MATH 355 Supplemental Notes Coordinatization
Coordinatization
In R3, we have the standard basis i, j and k. When we write a vector in coordinate form, say
3 v 2 , (1) “ »´ fi 5 — – fl it is understood as v 3i 2j 5k. “ ´ ` The numbers 3, 2 and 5 are the coordinates of v relative to the standard basis ⇠ i, j, k . It has p´ q “p q always been understood that a coordinate representation such as that in (1) is with respect to the ordered basis ⇠. A little thought reveals that it need not be so. One could have chosen the same basis elements in a di↵erent order, as in the basis ⇠ i, k, j . We employ notation indicating the 1 “p q coordinates are with respect to the di↵erent basis ⇠1:
3 v 5 , to mean that v 3i 5k 2j, r s⇠1 “ » fi “ ` ´ 2 —´ – fl reflecting the order in which the basis elements fall in ⇠1. Of course, one could employ similar notation even when the coordinates are expressed in terms of the standard basis, writing v for r s⇠ (1), but whenever we have coordinatization with respect to the standard basis of Rn in mind, we will consider the wrapper to be optional. r¨s⇠ Of course, there are many non-standard bases of Rn. In fact, any linearly independent collection of n vectors in Rn provides a basis. Say we take
1 1 1 4 ´ ´ » 0fi » 1fi » 1fi » 1fi u , u u u ´ . 1 “ 2 “ 3 “ 4 “ — 3 — 1 — 0 — 2 — —´ — — — 0 — 4 — 2 — 1 — — — —´ – fl – fl – fl – fl As per the discussion above, these vectors are being expressed relative to the standard basis of R4. A quick check of a related 4-by-4 determinant reveals these vectors are linearly independent, and hence form a basis of R4.
octave:80> A = [ 1 0 3 0; 1 1 1 4; 1 1 0 2; 4 12 1]’ ⌥ ´ ´ ´ ´ ´ ⌅ A =
11 14 ´ ´ 011 1 ´ 3 10 2 ´ 042 1 ´
2 MATH 355 Supplemental Notes Matrix of a Linear Transformation
octave:81> det(A) ans = 17 ´ ⌃ ⇧ Let us denote the ordered basis as ↵ u , u , u , u . Every vector of R4 can be expressed uniquely “p 1 2 3 4q as a linear combination of elements in the basis ↵. Gaussian elimination is a tool that reveals the weights.
Example 1:
Find the coordinatization relative to the basis ↵ of the vector v 8, 3, 13, 5 . “p ´ ´ q We can reduce the relevant augmented matrix to RREF:
octave:85> augA = [A [8; 3; 13; 5]] ⌥ ´ ´ ⌅ augA =
11 14 8 ´ ´ 011 1 3 ´ ´ 3 1 0 2 13 ´ 042 1 5 ´ ´
octave:86> rref(augA) ans =
10002 0100 1 ´ 00101 00013
⌃ ⇧ Thus, v 2u u u 3u , and “ 1 ´ 2 ` 3 ` 4 2 » 1fi v ´ . r s↵ “ — 1 — — 3 — – fl
Matrix of a Linear Transformation
Let T : be a linear transformation from the n-dimensional vector space having ordered V Ñ W V basis ↵ v , v2,...,v into to the m-dimensional vector space , having ordered basis “p 1 nq W “ w , w ,...,w . As we have seen, as soon as one knows the outputs from T for a basis, all of its p 1 2 mq
3 MATH 355 Supplemental Notes Matrix of a Linear Transformation behavior is known. Let us suppose we know the outputs of T on the basis ↵. More specifically, let us assume we know how each T v , an element of , is expressed relative to the basis . p kq W Specifically, we know the weights aik in the expression
a1k »a2k fi T v a w a w a w , so that T v . p kq“ 1k 1 ` 2k 2 `¨¨¨` mk m r p kqs “ . — . — —a — mk – fl Then the matrix of T in the ordered bases ↵ and is
a a a 11 12 ¨¨¨ 1n T v1 T v2 T vn »a21 a22 a2n fi T p q p q ¨¨¨ p q ¨¨¨ . (2) ↵ “ “ . . .. . «“ ‰ “ ‰ “ ‰ — . . . . “ ‰ ÓÓ Ó— —a a a — m1 m2 ¨¨¨ mn – fl Note that, if , it is still possible to have di↵erent bases ↵ and , but if the same basis V “ W ↵ is used twice, we can simplify the notation and write T ↵ as T ↵. You may realize from this discussion that there are many di↵erent matrix representations“ ‰ for“ ‰ the same linear transformation T, potentially one for each pair of bases ↵ and .
Example 2:
In R2 and R3 consider ordered bases ↵ v , v and w , w , w , respectively, where “p 1 2q “p 1 2 3q 4 0 1 2 1 ´ v1 , v2 ´ , and w1 » 2fi , w2 » 3fi , w3 » 2fi . “ «3 “ « 5 “ ´ “ “ 0 2 3 — —´ — – fl – fl – fl Suppose T : R2 R3 is linear, and produces the following images for the elements in ↵: Ñ 5 5
T v 1 and T v2 2 . p 1q“»´ fi p q“» fi 11 7 — —´ – fl – fl Find the matrix T ↵. “ ‰ Since the augmented matrices
4015 100 2 2 ´ 2321 010 1 , we have T v 1 , and »´ ´ fi „ » ´ fi p 1q “ »´ fi 0 2 3 11 001 3 3 — ´ — “ ‰ — – fl – fl – fl 4015 100 1 1 ´ 2322 010 2, so T v2 2 . »´ fi „ » fi p q “ » fi 0 237 001 1 1 — ´ ´ — ´ “ ‰ —´ – fl – fl – fl 4 MATH 355 Supplemental Notes Matrix of a Linear Transformation
Thus, 21 T v T v p 1q p 2q T ↵ » 12fi . “ « “ ´ “ ÓÓ‰ “ ‰ 3 1 “ ‰ — ´ – fl