Page 245 - 35Linear Algebra

13.2 Change of Basis 245

Meanwhile, we have:

X X X
j
0
k
0k
L(v ) = v k m = v j q m .
i i k i
k k j
Since the expression for a vector in a basis is unique, then we see that the
0
entries of MP are the same as the entries of QM . In other words, we see
that
−1
0
MP = QM 0 or M = Q MP.
Example 131 Let V be the space of polynomials in t and degree 2 or less and
2
L : V → R where

1 2 2 3
L(1) = L(t) = , L(t ) = .
2 1 3
From this information we can immediately read off the matrix M of L in the bases
2
2
S = (1, t, t ) and T = (e 1 , e 2 ), the standard basis for R , because
2
L(1), L(t), L(t ) = (e 1 + 2e 2 , 2e 1 + e 2 , 3e 1 + 3e 2 )

1 2 3 1 2 3
= (e 1 , e 2 ) ⇒ M = .
2 1 3 2 1 3
Now suppose we are more interested in the bases

1 2
0
0
2
0
0
2
S = (1 + t, t + t , 1 + t ) , T = , =: (w , w ) .
2 1 1 2
0
To compute the new matrix M of L we could simply calculate what L does the the
new input basis vectors in terms of the new output basis vectors:

1 2 2 3 1 3
2
2
L(1 + t), L(t + t ), L(1 + t )) = + , + , +
2 1 1 3 2 3
0
0
0
0
0
0
= (w + w , w + 2w , 2w + w )
1
2
1
2
2
1
1 1 2 1 1 2
0
0
0
= (w , w ) 1 2 1 ⇒ M = 1 2 1 .
1
2
Alternatively we could calculate the change of basis matrices P and Q by noting that
   
1 0 1 1 0 1
2
2
2
(1 + t, t + t , 1 + t ) = (1, t, t )  1 1 0  ⇒ P =  1 1 0 
0 1 1 0 1 1
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