00:01
Hello students, the question is, consider the inner product space v equal to rq with a standard inner product over r.
00:14
It is given that s equals to 1 to 1, 1 minus 2 minus 3, 3 5 minus 1 is a basis for v, applying ground smith process to s to find an orthonormal basis for v.
00:25
Gramsmouth orthogonalization process is let v be an inner product space and s equal to w1, w2 and wnb are linearly independent subset of v.
00:32
Define s -tash equal to v1 v2 to vn where v1 equal w1 and vk equals to wk minus summation j equals to 1 to k minus 1 inner product w k vj upon non vj square into vj for 2 less than equal to k less than equal to 1 then s -dash is an orthogonal set of non -zero vectors such that span s -dash equal to span s s -d equals to v1 v2 v3 take v1 equals to 1 to 1 then v2 equals to v2 v2 equals to 1 minus 2 minus 3 minus inner product 1 minus 2 minus 3 v1 121 upon non v1 square this equals 1 minus 2 minus 3 minus 1 into 1 minus 3 upon 1 2 1 2 1 upon 1 1 1 1 1 1 1 1 1 1 upon 1 .1 is 1 plus 2 into 2 is 4 plus 1 .1 to 1 minus 2, minus 3, minus 6, 1 to 1 upon equals to 1 minus 2, minus 2 minus 3, minus minus minus minus plus 1 to 1 equals to 1 plus 1 2 plus 2 is 0, minus 3 plus 1 to 1 is minus 2.
03:43
This is v2.
03:48
Now v3 equals to 5 minus 1 minus summation j equals to 1, 3 minus 1 which is 2, inner product 35 minus 1 vj upon non vj square into vj equals to 3 5 minus 1 minus 1 minus 0 inner product 35 minus 1 and j is 1 so v1 and v1 is equal to 1 to 1 2 v1 is 1 2 inner product 3 5 minus 1 comma v2 v2 is 2 0 minus 2 this equals 35 minus 1 minus 3 into 1 plus 3 into 1 plus 3 5 minus 3 plus 10 minus 1.
06:14
Into 1 .21 upon 1 .1.
06:23
1 .2 .2 is 1 plus 2 .4.
06:26
1 plus 4, 5 plus 1 .6 plus 06.
06:41
Plus 2.
06:43
8 into 2 minus 2.
06:54
Upon 2 into 2 is 4 plus 0 plus minus 2 into minus 2 is 4...