Matrix Representations of Linear Transformations and Changes of Coordinates

Matrix Representations of Linear Transformations and Changes of Coordinates 01 Subspaces and Bases 011 Definitions A subspace V of R n is a subset of R n that contains the zero element and is closed under addition and scalar multiplication: (1) 0 V (2) u, v V u + v V (3) u V and k R ku V Equivalently, V is a subspace if au + bv V for all a, b R and u, v V (You should try to prove that this is an equivalent statement to the first) Example 01 Let V {(t, 3t, 2t) t R} Then V is a subspace of R 3 : (1) 0 V because we can take t 0 (2) If u, v V, then u (s, 3s, 2s) and v (t, 3t, 2t) for some real numbers s and t But then u + v (s + t, 3s + 3t, 2s 2t) (s + t, 3(s + t), 2(s + t)) (t, 3t, 2t ) V where t s + t R (3) If u V, then u (t, 3t, 2t) for some t R, so if k R, then ku (kt, 3(kt), 2(kt)) (t, 3t, 2t ) V where t kt R Example 02 The unit circle S 1 in R 2 is not a subspace because it doesn t contain 0 (0, 0) and because, for example, (1, 0) and (0, 1) lie in S but (1, 0) + (0, 1) (1, 1) does not Similarly, (1, 0) lies in S but 2(1, 0) (2, 0) does not A linear combination of vectors v 1,, v k R n is the finite sum a 1 v 1 + + a k v k (01) which is a vector in R n (because R n is a subspace of itself, right?) The a i R are called the coefficients of the linear combination If a 1 a k 0, then the linear combination is said to be trivial In particular, considering the special case of 0 R n, the zero vector, we note that 0 may always be represented as a linear combination of any vectors u 1,, u k R n, 0u 1 + + 0u k 0 This representation is called the trivial representation of 0 by u 1,, u k If, on the other hand, there are vectors u 1,, u k R n and scalars a 1,, a n R such that a 1 u 1 + + a k u k 0 1

where at least one a i 0, then that linear combination is called a nontrivial representation of 0 Using linear combinations we can generate subspaces, as follows If S is a nonempty subset of R n, then the span of S is given by span(s) : {v R n v is a linear combination of vectors in S} (02) The span of the empty set,, is by definition span( ) : {0} (03) Remark 03 We showed in class that span(s) is always a subspace of R n (well, we showed this for S a finite collection of vectors S {u 1,, u k }, but you should check that it s true for any S) Let V : span(s) be the subspace of R n spanned by some S R n Then S is said to generate or span V, and to be a generating or spanning set for V If V is already known to be a subspace, then finding a spanning set S for V can be useful, because it is often easier to work with the smaller spanning set than with the entire subspace V, for example if we are trying to understand the behavior of linear transformations on V Example 04 x-y plane Let S be the unit circle in R 3 which lies in the x-y plane Then span(s) is the entire Example 05 Let S {(x, y, z) R 3 x y 0, 1 < z < 3} Then span(s) is the z-axis A nonempty subset S of a vector space R n is said to be linearly independent if, taking any finite number of distinct vectors u 1,, u k S, we have for all a 1,, a k R that a 1 u 1 + a 2 u 2 + + a k u k 0 a 1 a n 0 That is S is linearly independent if the only representation of 0 R n by vectors in S is the trivial one In this case, the vectors u 1,, u k themselves are also said to be linearly independent Otherwise, if there is at least one nontrivial representation of 0 by vectors in S, then S is said to be linearly dependent Example 06 The vectors u (1, 2) and v (0, 1) in R 2 are linearly independent, because if that is au + bv 0 a(1, 2) + b(0, 1) (0, 0) then (a, 2a b) (0, 0), which gives a system of equations: But the matrix by it gives which means a b 0 a 0 2a b 0 or 1 0 a 0 2 1 b 0 1 0 is invertible, in fact it is its own inverse, so that left-multiplying both sides 2 1 a b 1 0 0 1 a b 1 0 2 1 2 a b 1 0 2 1 0 0 0 0 2

