How do our representations change if we select another basis?

Transcription

1 CHAPTER 6 Linear Mappings and Matrices 99 THEOREM 6.: For any linear operators F; G AðV Þ, 6. Change of Basis mðg FÞ ¼ mðgþmðfþ or ½G FŠ ¼ ½GŠ½FŠ (Here G F denotes the composition of the maps G and F.) Let V be an n-dimensional vector space over a field K. We have shown that once we have selected a basis S of V, every vector v V can be represented by means of an n-tuple in K n, and every linear operator T in AðVÞ can be represented by an n n matrix over K. We ask the following natural question: How do our representations change if we select another basis? In order to answer this question, we first need a definition. DEFINITION: Let S ¼ fu ; u ;...; u n g be a basis of a vector space V; and let S 0 ¼ fv ; v ;...; v n g be another basis. (For reference, we will call S the old basis and S 0 the new basis.) Because S is a basis, each vector in the new basis S 0 can be written uniquely as a linear combination of the vectors in S; say, v ¼ a u þ a u þþa n u n v ¼ a u þ a u þþa n u n ::::::::::::::::::::::::::::::::::::::::::::::::: v n ¼ a n u þ a n u þþa nn u n Let P be the transpose of the above matrix of coefficients; that is, let P ¼½p ij Š, where p ij ¼ a ji. Then P is called the change-of-basis matrix (or transition matrix) from the old basis S to the new basis S 0. The following remarks are in order. Remark : The above change-of-basis matrix P may also be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the new basis vectors v i relative to the old basis S; namely, P ¼ ½v Š S ; ½v Š S ;...; ½v n Š S Remark : Analogously, there is a change-of-basis matrix Q from the new basis S 0 to the old basis S. Similarly, Q may be viewed as the matrix whose columns are, respectively, the coordinate column vectors of the old basis vectors u i relative to the new basis S 0 ; namely, Q ¼ ½u Š S 0; ½u Š S 0;...; ½u n Š S 0 Remark : Because the vectors v ; v ;...; v n in the new basis S 0 are linearly independent, the matrix P is invertible (Problem 6.8). Similarly, Q is invertible. In fact, we have the following proposition (proved in Problem 6.8). PROPOSITION 6.4: Let P and Q be the above change-of-basis matrices. Then Q ¼ P. Now suppose S ¼ fu ; u ;...; u n g is a basis of a vector space V, and suppose P ¼½p ij Š is any nonsingular matrix. Then the n vectors v i ¼ p i u i þ p i u þþp ni u n ; i ¼ ; ;...; n corresponding to the columns of P, are linearly independent [Problem 6.(a)]. Thus, they form another basis S 0 of V. Moreover, P will be the change-of-basis matrix from S to the new basis S 0.

2 00 CHAPTER 6 Linear Mappings and Matrices EXAMPLE 6.5 Consider the following two bases of R : S ¼ fu ; u g ¼ fð; Þ; ð; 5Þg and S 0 ¼ fv ; v g ¼ fð; Þ; ð; Þg (a) Find the change-of-basis matrix P from S to the new basis S 0. Write each of the new basis vectors of S 0 as a linear combination of the original basis vectors u and u of S. We have ¼ x ¼ x Thus, þ y 5 þ y 5 or or x þ y ¼ x þ 5y ¼ x þ y ¼ x þ 5y ¼ v ¼ 8u þ u v ¼ u þ 4u and hence; P ¼ 8 yielding x ¼ 8; y ¼ yielding x ¼ ; y ¼ 4 4 Note that the coordinates of v and v are the columns, not rows, of the change-of-basis matrix P. (b) Find the change-of-basis matrix Q from the new basis S 0 back to the old basis S. Here we write each of the old basis vectors u and u of S 0 as a linear combination of the new basis vectors v and v of S 0. This yields u ¼ 4v v u ¼ v 8v and hence; Q ¼ 4 8 As expected from Proposition 6.4, Q ¼ P. (In fact, we could have obtained Q by simply finding P.) EXAMPLE 6.6 Consider the following two bases of R : : E ¼ fe ; e ; e g ¼ fð; 0; 0Þ; ð0; ; 0Þ; ð0; 0; Þg and S ¼ fu ; u ; u g ¼ fð; 0; Þ; ð; ; Þ; ð; ; Þg (a) Find the change-of-basis matrix P from the basis E to the basis S. Because E is the usual basis, we can immediately write each basis element of S as a linear combination of the basis elements of E. Specifically, u ¼ð; 0; Þ ¼ e þ e 6 u ¼ð; ; Þ ¼e þ e þ e and hence; P ¼ u ¼ð; ; Þ ¼ e þ e þ e Again, the coordinates of u ; u ; u appear as the columns in P. Observe that P is simply the matrix whose columns are the basis vectors of S. This is true only because the original basis was the usual basis E. (b) Find the change-of-basis matrix Q from the basis S to the basis E. The definition of the change-of-basis matrix Q tells us to write each of the (usual) basis vectors in E as a linear combination of the basis elements of S. This yields e ¼ð; 0; 0Þ ¼ u þ u u 6 e ¼ð0; ; 0Þ ¼ u þ u and hence; Q ¼ 4 5 e ¼ð0; 0; Þ ¼ u u þ u 0 We emphasize that to find Q, we need to solve three systems of linear equations one system for each of e ; e ; e.

