Math 311 Lecture 15
DEFINITION. If W = {w1, w2, ..., wn} is a basis of V and v a
vector in V, then v can be written as a linear
combination c1w1+c2w2+...+cnwn in exactly one way.
Let [v]W = be the column vector of these
c1
c2
...
cn
coefficients. [v]W is the coordinate vector of v with
respect to W. Coordinate vectors are always columns.
DEFINITION. If W = {w1, ..., wn} and T = {t1, ..., tn} are bases
for a vector space V, let PW
á
T = ( [t1]W | ...| [tn]W ) be the
matrix whose ith column is [ti]W.
PW
á
T is the transition matrix from T to W.
LEMMA. For any matrices A, B, A=B iff for all v, Av = Bv.
PROOF. To prove that the first, second, ... columns are the
same, apply both matrices to [1, 0, ...]T, [0, 1, ...]T, ... .
LEMMA. D PW
á
T[v]T = [v]W for any vector vLV.
(E PW
á
TPT
á
S = PW
á
S.
(F PT
á
W = (PW
á
T)-1
PROOF. D Suppose [v]T = [a,b,c]T. Then v = at1+bt2+ct3.
Hence [v]W = [at1+bt2+ct3]W = a[t1]W+b[t2]W+c[t3]W
= ( [t1]W | ...| [tn]W )[a,b,c]T=PW
á
T[v]T.
(E For any v, PW
á
TPT
á
S[v]S = PW
á
T[v]T = [v]W = PW
á
S[v]S.
(F PW
á
T[v]T = [v]W. Hence [v]T = (PW
á
T)-1[v]W. Thus
PT
á
W[v]W = [v]T = (PW
á
T)-1[v]W. Thus PT
á
W = (PW
á
T)-1.
C Let V = R3 and let U = {u1, u2, u3} = {}
1
0
0
,
0
1
0
,
0
0
1
be the standard basis for R3. If v = [a, b, c]T, then
v .
=
a
b
c
=a
1
0
0
+b
0
1
0
+c
0
0
1
Hence for the standard basis U of Rn, v = [v]U .
CV = R3. W = {w1, w2, w3} = { }.
1
0
0
,
0
1
1
,
0
1
−1
PU
á
W = ([w1]U | [w2]U | [w3]U) = (w1 | w2 | w3) = .
10 0
01 1
01−1
D If [v]W = , what is v?
1
2
3
v = 1w1+2w2+3w3 = = .
1
1
0
0
+2
0
1
1
+3
0
1
−1
1
5
−1
Another way to get v is:
PU
á
W[v]W = = = [v]U = v .
10 0
01 1
01−1
1
2
3
1
5
−1
E If v = , what is [v]W? [v]W = iff
1
2
3
x
y
z
xw1+yw2+zw3 = v. We solve this two ways: L, LL.
L Translating this into a column vector problem gives
x
1
0
0
+y
0
1
1
+z
0
1
−1
=
1
2
3
Reducing the augmented matrix gives:
10 0 1
01 1 2
01−13
→
100 1
010 5
2
001−1
2
Hence x = 1, y = 5/2, z = -1/2 and so [v]w = .
1
5
2
−1
2
LL [v]W = PW
á
U[v]U = (PU
á
W)-1v
Calculating (PU
á
W)-1 gives
10 0 100
01 1 010
01−1001
→
10010 0
01001
21
2
00101
2−1
2
+Hence (PU
á
W)-1 = .
10 0
01
21
2
01
2−1
2
+Hence [v]W = (PU
á
W)-1v = (PU
á
W)-1 = .
1
2
3
1
5
2
−1
2
The examples above show that [v]U and PU
á
W are easy to
get for the standard basis. From them we can calculate
every coordinate vector and transition matrix.
LEMMA. For V=Rn for some n and U={[1,0,...]T, [0,1,... ]T,
... } the standard basis and W and T any other bases:
D [v]U = vE PU
á
W = (w1 | w2 | w3)
F PW
á
U = (PU
á
W)-1 G [v]W = PW
á
Uv
H PW
á
T = PW
á
UPU
á
T
PROOF. D, E, and F are as in the example. G PW
á
Uv =
PW
á
U[v]U = [v]W. H follows from Lemma E above.
CV = R2, S = { }, T = {}.
1
1
,
1
−1
0
1
,
−1
0
D Find PS
á
T. The definition is faster but we'll use the lemma.
PU
á
T = .
0−1
10
PS
á
U = (PU
á
S)-1 = =
11
1−1
−1
1
21
2
1
2−1
2
PS
á
T = PS
á
UPU
á
T = = .
1
21
2
1
2−1
2
⋅
0−1
10
1
2−1
2
−1
2−1
2
E If [v]T = , then [v]S = PS
á
T[v]T =
2
10
= .
1
2−1
2
−1
2−1
2
2
10
−4
−6