Example 07 The vectors (1, 2, 3), (4, 5, 6), (7, 8, 9) R 3 are not linearly independent because 1(1, 2, 3) 2(4, 5, 6) + 1(7, 8, 9) (0, 0, 0) That is, we have found a 1, b 2 and c 1, not all of which are zero, such that a(1, 2, 3) + b(4, 5, 6) + c(7, 8, 9) (0, 0, 0) Given S V, a nonzero vector v S is said to be an essentially unique linear combination of the vectors in S if, up to order of terms, there is one and only one way to express v as a linear combination of u 1,, u k S That is, if there are a 1,, a n, b 1,, b l R\{0} and distinct u 1,, u k S and distinct v 1,, v l S distinct, then, re-indexing the b i s if necessary, } { } v a 1 u 1 + + a n u k ai b i k l and for all i 1,, k b 1 v 1 + + b l v l u i v i If V is a subspace of R n, then a subset of V is called a basis for V if it is linearly independent and spans V We also say that the vectors of form a basis for V Equivalently, as explained in Theorem 011 below, is a basis if every nonzero vector v V is an essentially unique linear combination of vectors in Remark 08 In the context of inner product spaces V of inifinite dimension, there is a difference between a vector space basis, the Hamel basis of V, and an orthonormal basis for V, the Hilbert basis for V, because though the two always exist, they are not always equal unless dim(v ) < The dimension of a subspace V of R n is the cardinality of any basis for V, ie the number of elements in (which may in principle be infinite), and is denoted dim(v ) This is a well defined concept, by Theorem 013 below, since all bases have the same size V is finite-dimensional if it is the zero vector space {0} or if it has a basis of finite cardinality Otherwise, if it s basis has infinite cardinality, it is called infinite-dimensional In the former case, dim(v ) k < for some n N, and V is said to be k-dimensional, while in the latter case, dim(v ) κ, where κ is a cardinal number, and V is said to be κ-dimensional Remark 09 bases for R 2 Bases are not unique For example, {e 1, e 2 } and γ {(1, 1), (1, 0)} are both If V is finite-dimensional, say of dimension n, then an ordered basis for V a finite sequence or n-tuple (v 1,, v n ) of linearly independent vectors v 1,, v n V such that {v 1,, v n } is a basis for V If V is infinite-dimensional but with a countable basis, then an ordered basis is a sequence (v n ) n N such that the set {v n n N} is a basis for V 3

012 Properties of Bases Theorem 010 of the other v j Vectors v 1,, v k R n are linearly independent iff no v i is a linear combination Proof: Let v 1,, v k R n be linearly independent and suppose that v k c 1 v 1 + + c k 1 v k 1 (we may suppose v k is a linear combination of the other v j, else we can simply re-index so that this is the case) Then c 1 v 1 + + c k 1 v k 1 + ( 1)v k 0 But this contradicts linear independence, since 1 0 Hence v k cannot be a linear combination of the other v k By re-indexing the v i we can conclude this for all v i Conversely, suppose v 1,, v k are linearly dependent, ie there are scalars c 1,, c k R not all zero such that c 1 v 1 + + c k v k 0 Say c i 0 Then, ( v i c ) ( 1 v 1 + + c ) ( i 1 v i 1 + c ) ( i+1 v i+1 + + c ) k v k c i c i c i c i so that v i is a linear combination of the other v j This is the contrapositive of the equivalent statement, If no v i is a linear combination of the other v j, then v 1,, v k are linearly independent Theorem 011 Let V be a subspace of R n Then a collection {v 1,, v k } is a basis for V iff every vector v V has an essentially unique expression as a linear combination of the basis vectors v i Proof: Suppose is a basis and suppose that v has two representations as a linear combination of the v i : Then, v c 1 v 1 + + c k v k d 1 v 1 + + d k v k 0 v v (c 1 d 1 )v 1 + + (c k d k )v k so by linear independence we must have c 1 d 1 c k d k 0, or c i d i for all i, and so v has only one expression as a linear combination of basis vectors, up to order of the v i Conversely, suppose every v V has an essentially unique expression as a linear combination of the v i Then clearly is a spanning set for V, and moreover the v i are linearly independent: for note, since 0v 1 + + 0v k 0, by uniqueness of representations we must have c 1 v 1 + + c k v k 0 c 1 c k 0 Thus is a basis 4