3 CHAPTER 6 Linear Mappings and Matrices 0 Alternatively, we can find Q ¼ P by forming the matrix M ¼½P; IŠ and row reducing M to row canonical form: thus; M ¼ ¼½I; P Š Q ¼ P 6 ¼ (Here we have used the fact that Q is the inverse of P.) The result in Example 6.6(a) is true in general. We state this result formally, because it occurs often. PROPOSITION 6.5: The change-of-basis matrix from the usual basis E of K n to any basis S of K n is the matrix P whose columns are, respectively, the basis vectors of S. Applications of Change-of-Basis Matrix First we show how a change of basis affects the coordinates of a vector in a vector space V. The following theorem is proved in Problem 6.. THEOREM 6.6: Let P be the change-of-basis matrix from a basis S to a basis S 0 in a vector space V. Then, for any vector v V, we have P 0 ¼ and hence; P ¼ 0 Namely, if we multiply the coordinates of v in the original basis S by P, we get the coordinates of v in the new basis S 0. Remark : Although P is called the change-of-basis matrix from the old basis S to the new basis S 0, we emphasize that P transforms the coordinates of v in the original basis S into the coordinates of v in the new basis S 0. Remark : Because of the above theorem, many texts call Q ¼ P, not P, the transition matrix from the old basis S to the new basis S 0. Some texts also refer to Q as the change-of-coordinates matrix. We now give the proof of the above theorem for the special case that dim V ¼. Suppose P is the change-of-basis matrix from the basis S ¼ fu ; u ; u g to the basis S 0 ¼ fv ; v ; v g; say, v ¼ a u þ a u þ a a v ¼ b u þ b u þ b u and hence; P ¼ v ¼ c u þ c u þ c u a b c 4 a b c 5 a b c Now suppose v V and, say, v ¼ k v þ k v þ k v. Then, substituting for v ; v ; v from above, we obtain v ¼ k ða u þ a u þ a u Þþk ðb u þ b u þ b u Þþk ðc u þ c u þ c u Þ ¼ða k þ b k þ c k Þu þða k þ b k þ c k Þu þða k þ b k þ c k Þu

4 0 CHAPTER 6 Linear Mappings and Matrices Thus, k a k þ b k þ c k 0 ¼ 4 k 5 and ¼ 4 a k þ b k þ c k 5 k a k þ b k þ c k Accordingly, a b c P 0 ¼ 4 a b c 5 k a k þ b k þ c k 4 k 5 ¼ 4 a k þ b k þ c k 5 ¼ a b c k a k þ b k þ c k Finally, multiplying the equation ¼ P, by P, we get P ¼ P P 0 ¼ I 0 ¼ 0 The next theorem (proved in Problem 6.6) shows how a change of basis affects the matrix representation of a linear operator. THEOREM 6.: Let P be the change-of-basis matrix from a basis S to a basis S 0 in a vector space V. Then, for any linear operator T on V, ½TŠ S 0 ¼ P ½TŠ S P That is, if A and B are the matrix representations of T relative, respectively, to S and S 0, then B ¼ P AP EXAMPLE 6. Consider the following two bases of R : E ¼ fe ; e ; e g ¼ fð; 0; 0Þ; ð0; ; 0Þ; ð0; 0; Þg and S ¼ fu ; u ; u g ¼ fð; 0; Þ; ð; ; Þ; ð; ; Þg The change-of-basis matrix P from E to S and its inverse P were obtained in Example 6.6. (a) Write v ¼ð; ; 5Þ as a linear combination of u ; u ; u, or, equivalently, find. One way to do this is to directly solve the vector