus x4.
03:25
In order to prove that this element belongs to w, we have to show that x1 plus x3 should equals to x2 plus x4.
03:35
Clearly x1 is x2 and x3 is x4 which implies this is this condition holds.
03:44
Hence first property holds.
03:46
Now talk about the second, that is consider it.
03:49
This is the second property.
03:53
This is the second necessary and sufficient condition to prove a subset to be a subspace.
04:00
Alpha belongs to real.
04:03
Then alpha of x1, x2 should belongs to w only in the case when alpha of x1 is equal to alpha of xt.
04:13
Clearly this holds which implies that w is a subspace of a given vector space.
04:21
Now let's come to the next part.
04:30
In the fourth part, they are asking that w consists of all the vectors x such that x1 and x2 are rational numbers.
04:40
They both are the rational numbers.
04:44
Consider the contradiction, that is consider x1, x2 belonging to w which implies that x1 and x2 are rationales.
04:58
Now, consider any alpha belonging to r, consider here.
05:04
Here alpha to be irrational.
05:10
Then alpha of x1 and alpha of x2 can only belong to w if, can only belongs to w if alpha x1 and alpha x2 are rationals.
05:27
But here alpha is a irrational.
05:30
So alpha x1 and alpha x2 are irrational numbers.
05:42
Hence there is a contradiction which implies that there is a contradiction which implies that in this case w is not a subspace of given vector space.
06:02
It is not a subspace of r square under usual addition and multiplication.
06:06
Let's come to the next part.
06:10
In the fifth part they are asking that w consists of all those vectors such that its first coordinate is 0.
06:20
Consider x1, x2 and x3, x4 belonging to w with the condition given that x1 is 0 and x3 is 0, which implies that if we add them together, then x1 plus x3 and x2 plus x4, that is in this case it should be only belonging to w if x1 plus x3 is 0.
06:46
It is clearly 0 which implies first property holds now second.
06:57
For any alpha, now the second property for any alpha belonging to real, alpha of x1 and x2 belongs to w only if alpha of x1 is 0.
07:14
Since x1 and x2 belongs to w which implies that x1 is 0.
07:19
So alpha of x1 is also 0 which implies that in this case, w is a subspace of r square.
07:28
That is given vector space under usual addition and multiplication.
07:33
Now consider the next option, the next part.
07:38
In the next part, we are having the definition of w as it consists of all the vectors x such that more do of x1 plus more do of x2 to be of zero...