Page 189 - Advanced engineering mathematics

6.4 The Vector Space R n 169

If k and n are large, it may be difficult to tell whether a set of k vectors in R is linearly
n
independent or dependent. This task is simplified if the vectors are mutually orthogonal.

THEOREM 6.2

n
Let F 1 ,···F k be nonzero mutually orthogonal vectors in R . Then F 1 ,···F k are linearly
independent.
Proof Suppose
α 1 F 1 + α 2 F 2 + ··· + α k F k = O.
Take the dot product of this equation with F 1 :
α 1 F 1 · F 1 + α 2 F 1 · F 2 + ··· + α k F 1 · F k = O · F 1 = 0.
Because F 1 · F j = 0for j = 2,··· ,k, by the orthogonality of these vectors, this equation
reduces to
α 1 F 1 · F 1 = 0.
Then
2
α 1 F 1 = 0.
But F 1 is not the zero vector, so F 1 = 0 and therefore α 1 = 0. By using F j in place of F 1 in
this dot product, we conclude that each α j = 0. By (2) of Theorem 6.1, F 1 ,···F k are linearly
independent.

We would like to combine the notions of spanning set and linearly independence to define
vector spaces and subspaces as efficiently as possible. To this end, define a basis for a
subspace S of R to be a set of vectors that spans S and is linearly independent. In this
n
n
definition, S may be R .

EXAMPLE 6.15
3
3
3
The vectors i,j,k in R are linearly independent, and span R . These vectors form a basis for R .
n
In R , the standard unit vectors
e 1 =< 1,0,0,··· ,0 >,e 2 =< 0,1,0,··· ,0 >,··· ,e n < 0,0,··· ,0,1 >
form a basis.
EXAMPLE 6.16
n
Let S be the subspace of R consisting of all n− vectors with first component zero. Then
e 2 ,··· ,e n form a basis for S.

EXAMPLE 6.17
3
In R ,let M be the subspace of all vectors parallel to the plane x + y + z = 0. A point is on this
plane exactly when it has coordinates (x, y,−x − y). Therefore every vector in M has the form
< x, y,−x − y >. We can write this vector as

< x, y,−x − y >= x < 1,0,−1 > +y < 0,1,−1 >.

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October 14, 2010 14:21 THM/NEIL Page-169 27410_06_ch06_p145-186