equation v ¼ xu þ yu þ zu ; that is, x þ y þ z ¼ 4 5 ¼ x4 0 5 þ y4 5 þ z4 5 or y þ z ¼ 5 x þ y þ z ¼ 5 The solution is x ¼, y ¼ 5, z ¼ 4, so v ¼ u 5u þ 4u. On the other hand, we know that ½vŠ E ¼½; ; 5Š T, because E is the usual basis, and we already know P. Therefore, by Theorem 6.6, ¼ P ½vŠ E ¼ ¼ Thus, again, v ¼ u 5u þ 4u. (b) Let A ¼ 4 4 5, which may be viewed as a linear operator on R. Find the matrix B that represents A relative to the basis S

5 CHAPTER 6 Linear Mappings and Matrices 0 The definition of the matrix representation of A relative to the basis S tells us to write each of Aðu Þ, Aðu Þ, Aðu Þ as a linear combination of the basis vectors u ; u ; u of S. This yields Aðu Þ ¼ ð ; ; 5Þ ¼u 5u þ 6u 6 Aðu Þ¼ð; ; 9Þ ¼ u 4u þ 8u and hence; B ¼ Aðu Þ¼ð; 4; 5Þ ¼u 8e þ u 6 8 We emphasize that to find B, we need to solve three systems of linear equations one system for each of Aðu Þ, Aðu Þ, Aðu Þ. On the other hand, because we know P and P, we can use Theorem 6.. That is, B ¼ P AP ¼ ¼ This, as expected, gives the same result. 6.4 Similarity Suppose A and B are square matrices for which there exists an invertible matrix P such that B ¼ P AP; then B is said to be similar to A, or B is said to be obtained from A by a similarity transformation. We show (Problem 6.9) that similarity of matrices is an equivalence relation. By Theorem 6. and the above remark, we have the following basic result. THEOREM 6.8: Two matrices represent the same linear operator if and only if the matrices are similar. That is, all the matrix representations of a linear operator T form an equivalence class of similar matrices. A linear operator T is said to be diagonalizable if there exists a basis S of V such that T is represented by a diagonal matrix; the basis S is then said to diagonalize T. The preceding theorem gives us the following result. THEOREM 6.9: Let A be the matrix representation of a linear operator T. Then T is diagonalizable if and only if there exists an invertible matrix P such that P AP is a diagonal matrix. That is, T is diagonalizable if and only if its matrix representation can be diagonalized by a similarity transformation. We emphasize that not every operator is diagonalizable. However, we will show (Chapter 0) that every linear operator can be represented by certain standard matrices called its normal or canonical forms. Such a discussion will require some theory of fields, polynomials, and determinants. Functions and Similar Matrices Suppose f is a function on square matrices that assigns the same value to similar matrices; that is, f ðaþ ¼f ðbþ whenever A is similar to B. Then f induces a function, also denoted by f, on linear operators T in the following natural way. We define f ðtþ ¼f ð½tš S Þ where S is any basis. By Theorem 6.8, the function is well defined. The determinant (Chapter 8) is perhaps the most important example of such a function. The trace (Section.) is another important example of such a function.

