Page 128 - 35Linear Algebra
P. 128
128 Matrices
By linearity this specifies the action of L on any vector from V as
1 0 0
1 1 = (c 1 + c 2 ) 1 .
L c 1 + c 2
0 1 0
We had trouble expressing this linear operator as a matrix. Lets take input basis
1 0
1
,
B = =: (b 1 , b 2 ) ,
1
0 1
and output basis
1 0 0
,
,
1
E = .
0
0
0 0 1
Then
Lb 1 = 0e 1 + 1e 2 + 0e 3 ,
Lb 2 = 0e 1 + 1e 2 + 0e 3 ,
or
0 0 0 0
1
Lb 1 , Lb 2 ) = (e 1 , e 2 , e 3 ) , (e 1 , e 2 , e 3 ) = (e 1 , e 2 , e 3 ) 1 1 .
1
0 0 0 0
The matrix on the right is the matrix of L in these bases. More succinctly we could
write
0
x
L = (x + y)
1
y
B 0
E
0 0
and thus see that L acts like the matrix 1 1 .
0 0
Hence
0 0
x x
L = 1 1 ;
y y
B 0 0
E
given input and output bases, the linear operator is now encoded by a matrix.
This is the general rule for this chapter:
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