Page 243 - 35Linear Algebra

13.2 Change of Basis 243

i
Here, the p are constants, which we can regard as entries of a square ma-
k
i
trix P = (p ). The matrix P must have an inverse since we can also write
k
0
each v j uniquely as a linear combination of the v ;
k
X
0
k
v j = v q .
k j
k
Then we can write
X X
i k
v j = v i p q .
k j
k i
P
i k
But p q is the k, j entry of the product matrix PQ. Since the expression
k k j
for v j in the basis S is v j itself, then PQ maps each v j to itself. As a result,
each v j is an eigenvector for PQ with eigenvalue 1, so PQ is the identity, i.e.
PQ = I ⇔ Q = P −1 .

The matrix P is called a change of basis matrix. There is a quick and
dirty trick to obtain it; look at the formula above relating the new basis
0
0
0
vectors v , v , . . . v to the old ones v 1 , v 2 , . . . , v n . In particular focus on v 0 1
2
n
1
for which
 
p 1
1
p
 2
0

v = v 1 , v 2 , · · · , v n  .  .
 1
.
1
 . 
p n
1
This says that the first column of the change of basis matrix P is really just
0
the components of the vector v in the basis v 1 , v 2 , . . . , v n .
1
The columns of the change of basis matrix are the components
of the new basis vectors in terms of the old basis vectors.
0
0
0
Example 130 Suppose S = (v , v ) is an ordered basis for a vector space V and that
2
1
with respect to some other ordered basis S = (v 1 , v 2 ) for V
1 ! 1 !
√ √
0
0
v = 2 and v = 3 .
1 √ 1 2 − √ 1
2 S 3 S
243