6 04 CHAPTER 6 Linear Mappings and Matrices EXAMPLE 6.8 Consider the following linear operator F and bases E and S of R : Fðx; yþ ¼ðx þ y; 4x 5yÞ; E ¼ fð; 0Þ; ð0; Þg; S ¼ fð; Þ; ð; 5Þg By Example 6., the matrix representations of F relative to the bases E and S are, respectively, A ¼ 5 9 and B ¼ Using matrix A, we have (i) Determinant of F ¼ detðaþ ¼ 0 ¼ ; (ii) Trace of F ¼ trðaþ ¼ 5 ¼ : On the other hand, using matrix B, we have (i) Determinant of F ¼ detðbþ ¼ 860 þ 88 ¼ ; (ii) Trace of F ¼ trðbþ ¼5 55 ¼. As expected, both matrices yield the same result. 6.5 Matrices and General Linear Mappings Last, we consider the general case of linear mappings from one vector space into another. Suppose V and U are vector spaces over the same field K and, say, dim V ¼ m and dim U ¼ n. Furthermore, suppose S ¼ fv ; v ;...; v m g and S 0 ¼ fu ; u ;...; u n g are arbitrary but fixed bases, respectively, of V and U. Suppose F: V! U is a linear mapping. Then the vectors Fðv Þ, Fðv Þ;...; Fðv m Þ belong to U, and so each is a linear combination of the basis vectors in S 0 ; say, Fðv Þ¼ a u þ a u þþa n u n Fðv Þ¼ a u þ a u þþa n u n ::::::::::::::::::::::::::::::::::::::::::::::::::::::: Fðv m Þ¼a m u þ a m u þþa mn u n DEFINITION: The transpose of the above matrix of coefficients, denoted by m S;S 0ðFÞ or ½FŠ S;S 0, is called the matrix representation of F relative to the bases S and S 0. [We will use the simple notation mðfþ and ½FŠ when the bases are understood.] The following theorem is analogous to Theorem 6. for linear operators (Problem 6.6). THEOREM 6.0: For any vector v V, ½FŠ S;S 0 ¼½FðvÞŠ S 0. That is, multiplying the coordinates of v in the basis S of V by ½FŠ, we obtain the coordinates of FðvÞ in the basis S 0 of U. Recall that for any vector spaces V and U, the collection of all linear mappings from V into U is a vector space and is denoted by HomðV; UÞ. The following theorem is analogous to Theorem 6. for linear operators, where now we let M ¼ M m;n denote the vector space of all m n matrices (Problem 6.6). THEOREM 6.: The mapping m: HomðV; U Þ! M defined by mðfþ ¼ ½FŠ is a vector space isomorphism. That is, for any F; G HomðV; UÞ and any scalar k, (i) (ii) (iii) mðf þ GÞ ¼mðFÞ þmðgþ or ½F þ GŠ ¼½FŠ þ½gš mðkfþ ¼kmðFÞ or ½kFŠ ¼k½FŠ m is bijective (one-to-one and onto).

7 CHAPTER 6 Linear Mappings and Matrices 05 Our next theorem is analogous to Theorem 6. for linear operators (Problem 6.6). THEOREM 6.: Let S; S 0 ; S 00 be bases of vector spaces V; U; W, respectively. Let F: V! U and G U! W be linear mappings. Then ½G FŠ S;S 00 ¼½GŠ S 0 ;S 00½FŠ S;S 0 That is, relative to the appropriate bases, the matrix representation of the composition of two mappings is the matrix product of the matrix representations of the individual mappings. Next we show how the matrix representation of a linear mapping F: V! U is affected when new bases are selected (Problem 6.6). THEOREM 6.: Let P be the change-of-basis matrix from a basis e to a basis e 0 in V, and let Q be the change-of-basis matrix from a basis f to a basis f 0 in U. Then, for any linear map F: V! U, ½FŠ e 0 ; f 0 ¼ Q ½FŠ e; f P In other words, if A is the matrix representation of a linear mapping F relative to the bases e and f, and B is the matrix representation of F relative to the bases e 0 and f 0, then B ¼ Q AP Our last theorem, proved in Problem 6.6, shows that any linear mapping from one vector space V into another vector space U can be represented by a very simple matrix. We note that this theorem is analogous to Theorem.8 for m n matrices. THEOREM 6.4: Let F: V! U be linear and, say, rankðfþ ¼r. Then there exist bases of V and U such that the matrix representation of F has the form A ¼ I r where I r is the r-square identity matrix. The above matrix A is called the normal or canonical form of the linear map F. SOLVED PROBLEMS Matrix Representation of Linear Operators 6.. Consider the linear mapping F: R! R defined by Fðx; yþ ¼ðx þ 4y; x 5yÞ and the following bases of R : E ¼ fe ; e g ¼ fð; 0Þ; ð0; Þg and S ¼ fu ; u g ¼ fð; Þ; ð; Þg (a) Find the matrix A representing F relative to the basis E. (b) Find the matrix B representing F relative to the basis S. (a) Because E is the usual basis, the rows of A are simply the coefficients in the components of Fðx; yþ; that is, using ða; bþ ¼ae þ be, we have Fðe Þ¼Fð; 0Þ ¼ð; Þ ¼ e þ e and so A ¼ 4 Fðe Þ¼Fð0; Þ ¼ð4; 5Þ ¼4e 5e 5 Note that the coefficients of the basis vectors are written as columns in the matrix representation.

