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LECTURE 15
Coordinatization and Change of Basis
- Review: Coordinatization of Vectors
Suppose B = {u 1 , u 2 ,... , un} is a basis of Rn. Then, by definition, for any vector v ∈ Rn, there is a unique choice of coefficients c 1 ,... , cn such that v = c 1 u 1 + · · · + cnun We can thus take the (ordered set of) n numbers c 1 ,... , cn and assemble them into a vector. We call the result the coordinate vector of u with respect to the basis B, and denote it by vB : vB = [c 1 ,... , cn] Example 15.1. Find the coordinate vector of [1, 2 , −2] with respect to the basis {[1, 1 , 1] , [1, 2 , 0] , [1, 0 , 1]} of R^3.
(^) + c 2
(^) + c 3
which is the linear system corresponding to the following augmented matrix:
This matrix row reduces to the following reduced row echelon form:
And so the solution is c 1 = − 4 , c 2 = 3 , c 3 = 2 So vB = [− 4 , 3 , 2]
Summary: To find the coordinate vector vB of a vector v ∈ Rn^ with respect to an ordered basis B = {u 1 ,... , un} , Fact 15.2. (1) Form the augmented matrix [u 1 ,... , un | v] using the vectors u 1 ,... , un as the first n columns, and the vector v as the last column. (2) Row reduce the augmented matrix to reduced row-echelon form [I | vB ]. The vector vB in the last column will be the desired coordinate vector.
The method described above is good for finding the coordinate vector of a particular vector v with respect to the basis B = {u 1 ,... , un}, but what if we had needed to find the coordinate vectors for a whole bunch 1
of vectors v or even an arbitrary vector v in Rn. Well, note that the left hand side of the condition we wish to satisfy c 1 u 1 + c 2 u 2 + · · · + cnun = v can be written MB c = v where M is the matrix constructed by using the basis vectors v 1 ,... , vn as columns and c is the column vector constructed out of the coefficients c 1 ,... , cn when we apply the identity
MB c =
u 1 · · · un ↓ ↓
c 1 ... cn
= c 1
u 1 ↓
(^) + c 2
u 2 ↓
(^) + · · · + cn
un ↓
Thus, Fact 15.3. The problem of finding the coordinate vector vB for a vector v with respect to a basis B = {u 1 ,... , un} is equivalent to solving the following linear system MB x = v
Now note that since the elements of a basis are, practically by definition, linearly independent, the matrix MB constructed as above from a basis B will always be invertible. Thus, for any vector v we can obtain the coordinate vector vB of v with respect to B by setting vB = (MB )−^1 v.
Reinterpretation:
Think of the original vector v = [v 1 ,... , vn] as the coordinate vector of v with respect to the standard basis Bstd = {e 1 ,... , en} of Rn^ : v = [v 1 ,... , vn] = v 1 [1, 0 ,... , 0] + v 2 [0, 1 , 0 ,... , 0] + · · · vn [0,... , 0 , 1] = v 1 e 1 + v 2 e 2 + · · · + vnvn =⇒ vBstd = [v 1 ,... , vn] Then the matrix (MB )−^1 converts the coordinate vector for v with respect to the standard basis to the coordinate vector for v with respecto to the basis B: vB = (MB )−^1 v = (MB )−^1 vBstd
- Change of Basis
Suppose now we have two different, non-standard, bases for Rn^ : B = {b 1 ,... , bn} B′^ = {b′ 1 ,... , b′ n} Let v be an arbitrary vector in Rn. How might we convert the coordinate vector vB for v with respect to the basis B = {b 1 ,... , bn} directly to the coordinate vector vB′ for v with respect to the basis B′. Well, let
MB =
b 1 · · · bn | · · · |
, MB′ =
b′ 1 · · · b′ n | · · · |
so that vB = (MB )−^1 v vB′^ = (MB′^ )−^1 v
such that () c 1 b 1 + · · · + cnbn = v = c′ 1 b′ 1 + · · · + c′ nb′ n Suppose we know the coordinate vectors of b′ 1 ,... , b′ n with respect to the basis B. b′ 1 = b′ 11 b 1 + · · · b′ 1 nbn ... b′ n = b′ n 1 b 1 + · · · + b′ nnbn Then, since the vectors b 1 ,... , bn are all linearly independent, the linear system () c 1 b 1 + · · · + cnbn = c′ 1 (b′ 11 b 1 + · · · b′ 1 nbn) + · · · + c′ n (b′ n 1 b 1 + · · · b′ nnbn) leads to c 1 = (c′ 1 b′ 11 + · · · + c′ nb′ n 1 ) c 2 = (c′ 1 b′ 21 + · · · + c′ nb′ n 2 ) ... cn = (c′ 1 b 1 n + · · · + c′ nb′ nn) or
vB =
c 1 ... cn
b′ 11 b′ n 1 b′ 1 n bnn′
c′ 1 ... c′ n
= CBB′ vB′
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