Page 220 - 35Linear Algebra

220 Basis and Dimension

P 1 (t) = W, let’s choose the same ordered basis B = (1 − t, 1 + t) for V and W.

0

L(1 − t) = (1 − 1)t = 0 = (1 − t) · 0 + (1 + t) · 0 = 1 − t, 1 + t
0

−1

L(1 + t) = (1 + 1)t = 2t = (1 − t) · −1 + (1 + t) · 1 = 1 − t, 1 + t
1

a 0 −1 a
⇒ L = .
b 0 1 b
B B
n
When the vector space is R and the standard basis is used, the problem
of finding the matrix of a linear transformation will seem almost trivial. It
is worthwhile working through it once in the above language though.

n
Example 124 Any vector in R can be written as a linear combination of the standard
(ordered) basis (e 1 , . . . e n ). The vector e i has a one in the ith position, and zeros
everywhere else. I.e.

     
1 0 0
0
0
1
     
e 1 =  . , e 2 =  . , . . . , e n =  . .
 
 
 
. . .
 .   .   . 
0 0 1
n
n
Then to find the matrix of any linear transformation L: R → R , it suffices to know
what L(e i ) is for every i.
For any matrix M, observe that Me i is equal to the ith column of M. Then if the
n
ith column of M equals L(e i ) for every i, then Mv = L(v) for every v ∈ R . Then
the matrix representing L in the standard basis is just the matrix whose ith column
is L(e i ).
For example, if
           
1 1 0 2 0 3
4
6
1
0
5
0
L   =   , L   =   , L   =   ,
0 7 0 8 1 9
then the matrix of L in the standard basis is simply
 
1 2 3
4 5 6 .
 
7 8 9
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