Page 264 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes = (7, 0, –1)
2
= (1, 5, 4)
3
3
in R equipped with the standard inner product. Applying the Gram-Schmidt process to , , ,
1 2 3
we obtain the following vectors.
= (4, 0, 3)
1
(7, 0, 1|4, 0, 3)
= (7, 0, –1) – (4, 0, 3)
2
25
= (7, 0, – 1) – (4, 0, 3) = (3, 0, – 4)
(1, 5, 4|3, 0, 4) (1, 5, 4|4, 0, 3)
= (1, 5, 4) – (3, 0, 4) (4, 0, 3)
3 25 25

13 16
= (1, 5, 4) + (3, 0, 4) (4, 0, 3)
25 25
= (0, 5, 0)
These vectors are evidently non-zero and mutually orthogonal. Hence ( , , ) is an orthogonal
1 2 3
3
3
basis for R . To express an arbitrary vector (x , x , x ,) in R as a linear combination of , , ,
1 2 3 1 2 3
it is not necessary to solve any linear equation. For it suffices to use (8).
Thus
x
3x 4x (3x 4 ) x
(x , x , x ) = 3 1 1 1 3 2 2 3
1 2 3 25 25 5
as is readily verified. In particular,

13 9 2
(1, 2, 3) = (4, 0, 3) (3, 0, 4) (0, 5, 0)
25 25 5
To put this point in another way, what we have shown in the following: The basis (f , f , f ) of (R )
3
1 2 3
which is dual to basis ( , , ) is defined explicitly by the equations
1 2 3
4x 3x
f (x , x , x ) = 1 3
1 1 2 3
25
3x 4x
f (x , x , x ) = 1 3
2 1 2 3
25
x
f (x , x , x ) = 2 ,
3 1 2 3
5
and these equations may be written more generally in the form
,
( , x x | )
x
1 2 3 j
f (x , x x ) = 2 .
j 1 2 3
j
Finally, note that from , , we get the orthonormal basis
1 2 3
1 1
(4, 0, 3), (3, 0, 4), (0, 1, 0).
5 5

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