Page 136 - A Course in Linear Algebra with Applications

120 Chapter Five: Basis and Dimension

equal to 0. The vectors Xi,X2, • • • ,X n~ r are linearly inde-
pendent, just as in the example, because of the arrangement
of O's and l's among their entries. It follows that a basis of
the null space of A is {Xi,X 2, • • ., X n- r}. We can therefore
state:

Theorem 5.1.7
Let A be a matrix with n columns and suppose that the number
of pivots in the reduced row echelon form of A is r. Then the
null space of A has dimension n — r.

Coordinate column vectors
Let V be a vector space with an ordered basis
{vi,..., v n }; this means that the basis vectors are to be writ-
ten in the prescribed order. We have seen in 5.1.3 that each
vector v of V has a unique expression in terms of the basis,

V = CiVi -i h C nV n
say. Thus v is completely determined by the scalars c x,..., c n .
We call the column
/ C ! \

\c J
n
the coordinate vector of v with respect to the ordered basis
{vi,..., v n }. Thus each vector in the abstract vector space V
is represented by an n-column vector. This provides us with
a concrete way of representing abstract vectors.

Example 5.1.5
Find the coordinate vector of ( j with respect to the ordered

basis of R 2 consisting of the vectors

( ! ) • ( ! ) •