Question

5) (20 points) a) Show that the vectors x1 = (1, 1, 0)T , x2 = (1, 0, 1)T , x3 = (1, 0, 0)T are linearly independent. Do they form a basis of R3 ? Explain.

b) Find an orthonormal basis of R3 using x1 = (1, 1, 0)T , x2 = (1, 0, 1)T and x3 = (1, 0, 0)T .

Answer 1

Amy. (a) given that the vectors xp = (1,1,0)", k, = (1,0,1) *3 = (1,00) T > To show : x, xq, X3 are L.I. Now, x , xz are Lo I. if + de tu O do [] has only the trivial solution &, = 2 = 23 = 0 -[]=[] =) ㅗ L.o L. L 1 ㅗ 0 to ㅗ 0 I o I olo o t 0 0 0 o 1 0 Augumented I А matrix 2, + 2 + 2 = 0 i 22 +2₂=0 (1) +63 =0 fromci, da to = 0 of from cis, se totoo 2 = 0 21=0 Hence, a da and az are Linearly independent. Now, x = (1, 1,0), x = (1,0,1), x= (1,0,0) L I I I 1 1 L 0 I ㅗ L 1 3102 L 1 0 there are three pivot columns, therefore the given set of vectors form a basis for 1R3.

T x = (1,0), x = (0,1), &z= (1,0,0) To find : orthorrormal basis of 123 using x,, ,, &3 x = L Hy 6 1 33 O met ui, & and Uz are ortho normal basis. by Gram - Schmidt orthogonalization process - xx.x3 = [![:] 1x1 t Lxo toxl =1 let u = x l 1 L " = 0 - [+] - 1:10 [] (3] ㅗ (AxL+ Oxt + txo) (1*1+1x++oxo) L A 4/2 1 2 L = 0 - 1/2 O we #3 xz. V, v xz. da I. "2 6) [8][ lilli -1/2 4 2 1 -½ 1 (tx1 + 0xlt oxo) (1x1 + 1X1 +010) 6] ("xtox tox 1) (+**1*} + 1x1 2 - 1 18] - ²1:]- 3 1 - 2

- % -16 3 01-09].[ 13 43 1 - ₂ - o-} +/ - / -1/3 -1/3 3 Hence orthomormal basis are 1 1/2 13 1 al2 1/3 And I -1/3