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Coordinatization 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 — ffi – 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 —´ ffi – 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 “ — 3ffi — 1ffi — 0ffi — 2ffi — ffi —´ ffi — ffi — ffi — 0ffi — 4ffi — 2ffi — 1ffi — ffi — ffi — ffi —´ ffi – 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↵ “ — 1ffi — ffi — 3ffi — ffi – 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 β “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 β. 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 kqsβ “ . — . ffi — ffi —a ffi — mkffi – fl Then the matrix of T in the ordered bases ↵ and β is a a a 11 12 ¨¨¨ 1n β T v1 T v2 T vn »a21 a22 a2n fi T p q β p q β ¨¨¨ p q β ¨¨¨ . (2) ↵ “ “ . .. «“ ‰ “ ‰ “ ‰ ff — . ffi “ ‰ ÓÓ Ó— ffi —a a a ffi — m1 m2 ¨¨¨ mnffi – fl Note that, if , it is still possible to have di↵erent bases ↵ and β, 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 β. Example 2: In R2 and R3 consider ordered bases ↵ v , v and β 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 . “ «3ff “ « 5ff “ ´ “ “ 0 2 3 — ffi —´ ffi — ffi – 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 — ffi —´ ffi – fl – fl β 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 β “ »´ fi 0 2 3 11 001 3 3 — ´ ffi — ffi “ ‰ — ffi – fl – fl – fl 4015 100 1 1 ´ 2322 010 2, so T v2 2 . »´ fi „ » fi p q β “ » fi 0 237 001 1 1 — ´ ´ ffi — ´ ffi “ ‰ —´ ffi – fl – fl – fl 4 MATH 355 Supplemental Notes Matrix of a Linear Transformation Thus, 21 T v T v β p 1q β p 2q β T ↵ » 12fi . “ « ff “ ´ “ ÓÓ‰ “ ‰ 3 1 “ ‰ — ´ ffi – fl β One expects that, having found the matrix T , it can be used to find T v for any input vector ↵ r p qsβ v. Notice that one obtains “ ‰ 21 2 1 T v1 β » 12fi » 1fi . r p qs “ ´ «0ff “ ´ 3 1 3 — ´ ffi — ffi – fl – fl That is, the result T v comes from matrix multiplication after expressing v as 1, 0 , which is r p 1qsβ 1 p q its coordinatization relative to the basis ↵. Speaking generally, it is the case that β T v T v . (3) p q β “ ↵ ↵ “ ‰ “ ‰ “ ‰ Example 3: rd 3, the vector subspace of R consisting of at-most-3 -degree polynomials, has ordered P 2 3 Cp q basis ↵ 1, x, x , x . Vectors/members of 3 can be coordinatized with respect to this basis. “p q P For instance, the polynomial 7 » 1fi p x 2x3 x 7 has coordinates p ´ . p q“ ´ ` ↵ “ — 0ffi “ ‰ — ffi — 2ffi — ffi – fl The operation of di↵erentiation, d dx, is linear, and for each input p x from 3 the image/result { p q P d dx p p1 x is in 2 and can be represented by coordinates relative to the ordered basis { p q“2 p q P β 1, x, x of 2. We have “p q P 0100 d dx 1 d dx x d dx x2 d dx x3 β { p q β { p q β { p q β { p q β d dx ↵ »0020fi . { “ « ff “ “ ÓÓ‰ “ ‰ “ Ó‰ “ Ó‰ 0003 “ ‰ — ffi – fl 3 Using this, you can find the derivative of any polynomial in 3. For p x 2x x 7, for P p q“ ´ ` instance, its derivative is 7 0100 1 β » 1fi ´ d dx p 0020 ´ 0 , { ↵ ↵ “ » fi “ » fi — 0ffi “ ‰ “ ‰ 0003— ffi 6 — ffi — 2ffi — ffi – fl — ffi – fl – fl which should be understood as p x 6x2 1. 1p q“ ´ 5 MATH 355 Supplemental Notes Change of Basis An important result is the following, which affirms that matrix multiplication is defined in the right way. Theorem 1: Suppose , , and are finite-dimensional vector spaces with ordered bases U V W ↵, β and γ, respectively. Suppose that S: and T : are linear transformations. U Ñ V V Ñ W Then The composition T S given by T S x T S x is a linear transformation from • ˝ p ˝ qp q“ p p qq to . U W The matrix of T S in the ordered bases ↵ and γ is the product of the matrix of S in • ˝ the ordered bases ↵ and β with the matrix of T in the ordered bases β and γ; that is γ γ β T S T S . ˝ ↵ “ β ↵ “ ‰ “ ‰ “ ‰ Change of Basis A special instance is the identity operator id: Rn Rn, given by id x x. If ⇠ represents the Ñ p q“ standard basis for Rn, and ↵ v ,...,v is another ordered basis, then “p 1 nq ⇠ v1 v2 vn id ↵ ¨¨¨ . “ « ff “ ‰ ÓÓ Ó We call this a change-of-basis, or transition matrix, as it maps a vector from its coordinate representation relative to basis ↵ to its representation in another (in this case, the standard) basis, ⇠: ⇠ v id v . ⇠ “ ↵ ↵ “ ‰ “ ‰ “ ‰ What this means is that any nonsingular matrix X can be viewed as a change-of-basis matrix, mapping from coordinates relative to the ordered basis ↵ consisting of the columns of X to standard coordinates. The matrix which maps from standard coordinates back to coordinates in the ordered ⇠ ↵ basis ↵ is naturally X 1. That is, if id X, then id X 1. ´ ↵ “ ⇠ “ ´ As with general linear maps, one might“ ‰ use two nonstandard“ ‰ bases ↵ v ,...,v and β for Rn. “p 1 nq The matrix of the identity map id: Rn Rn in the ordered bases ↵ and β, since id v v , is Ñ p jq“ j 6 MATH 355 Supplemental Notes Similarity found by tailoring the formula (2) appropriately: v v v β 1 β 2 β ¨¨¨ n β id ↵ .
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