Theorem 012 (Replacement Theorem) Let V be a subspace of R n and let v 1,, v p and w 1,, w q be vectors in V If v 1,, v p are linearly independent and w 1,, w q span V, then p q Proof: Let A w 1 w q R n q be the matrix whose columns are the w j and let B v 1 v p R n p be the matrix whose columns are the v k Then note that {v 1,, v p } V span(w 1,, w q ) im A Thus, there exist u 1,, u p R q such that Au i v i Consequently, B v 1 v p Au 1 u p A u 1 u p AC where C u 1 u p R q p Now, since v 1 v p are linearly independent, c 1 v 1 + +c p v p 0 implies all c i 0, ie Bc 0 implies c 0, or ker B {0} But you will notice that ker C ker B, since if x ker C, then the fact that B AC implies Bx (AC)x A(Cx) A0 0, or x ker C Since ker B {0}, this means that ker C {0} as well But then C must have at least as many rows as columns, ie p q, because rref(c) must have the form submatrix, but at least with I p in the top portion Ip O, possibly with no O Theorem 013 Let V be a subspace for R n Then all bases for V have the same size Proof: By the previous theorem two bases {v 1,, v p } and γ {w 1,, w q } for V both span V and both are linearly independent, so we have p q and p q Therefore p q Corollary 014 All bases for R n have n vectors Proof: Notice that ρ {e 1,, e n } forms a basis for R n : first, the elementary vectors e i span R n, since if x (a 1,, a n ) R n, then x (a 1,, a n ) a 1 (1, 0,, 0) + a 2 (0, 1, 0,, 0) + + a n (0,, 0, 1) Also, e 1,, e n are linearly independent, for if 0 c 1 e 1 + + c n e n a 1 e 1 + a 2 e 2 + + a n e n span(e 1,, e n ) 1 e 1 e n c I n c c n then c (c 1,, c n ) ker I n {0}, so c 1 c n 0 Since ρ n, all bases for R n satisfy n be the previous theorem Theorem 015 (Characterizations of Bases) If V is a subspace of R n and dim(v ) k, then (1) There are at most k linearly independent vectors in V Consequently, a basis is a maximal linearly independent set in V (2) At least k vectors are needed to span V Thus a basis is a minimal spanning set (3) If k vectors in V are linearly independent, then they form a basis for V (4) If k vectors span V, then they form a basis for V 5

Proof: (1) If v 1,, v p V are linearly independent and w 1,, w k V form a basis for V, then p k by the Replacement Theorem (2) If v 1,, v p V span V and w 1,, w k V form a basis for V then again we must have k p by the Replacement Theorem (3) If v 1,, v k V are linearly independent, we must show they also span V Pick v V and note that by (1) the vectors v 1,, v k, V are linearly dependent, because there are k + 1 of them (4) If v 1,, v k, V span V but are not linearly independent, then say v k span(v 1,, v k 1 ) But in this case V span(v 1,, v k 1 ), contradicting (2) Theorem 016 If A R m n, then dim(im A) rank A Proof: This follows from Theorem 015 in Systems of Linear Equations, since if B rref(a), then rank A rank B # of columns of the form e i in B # of nonredundant vectors in A Theorem 017 (Rank-Nullity Theorem) If A R m n, then dim(ker A) + dim(im A) n (04) or null A + rank A n (05) Proof: If B rref(a), then dim(ker A) n # of leading 1s n rank A 6

