Page 244 - 35Linear Algebra

244 Diagonalization

This means

1 ! 1 !
√ √
0 2 v 1 + v 2 0 3 v 1 − v 2
v = v 1 , v 2 √ 1 = √ and v = v 1 , v 2 1 = √ .
2
1
2 2 − √ 3 3
0
0
The change of basis matrix has as its columns just the components of v and v ;
1 2
1 1 !
√ √
P = 1 2 1 3 .
√ − √
2 3
Changing basis changes the matrix of a linear transformation. However,
as a map between vector spaces, the linear transformation is the same
no matter which basis we use. Linear transformations are the actual
objects of study of this book, not matrices; matrices are merely a convenient
way of doing computations.

Change of Basis Example

Lets now calculate how the matrix of a linear transformation changes
i
when changing basis. To wit, let L: V −→ W with matrix M = (m ) in the
j
ordered input and output bases S = (v 1 , . . . , v n ) and T = (w 1 , . . . , w m ) so
X
k
L(v i ) = w k m .
i
k
0
0
0
0
0
0
Now, suppose S = (v , . . . , v ) and T = (w , . . . , w ) are new ordered input
1 n 1 m
0 k
0
and out bases with matrix M = (m ). Then
i
X
0k
0
L(v ) = w k m .
i i
k
i
Let P = (p ) be the change of basis matrix from input basis S to the basis
j
j
0
S and Q = (q ) be the change of basis matrix from output basis T to the
k
0
basis T . Then:
!
X X X X
0
k i
i
L(v ) = L v i p i = L(v i )p = w k m p .
j j j i j
i i i k
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