Page 218 - 35Linear Algebra

218 Basis and Dimension

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in the unknowns x . For this, we need to find a unique solution for the linear
system MX = w.
Thus, we need to show that M −1 exists, so that

X = M −1 w
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is the unique solution we desire. Then we see that S is a basis for R if and
only if det M 6= 0.

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Theorem 11.1.1. Let S = {v 1 , . . . , v m } be a collection of vectors in R .
Let M be the matrix whose columns are the vectors in S. Then S is a basis
for V if and only if m is the dimension of V and

det M 6= 0.

Remark Also observe that S is a basis if and only if RREF(M) = I.

Example 122 Let

1 0 1 1
S = , and T = , .
0 1 1 −1

1 0
2
Then set M S = . Since det M S = 1 6= 0, then S is a basis for R .
0 1

1 1
2
Likewise, set M T = . Since det M T = −2 6= 0, then T is a basis for R .
1 −1
11.2 Matrix of a Linear Transformation (Redux)

Not only do bases allow us to describe arbitrary vectors as column vectors,
they also permit linear transformations to be expressed as matrices. This
is a very powerful tool for computations, which is covered in chapter 7 and
reviewed again here.
Suppose we have a linear transformation L: V → W and ordered input
and output bases E = (e 1 , . . . , e n ) and F = (f 1 , . . . , f m ) for V and W re-
spectively (of course, these need not be the standard basis–in all likelihood
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V is not R ). Since for each e j , L(e j ) is a vector in W, there exist unique
i
numbers m such that
j
 
m 1
j
m
1
.
L(e j ) = f 1 m + · · · + f m m = (f 1 , . . . , f m )  .  .
 . 
j j
m m
j
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