02 Coordinate Representations of Vectors and Matrix Representations of Linear Transformations 021 Definitions If (v 1,, v k ) is an ordered basis for a subspace V of R n, then we know that for any vector v V there are unique scalars a 1,, a k R such that v a 1 v 1 + + a k v k The coordinate vector of v R n with respect to, or relative to, is defined to be the (column) vector in R k consisting of the scalars a i : a 1 a 2 x : a k and the coordinate map, also called the standard representation of V with respect to, is given by (06) φ : V R k (07) φ (x) x (08) Example 018 Let v (5, 7, 9) R 3 and let (v 1, v 2 ) be the ordered basis for V span(v 1, v 2 ), where v 1 (1, 1, 1) and v 2 (1, 2, 3) Can you express v as a linear combination of v 1 and v 2? In other words, does v lie in V? If so, find v Solution: To find out whether v lies in V, we must see if there are scalars a, b R such that v av 1 + bv 2 Note, that if we treat v and the v i as column vectors we get a matrix equation: a v av 1 + bv 2 v 1 v 2 b or 5 1 1 7 1 2 a b 9 1 3 This is a system of equations, Ax b with 1 1 A 1 2, x 1 3 a b, b Well, the agmented matrix A b reduces to rref(a b) as follows: 1 1 5 1 0 3 1 2 7 0 1 2 1 3 9 0 0 0 This means that a 3 and b 2, so that v 3v 1 + 2v 2 5 7 9 and v lies in V, and moreover v 3 2 7

In general, if (v 1,, v k ) is a basis for a subspace V of R n and v V, then the coordinate map will give us a matrix equation if we treat all the vectors v, v 1,, v k as column vectors: or v a 1 v 1 + + a k v k v 1 v k v Bv where B v 1 v k, the n k matrix with columns v j If V R n, then B will be an n n matrix whose columns are linearly independent Therefore, im B R n, so that by the Rank-Nullity Theorem ker B {0}, which means B represents an injective and surjective linear transformation, and is therefore invertible In this case, we can solve for v rather easily: a 1 a k v B 1 v (09) Let V and W be finite dimensional subspaces of R n and R m, respectively, with ordered bases (v 1,, v k ) and γ (w 1,, w l ), respectively If there exist (and there do exist) unique scalars a ij R such that l T (v j ) a ij w i for j 1,, k (010) i1 then the matrix representation of a linear transformation T L(V, W ) in the ordered bases and γ is the l k matrix A defined by A ij a ij, A T γ : T (v 1 ) γ T (v 2 ) γ T (v k ) γ a 11 a 21 a l1 a 12 a 22 a l2 a 1k a 2k a lk a 11 a 1k a l1 a lk Note that T (v) γ ϕ γ (T (v)) (ϕ γ T )(v) Notation 019 If V W and γ, we write T instead of T Example 020 Let T L(R 2, R 3 ) be given by T (x, y) (x + y, 2x y, 3x + 5y) In terms of matrices and column vectors T behaves as follows: 1 1 x T 2 1 x y y 3 5 But this matrix, call it A, is actually the representation of T with respect to the standard ordered bases ρ 2 (e 1, e 2 ) and ρ 3 (e 1, e 2, 3 ), that A T ρ3 ρ 2 What if we were to choose different bases for R 2 and R 3? Say, ( (1, 1), (0, 1) ), γ ( (1, 1, 1), (1, 0, 1), (0, 0, 1) ) 8