8 06 CHAPTER 6 Linear Mappings and Matrices (b) First find Fðu Þ and write it as a linear combination of the basis vectors u and u. We have Fðu Þ¼Fð; Þ ¼ð; 8Þ ¼xð; Þþyð; Þ; Solve the system to obtain x ¼ 49, y ¼ 0. Therefore, Fðu Þ¼ 49u þ 0u and so x þ y ¼ x þ y ¼ 8 Next find Fðu Þ and write it as a linear combination of the basis vectors u and u. We have Fðu Þ¼Fð; Þ ¼ð8; Þ ¼xð; Þþyð; Þ; Solve for x and y to obtain x ¼ 6, y ¼ 4. Hence, and so Fðu Þ¼ 6u þ 4u 49 6 Write the coefficients of u and u as columns to obtain B ¼ 0 4 x þ y ¼ 8 x þ y ¼ (b 0 ) Alternatively, one can first find the coordinates of an arbitrary vector ða; bþ in R relative to the basis S. We have ða; bþ ¼xð; Þþyð; Þ ¼ðx þ y; x þ yþ; and so Solve for x and y in terms of a and b to get x ¼ a þ b, y ¼ a b. Thus, ða; bþ ¼ ð a þ bþu þða bþu x þ y ¼ a x þ y ¼ b Then use the formula for ða; bþ to find the coordinates of Fðu Þ and Fðu Þ relative to S: Fðu Þ¼Fð; Þ ¼ð; 8Þ ¼ 49u þ 0u 49 6 and so B ¼ Fðu Þ¼Fð; Þ ¼ð8; Þ ¼ 6u þ 4u Consider the following linear operator G on R and basis S: Gðx; yþ ¼ðx y; 4x þ yþ and S ¼ fu ; u g ¼ fð; Þ; ð; 5Þg (a) Find the matrix representation ½GŠ S of G relative to S. (b) Verify ½GŠ S ¼½GðvÞŠ S for the vector v ¼ð4; Þ in R. First find the coordinates of an arbitrary vector v ¼ða; bþ in R relative to the basis S. We have a ¼ x þ y x þ y ¼ a ; and so b 5 x þ 5y ¼ b Solve for x and y in terms of a and b to get x ¼ 5a þ b, y ¼ a b. Thus, ða; bþ ¼ ð 5a þ bþu þða bþu ; and so ½vŠ ¼ ½ 5a þ b; a bš T (a) Using the formula for ða; bþ and Gðx; yþ ¼ðx y; 4x þ yþ, we have Gðu Þ¼Gð; Þ ¼ ð 9; Þ ¼u 0u 0 and so ½GŠ Gðu Þ¼Gð; 5Þ ¼ ð ; Þ ¼0u 6u S ¼ 0 6 (We emphasize that the coefficients of u and u are written as columns, not rows, in the matrix representation.) (b) Use the formula ða; bþ ¼ ð 5a þ bþu þða bþu to get v ¼ð4; Þ ¼ 6u þ 5u GðvÞ ¼Gð4; Þ ¼ð0; Þ ¼ u þ 80u Then ¼ ½ 6; 5Š T and ½GðvÞŠ S ¼ ½ ; 80Š T

9 CHAPTER 6 Linear Mappings and Matrices 0 Accordingly, ½GŠ S ¼ (This is expected from Theorem 6..) ¼ Consider the following matrix A and basis S of R : A ¼ 4 and S ¼ fu 5 6 ; u g ¼ ¼½GðvÞŠ 80 S ; The matrix A defines a linear operator on R. Find the matrix B that represents the mapping A relative to the basis S. First find the coordinates of an arbitrary vector ða; bþ T with respect to the basis S. We have a b ¼ x þ y or x þ y ¼ a x y ¼ b Solve for x and y in terms of a and b to obtain x ¼ a þ b, y ¼ a b. Thus, ða; bþ T ¼ða þ bþu þ ð a bþu Then use the formula for ða; bþ T to find the coordinates of Au and Au relative to the basis S: Au ¼ 4 ¼ 6 ¼ 6u þ 9u 5 6 Au ¼ 4 ¼ ¼ 5u þ u 5 6 Writing the coordinates as columns yields B ¼ Find the matrix representation of each of the following linear operators F on R relative to the usual basis E ¼ fe ; e ; e g of R ; that