Page 186 - Advanced engineering mathematics

166 CHAPTER 6 Vectors and Vector Spaces

In the plane and in 3-space, it is easy to visualize all of the subspaces in addition to the entire
space and the trivial subspace.
2
First consider R and look at a straight line y = mx through the origin. Every point on this
line has the form (x,mx). With i =< 1,0 > and j =< 0,1 >, every vector xi + mxj, with second
component m times the first, is along this line. Further, any sum of two vectors x 1 i + mx 1 j and
x 2 i + mx 2 j has this form, as does any multiple of such a vector by a real number. Therefore the
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vectors xi + mxj form a subspace of R .
So far we have excluded the vertical axis, which is also a line through the origin, but does
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not have finite slope. However, all vectors parallel to the vertical axis also form a subspace of R ,
being scalar multiples of j.
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Every line through the origin therefore determines a subspace of R , consisting of all vectors
parallel to this line.
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Are there any other subspaces of R that we have missed?
Suppose S is a nontrivial subspace containing two vectors ai + bj and ci + dj that are not on
the same line through the origin. Then ad − bc = 0, because the lines along these vectors have
different slopes. We claim that this forces every 2-vector xi + yj to be in S. To verify this, we
will solve for numbers α and β such that
xi + yj = α(ai + bj) + β(ci + dj).

This requires that
αa + βc = x, and
αb + βd = y.

But these equations have the solutions
dx − cy ay − bx
α = and β = .
ad − bc ad − bc
2
Therefore every 2-vector xi + yj in R is of the form
α(ai + bj) + β(ci + dj)
2
2
2
hence is in S. In this event S = R . We therefore know all of the subspaces of R .Theyare R ,
the trivial subspace {< 0,0 >} and, for any line L through the origin, all vectors parallel to L.
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By similar reasoning, there are exactly four kinds of subspaces of R . These are R ,the
3
trivial subspace containing just the zero vector, the subspace of all vectors on any given line
through the origin, and the subspace of all vectors lying on any given plane through the origin.
n
A linear combination of k vectors F 1 ,··· ,F k in R is a sum of the form
α 1 F 1 + α 2 F 2 + ··· + α k F k .

in which each α j is a real number.
n
The span of vectors F 1 ,F 2 ,··· ,F k in R consists of all linear combinations of these
vectors, that is, of all vectors of the form
α 1 F 1 + α 2 F 2 + ··· + α k F k .

n
n
From Example 6.11, the span of any set of vectors in R is a subspace of R . We say that these
vectors form a spanning set for this subspace.
Every nontrivial subspace has many spanning sets.

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October 14, 2010 14:21 THM/NEIL Page-166 27410_06_ch06_p145-186