How would T look with respect to these bases? Let us first find the coordinate representations of T (1, 1) and T (0, 1) with resepct to γ: Note, T (1, 1) (2, 1, 8) and T (0, 1) ( 1, 1, 5), and to find (2, 1, 8) γ and ( 1, 1, 5) we have to solve the equations: 1 1 0 2 1 0 0 1 1 1 1 8 γ 2 1 8 and If B is the matrix 1 1 0 1 0 0, then B 1 0 1 0 1 1 0, so 1 1 1 1 0 1 2 0 1 0 2 1 1 1 1 1 0 1 1 and 1 8 1 0 1 8 6 5 γ Let us verify this: 1 1 0 1 1 0 0 1 1 1 1 5 γ γ 1 1 5 0 1 0 1 1 1 1 0 1 2 1 0 1 5 4 1(1, 1, 1) + 1(1, 0, 1) + 6(0, 0, 1) (2, 1, 8) and 1(1, 1, 1) 2(1, 0, 1) 4(0, 0, 1) ( 1, 1, 5) so indeed we have found T (1, 1) γ and T (0, 1) γ, and therefore T γ : T (1, 1) γ T (0, 1) γ 1 1 1 2 1 1 1 2 6 4 6 4 022 Properties of Coordinate and Matrix Representations Theorem 021 (Linearity of Coordinates) all x, y V and all k R we have Let be a basis for a subspace V of R n Then, for (1) x + y x + y (2) kx kx Proof: (1) On the one hand, x + y Bx + y, and on the other x Bx and y By, so x + y Bx + By Thus, Bx + y x + y Bx + By B(x + y ) so that, subtracting the right hand side from both sides, we get B ( x + y (x + y ) ) 0 Now, B s columns are basis vectors, so they are linearly independent, which means Bx 0 has only the trivial solution, because if (v 1,, v k ) and x (x 1,, x k ), then 0 x 1 v 1 + + x k v k Bx x 1 x k 0, or x 0 But this means the kernel of B is {0}, so that x + y (x + y ) 0 or x + y x + y The proof of (2) follows even more straighforwardly: First, note that if x a 1 x a 1 v 1 + + a k v k, so that kx ka 1 v 1 + + ka k v k, and therefore a k T, then kx ka 1 ka n T ka 1 a k T kx 9

Corollary 022 The coordinate maps ϕ are linear, ie ϕ L(V, R k ), and further they are isomorphisms, that is they are invertible, and so ϕ GL(V, R k ) Proof: Linearity was shown in the previous theorem To see that ϕ is an isomorphism, note first that ϕ takes bases to bases: if (v 1,, v k ) is a basis for V, then ϕ (v i ) e i, since v i 0v 1 + + 1v i + + 0v k Thus, it takes to the standard basis ρ k (e 1,, e k ) for R k Consequently, it is surjective, because if x (x 1,, x k ) R k, then x x 1 e 1 + + x k e k x 1 ϕ (v 1 ) + + x k ϕ (v k ) ϕ (x 1 v 1 + + x k v k ) If we let v x 1 v 1 + + x k v k, then we see that v V satisfies ϕ (v) x But ϕ is also injective: if v ker ϕ, then v a 1 v 1 + + a k v k, so that 0 ϕ (v) ϕ (a 1 v 1 + + a k v k ) a 1 ϕ (v 1 ) + + a k ϕ (v k ) a 1 e 1 + + a k e k By the linear independence of the e i we must have a 1 a k 0, and so v 0 Thus, ϕ is also injective Theorem 023 Let V and W be finite-dimensional subspaces of R n having ordered bases (v 1,, v k ) and γ (w 1,, w l ), respectively, and let T L(V, W ) Then for all v V we have T (v) γ T γ v (011) In other words, if D T γ is the matrix representation of T in and γ coordinates, with T D L(R k, R l ) the corresponding matrix multiplication map, if A T ρ l ρ k is the matrix representation of T in standard coordinates, with corresponding T A L(R k, R l ), and if φ GL(V, R k ) and φ γ GL(W, R l ) are the respective coordinate maps, with matrix representations B 1 and C 1, respectively, where B v 1 v k and C w1 w l, then φ γ T T D φ or, in terms of matrices, C 1 A DB 1 (012) and the following diagrams commute: V T W V A W φ R k φ γ R l T D or, in terms of matrices, B 1 R k C 1 D R l Proof: If (v 1,, v k ) is an odered basis for V and γ (w 1,, w l ) is an ordered basis for W, then let a 11 a 1kn a 11 a 1k T γ T (v 1 ) γ T (v k ) γ a l1 a lk a l1 a lk Now, for all u V there are unique b 1,, b n R such that u b 1 v 1 + + b k v k Therefore, T (u) T (b 1 v 1 + + b k v k ) b 1 T (v 1 ) + + b k T (v k ) 10