is, find ½FŠ ¼½FŠ E : (a) F defined by Fðx; y; zþ ¼ðxþy z; 4x 5y 6z; x þ 8y þ 9z). (b) F defined by the matrix A ¼ (c) F defined by Fðe Þ¼ð; ; 5Þ; Fðe Þ¼ð; 4; 6Þ, Fðe Þ¼ð; ; Þ. (Theorem 5. states that a linear map is completely defined by its action on the vectors in a basis.) (a) Because E is the usual basis, simply write the coefficients of the components of Fðx; y; zþ as rows: ½FŠ ¼ (b) Because E is the usual basis, ½FŠ ¼A, the matrix A itself. (c) Here Fðe Þ¼ð; ; 5Þ ¼ e þ e þ 5e Fðe Þ¼ð; 4; 6Þ ¼e þ 4e þ 6e Fðe Þ¼ð; ; Þ ¼e þ e þ e and so ½FŠ ¼ That is, the columns of ½FŠ are the images of the usual basis vectors Let G be the linear operator on R defined by Gðx; y; zþ ¼ðy þ z; x 4y; xþ. (a) Find the matrix representation of G relative to the basis S ¼ fw ; w ; w g ¼ fð; ; Þ; ð; ; 0Þ; ð; 0; 0Þg (b) Verify that ½GŠ½vŠ ¼½GðvÞŠ for any vector v in R.

10 08 CHAPTER 6 Linear Mappings and Matrices First find the coordinates of an arbitrary vector ða; b; cþ R with respect to the basis S. Write ða; b; cþ as a linear combination of w ; w ; w using unknown scalars x; y, and z: ða; b; cþ ¼xð; ; Þþyð; ; 0Þþzð; 0; 0Þ ¼ðx þ y þ z; x þ y; xþ Set corresponding components equal to each other to obtain the system of equations x þ y þ z ¼ a; x þ y ¼ b; x ¼ c Solve the system for x; y, z in terms of a; b, c to find x ¼ c, y ¼ b c, z ¼ a b. Thus, ða; b; cþ ¼cw þðb cþw þða bþw, or equivalently, ½ða; b; cþš ¼ ½c; b c; a bš T (a) Because Gðx; y; zþ ¼ðy þ z; x 4y; xþ, Gðw Þ¼Gð; ; Þ ¼ð; ; Þ ¼w 6x þ 6x Gðw Þ¼Gð; ; 0Þ ¼ð; ; Þ ¼w 6w þ 5w Gðw Þ¼Gð; 0; 0Þ ¼ð0; ; Þ ¼w w w Write the coordinates Gðw Þ, Gðw Þ, Gðw Þ as columns to get ½GŠ ¼ (b) Write GðvÞ as a linear combination of w ; w ; w, where v ¼ða; b; cþ is an arbitrary vector in R, GðvÞ ¼Gða; b; cþ ¼ðb þ c; a 4b; aþ ¼aw þ ð a 4bÞw þ ð a þ 6b þ cþw or equivalently, Accordingly, ½GðvÞŠ ¼ ½a; a 4b; a þ 6b þ cš T ½GŠ½vŠ ¼ c b c a b 5 ¼ 4 a a 4b a þ 6b þ c 5 ¼½GðvÞŠ 6.6. Consider the following matrix A and basis S of R : 8 < A ¼ 4 05 and S ¼ fu ; u ; u g ¼ 4 5; : ; 9 = 4 5 ; The matrix A defines a linear operator on R. Find the matrix B that represents the mapping A relative to the basis S. (Recall that A represents itself relative to the usual basis of R.) First find the coordinates of an arbitrary vector ða; b; cþ in R with respect to the basis S. We have a 0 4 b 5 ¼ x4 5 þ y4 5 þ z4 5 or c x þ z ¼ a x þ y þ z ¼ b x þ y þ z ¼ c Solve for x; y; z in terms of a; b; c to get x ¼ a þ b c; y ¼ a þ b c; z ¼ c b thus; ða; b; cþ T ¼ða þ b cþu þ ð a þ b cþu þðc bþu