so that by the linearity of ϕ, T (u) γ φ γ ( T (u) ) φ γ ( b1 T (v 1 ) + + b k T (v k ) ) b 1 φ γ ( T (v1 ) ) + + b k φ γ ( T (vk ) ) b 1 T (v 1 ) γ + + b k T (v k ) γ T (v 1 ) γ T (v k ) γ T γ u This shows that φ γ T T D φ, since T (u) γ (ϕ γ T )(u) and T γ u (T D ϕ )(u) Finally, since ϕ (x) B 1 x and ϕ γ (y) C 1 y, we have the equivalent statement C 1 A DB 1 b 1 b k Remark 024 Two square matrices A, B R n n are said to be similar, denoted A B, if there exists an invertible matrix P GL(n, R) such that A P BP 1 Similarity is an equivalence relation (it is reflexive, symmetric and transitive): (1) (reflexivity) A A because we can take P I n, so that A I n AI 1 n (2) (symmetry) If A B, then there is an invertible P such that A P BP 1 But then leftmultiplying by P 1 and right-multiplying P gives P 1 AP B, so since P 1 is invertible, we have B A (3) (transitivity) If A B and B C, then there are invertible matrices P and Q such that A P BP 1 and B QCQ 1 Plugging the second into the first gives A P BP 1 P (QCQ 1 )P 1 (P Q)C(Q 1 P 1 ) (P Q)C(P Q) 1 so since P Q is invertible, with inverse Q 1 P 1, A C In the previous theorem, if we take W V and γ, we ll have B C, so that which shows that A D, ie B 1 A DB 1, or A BDB 1 T ρ T Since similarity is an equivalence relation, if and γ are any two bases for V, then T T ρ and T ρ T γ T T γ This demonstrates that if we represent a linear transformation T L(V, V ) with respect to two different bases, then the corresponding matrices are similar T T γ Indeed, the invertible matrix P is B 1 C, because T B 1 T ρ B and T ρ CT γ C 1, so that T B 1 T ρ B B 1 CT γ C 1 B (B 1 C)T γ (B 1 C) 1 We will show below that the converse is also true: if two matrices are similar, then they represent the same linear transformation, possibly with respect to different bases! 11

Theorem 025 If V, W, Z are finite-dimensional subspaces of R n, R m and R p, respectively, with ordered bases α, and γ, respectively, and if T L(V, W ) and U L(W, Z), then U T γ α U γ T α (013) Proof: By the previous two theorems we have U T γ α (U T )(v 1 ) γ (U T )(v k ) γ U(T (v 1 )) γ U(T (v k )) γ U γ T (v 1) U γ T (v k) U γ T (v 1 ) T (v k ) U γ T α which completes the proof Corollary 026 If V is an k-dimensional subspace of R n with an ordered basis and if I L(V, V ) is the identity operator, then I I k R k k Proof: Given any T L(V ), we have T I T, so that T I T I T, and similarly T T I Hence, taking T I we will have I 2 I Note also that I 1 I, because I 1 I I I 1 I I 1 I But then I I n, because any A R n n that is invertible and satisfies A 2 A will satisfy A I n : A AI n A(AA 1 ) (AA)A 1 A 2 A 1 AA 1 I k An alternative proof follows directly from the definition, since I(v i ) v i, so that I(v i ) e i, whence I I(v 1 ) I(v k ) e1 e k I k Theorem 027 If V and W are subspaces of R n with ordered bases (b 1,, b k ) and γ (c 1,, c l ), respectively, then the function Φ : L(V, W ) R l k (014) Φ(T ) T γ (015) is an isomorphism, that is Φ GL ( L(V, W ), R l k), and consequently the space of linear transformations is isomorphic to the space of all l k matrices: L(V, W ) R l k (016) Proof: First, Φ is linear: there exist scalars r ij, s ij R, for i 1,, l and j 1,, k, such that for any T L(V, W ) we have T (b 1 ) r 11 c 1 + + r l1 c l T (b k ) r 1k c 1 + + r lk c l Hence, for all s, t R and j 1,, n we have and U(b 1 ) s 11 c 1 + + s l1 c l U(b k ) s 1k c 1 + + s lk c l l l l (st + tu)(b j ) st (b j ) + tu(b j ) s r ij c i + t s ij c i (sr ij + ts ij )c i i1 i1 i1 12