11 CHAPTER 6 Linear Mappings and Matrices 09 Then use the formula for ða; b; cþ T to find the coordinates of Au, Au, Au relative to the basis S: Aðu Þ¼Að; ; Þ T ¼ð0; ; Þ T ¼ u þ u þ u Aðu Þ¼Að; ; 0Þ T ¼ ð ; ; Þ T ¼ 4u u þ u so B ¼ Aðu Þ¼Að; ; Þ T ¼ð0; ; Þ T ¼ u u þ u For each of the following linear transformations (operators) L on R, find the matrix A that represents L (relative to the usual basis of R ): (a) L is defined by Lð; 0Þ ¼ð; 4Þ and Lð0; Þ ¼ð5; 8Þ. (b) L is the rotation in R counterclockwise by 90. (c) L is the reflection in R about the line y ¼ x. (a) Because fð; 0Þ; ð0; Þg is the usual basis of R, write their images under L as columns to get A ¼ (b) Under the rotation L, we have Lð; 0Þ ¼ð0; Þ and Lð0; Þ ¼ ð ; 0Þ. Thus, A ¼ 0 0 (c) Under the reflection L, we have Lð; 0Þ ¼ð0; Þ and Lð0; Þ ¼ ð ; 0Þ. Thus, A ¼ The set S ¼ fe t, te t, t e t g is a basis of a vector space V of functions f : R! R. Let D be the differential operator on V; that is, Dð f Þ¼df =dt. Find the matrix representation of D relative to the basis S. Find the image of each basis function: Dðe t Þ ¼ e t ¼ ðe t Þþ0ðte t Þþ0ðt e t Þ Dðte t Þ ¼ e t þ te t ¼ ðe t Þþðte t Þþ0ðt e t Þ Dðt e t Þ¼te t þ t e t ¼ 0ðe t Þþðte t Þþðt e t Þ and thus; 0 ½DŠ ¼ Prove Theorem 6.: Let T: V! V be a linear operator, and let S be a (finite) basis of V. Then, for any vector v in V, ½TŠ S ¼½TðvÞŠ S. Suppose S ¼ fu ; u ;...; u n g, and suppose, for i ¼ ;...; n, Then ½TŠ S is the n-square matrix whose jth row is Tðu i Þ¼a i u þ a i u þþa in u n ¼ Pn j¼ a ij u j ða j ; a j ;...; a nj Þ ðþ Now suppose v ¼ k u þ k u þþk n u n ¼ Pn i¼ k i u i Writing a column vector as the transpose of a row vector, we have ¼½k ; k ;...; k n Š T ðþ

12 0 CHAPTER 6 Linear Mappings and Matrices Furthermore, using the linearity of T, TðvÞ ¼T Pn i¼ ¼ Pn P n j¼ i¼ k i u i a ij k i ¼ Pn i¼ u j ¼ Pn Thus, ½TðvÞŠ S is the column vector whose jth entry is k i Tðu i Þ¼ Pn j¼ i¼ k i P n j¼ a ij u j ða j k þ a j k þþa nj k n Þu j a j k þ a j k þþa nj k n ðþ On the other hand, the jth entry of ½TŠ S is obtained by multiplying the jth row of ½TŠ S by that is () by (). But the product of () and () is (). Hence, ½TŠ S and ½TðvÞŠ S have the same entries. Thus, ½TŠ S ¼½TðvÞŠ S Prove Theorem 6.: Let S ¼ fu ; u ;...; u n g be a basis for V over K, and let M be the algebra of n-square matrices over K. Then the mapping m: AðVÞ! M defined by mðtþ ¼½TŠ S is a vector space isomorphism. That is, for any F; G AðVÞ and any k K, we have (i) ½F þ GŠ ¼ ½FŠ þ ½GŠ, (ii) ½kFŠ ¼ k½fš, (iii) m is one-to-one and onto. (i) Suppose, for i ¼ ;...; n, Fðu i Þ¼ Pn j¼ a ij u j and Gðu i Þ¼ Pn Consider the matrices A ¼½a ij Š and B ¼½b ij Š. Then ½FŠ ¼A T and ½GŠ ¼B T. We have, for i ¼ ;...; n, Because A þ B is the matrix ða ij þ b ij Þ, we have (ii) Also, for i ¼ ;...; n; Because ka is the matrix ðka ij Þ, we have (iii) j¼ b ij u j ðf þ GÞðu i Þ¼Fðu i ÞþGðu i Þ¼ Pn ða ij þ b ij Þu j j¼ ½F þ GŠ ¼ðA þ BÞ T ¼ A T þ B T ¼½FŠþ½GŠ ðkfþðu i Þ¼kFðu i Þ¼k Pn j¼ a ij u j ¼ Pn ðka ij Þu j j¼ ½kFŠ ¼ðkAÞ T ¼ ka T ¼ k½fš