As a consequence of this and the rules of matrix addition and scalar multiplication we have Φ(sT + tu) st + tu γ sr 11 + ts 11 sr 1k + ts 1k sr l1 + ts l1 sr lk + ts lk r 11 r 1k s 11 s 1k s + t r l1 r lk s l1 s lk st γ + tuγ sφ(t ) + tφ(u) Moreover, Φ is bijective, since for all A R l k there is a unique linear transformation T L(V, W ) such that Φ(T ) A, because there exists a unique T L(V, W ) such that T (b j ) A 1j c 1 + + A kj c l for j 1,, k This makes Φ onto, and also 1-1 because ker(φ) {T 0 }, the zero operator, because for O R l k there is only T 0 L(V, W ) satisying Φ(T 0 ) O, because T (b j ) 0c 1 + + 0c l 0 for j 1,, k defines a unique transformation, T 0 Theorem 028 Let V and W be subspaces of R n of the same dimension k, with ordered bases (b 1,, b k ) and γ (c 1,, c k ), respectively, and let T L(V, W ) Then T is an isomporhpism iff T γ is invertible, that is T GL(V, W ), iff T γ GL(k, R) In this case T 1 γ ( T γ ) 1 (017) Proof: If T GL(V, W ) and dim(v ) dim(w ) k, then T 1 L(W, V ), and T T 1 I W and T 1 T I V, so that by Theorem 025 and Corollary 026 we have T γ T 1 γ T T 1 γ I W γ I n I V T 1 T T 1 γt γ so that T γ is invertibe with inverse T 1 γ, and by the uniqueness of the multiplicative inverse in R k k, which follows from the uniqueness of T 1 A L(Rk, R k ), we have T 1 γ ( T γ Conversely, if A T γ is invertible, there is a n n matrix B such that AB BA I n Define U L(W, V ) on the basis elements as follows, U(c j ) v j n i1 B ijb j, and extend U by linearity Then B U γ To show that U T 1, note that ) 1 U T U γt γ BA I n I V and T U γ T γ U γ AB I n I W γ But since Φ GL ( L(V, W ), R k k) is an isomorphism, and therefore 1-1, we must have that U T I V and T U I W By the uniqueness of the inverse, however, U T 1, and T is an isomorphism 13

03 Change of Coordinates 031 Definitions We now define the change of coordinates, or change of basis, operator If V is a k-dimensional subspace of R n and (b 1,, b k ) and γ (c 1,, c k ) are two ordered bases for V, then the coordinate maps φ, φ γ GL(V, R k ), which are isomorphisms by Corollary 022 above, may be used to define a change of coordinates operator φ,γ GL(R k, R k ) changing coordinates into γ coordinates, that is having the property We define the operator as follows: φ,γ (v ) v γ (018) φ,γ : φ γ φ 1 (019) The relationship between these three functions is illustrated in the following commutative diagram: R k φ γ φ 1 R k φ As we will see below, the change of coordinates operator has a matrix representation M,γ φ,γ ρ φ γ ρ b 1 γ b 2 γ b n γ V φ γ (020) 032 Properties of Change-of-Coordinate Maps and Matrices Theorem 029 (Change of Coordinate Matrix) Let V be a k-dimensional subspace of R n and let (b 1,, b k ) and γ (c 1,, c k ) be two ordered bases for V Since φ and φ γ are isomorphisms, the following diagram commutes, R k φ γ φ 1 R k φ V and the change of basis operator φ,γ : φ γ φ 1 L(R k, R k ), changing coordinates into γ coordinates, is an isomorphism It s matrix representation, φ γ M,γ φ,γ ρ R k k (021) where ρ (e 1,, e n ) is the standard ordered basis for R k, is called the change of coordinate matrix, and it satisfies the following conditions: 1 M,γ φ,γ ρ φ γ ρ b 1 γ b 2 γ b k γ 2 v γ M,γ v, v V 3 M,γ is invertible and M 1,γ M γ, φ ρ γ 14

Proof: The first point is shown as follows: M,γ φ,γ ρ φ,γ (e 1 ) φ,γ (e 2 ) φ,γ (e k ) (φ γ φ 1 )(e 1) (φ γ φ 1 )(e 2) (φ γ φ 1 )(e k) (φ γ φ 1 )(b 1 ) (φ γ φ 1 )(b 2 ) (φ γ φ 1 )(b k ) φ γ (b 1 ) φ γ (b 2 ) φ γ (b k ) b 1 γ b 2 γ b n γ φ γ or, alternatively, by Theorem 025 and Theorem 028 we have that M,γ φ,γ ρ φ γ φ 1 The second point follows from Theorem 023, since ρ φ γ ρ φ 1 ρ φ γ ρ γ(φ ρ ) 1 φ γ ρ I 1 n φ γ ρ I n φ γ ρ implies that φ γ (v) (φ γ I)(v) (φ γ (φ 1 φ ))(v) ((φ γ φ 1 ) φ )(v) v γ φ γ (v) ρ ((φ γ φ 1 ) φ )(v) ρ φ γ φ 1 ρφ (v) ρ φ,γ ρ v M,γ v And the last point follows from the fact that φ and φ γ are isomorphism, so that φ,γ is an isomorphism, and hence φ 1,γ L(Rk ) is an isomorphism, and because the diagram above commutes we must have φ 1,γ (φ γ φ 1 ) 1 φ φ 1 γ φ,γ so that by (1) or alternatively by Theorem 028 M 1,γ φ 1,γ ρ φ γ, ρ M γ, M 1,γ (φ,γ ρ ) 1 φ 1,γ ρ φ γ, ρ φ ρ γ M γ, Corollary 030 (Change of Basis) Let V and W be subspaces of R n and let (, γ) and (, γ ) be pairs of ordered bases for V and W, respectively If T L(V, W ), then where M γ,γ T γ M γ,γ T γ M, (022) M γ,γ T γ M 1, (023) and M, are change of coordinate matrices That is, the following diagram commutes R k T A R k φ, φ γ,γ φ R k T A φ φ γ R k V T W φ γ 15

Proof: This follows from the fact that if (b 1,, b k ), (b 1,, b k ), γ (c 1,, c l ), γ (c 1,, c l ), then for each i 1,, k we have so T (b i) γ (φ γ,γ T φ 1, )(b i) γ T γ φ γ,γ T φ 1, γ φ γ,γ ρm T γ φ 1, ρn φ γ,γ ρm T γ (φ, ρ n ) 1 M γ,γ T γ M 1, which completes the proof If V is a subspaces of R n with or- Corollary 031 (Change of Basis for a Linear Operator) dered bases and γ, and T L(V ), then T γ M,γ T M 1,γ (024) where M,γ is the change of coordinates matrix Corollary 032 If we are given any two of the following: (1) A R n invertible (2) an ordered basis for R n (3) an ordered basis γ for R n The third is uniquely determined by the equation A M,γ, where M,γ is the change of coordinates matrix of the previous theorem Proof: If we have A M,γ φ γ b 1 γ b 2 γ b n γ, suppose we know A and γ Then, b i γ is given by A, so b i A i1 c 1 + + A in c n, so is uniquely determined If and γ are given, then by the previous theorem M,γ is given by M,γ b 1 γ b 2 γ b n γ Lastly, if A and are given, then γ is given by the first case applied to A 1 M γ, 16