Finally, m is one-to-one, because a linear mapping is completely determined by its values on a basis. Also, m is onto, because matrix A ¼½a ij Š in M is the image of the linear operator, Thus, the theorem is proved. Fðu i Þ¼ Pn j¼ a ij u j ; i ¼ ;...; n 6.. Prove Theorem 6.: For any linear operators G; F AðVÞ, ½G FŠ ¼½GŠ½FŠ. Using the notation in Problem 6.0, we have P n ðg FÞðu i Þ¼GðFðu i ÞÞ ¼ G a ij u j ¼ Pn a ij Gðu j Þ j¼ j¼ ¼ Pn P a n ij b jk u k ¼ Pn P n a ij b jk u k j¼ k¼ k¼ j¼ Recall that AB is the matrix AB ¼½c ik Š, where c ik ¼ P n j¼ a ijb jk. Accordingly, The theorem is proved. ½G FŠ ¼ðABÞ T ¼ B T A T ¼½GŠ½FŠ

13 CHAPTER 6 Linear Mappings and Matrices 6.. Let A be the matrix representation of a linear operator T. Prove that, for any polynomial f ðtþ, we have that f ðaþ is the matrix representation of f ðtþ. [Thus, f ðtþ ¼0 if and only if f ðaþ ¼0.] Let f be the mapping that sends an operator T into its matrix representation A. We need to prove that fð f ðtþþ ¼ f ðaþ. Suppose f ðtþ ¼a n t n þþa t þ a 0. The proof is by induction on n, the degree of f ðtþ. Suppose n ¼ 0. Recall that fði 0 Þ¼I, where I 0 is the identity mapping and I is the identity matrix. Thus, fð f ðtþþ ¼ fða 0 I 0 Þ¼a 0 fði 0 Þ¼a 0 I ¼ f ðaþ and so the theorem holds for n ¼ 0. Now assume the theorem holds for polynomials of degree less than n. Then, because f is an algebra isomorphism, and the theorem is proved. Change of Basis fð f ðtþþ ¼ fða n T n þ a n T n þþa T þ a 0 I 0 Þ ¼ a n fðtþfðt n Þþfða n T n þþa T þ a 0 I 0 Þ ¼ a n AA n þða n A n þþa A þ a 0 IÞ¼f ðaþ The coordinate vector in this section will always denote a column vector; that is, 6.. Consider the following bases of R : ¼½a ; a ;...; a n Š T E ¼ fe ; e g ¼ fð; 0Þ; ð0; Þg and S ¼ fu ; u g ¼ fð; Þ; ð; 4Þg (a) Find the change-of-basis matrix P from the usual basis E to S. (b) Find the change-of-basis matrix Q from S back to E. (c) Find the coordinate vector ½vŠ of v ¼ð5; Þ relative to S. (a) Because E is the usual basis, simply write the basis vectors in S as columns: P ¼ 4 (b) Method. Use the definition of the change-of-basis matrix. That is, express each vector in E as a linear combination of the vectors in S. We do this by first finding the coordinates of an arbitrary vector v ¼ða; bþ relative to S. We have x þ y ¼ a ða; bþ ¼xð; Þþyð; 4Þ ¼ðxþy; x þ 4yÞ or x þ 4y ¼ b Solve for x and y to obtain x ¼ 4a b, y ¼ a þ b. Thus, v ¼ð4a bþu þ ð a þ bþu and ¼ ½ða; bþš S ¼½4a b; a þ bš T Using the above formula for and writing the coordinates of the e i as columns yields e ¼ð; 0Þ ¼ 4u u 4 and Q ¼ e ¼ð0; Þ ¼ u þ u Method. Thus, Because Q ¼ P ; find P, say by using the formula for the inverse of a matrix. P ¼ 4 (c) Method. Write v as a linear combination of the vectors in S, say by using the above formula for v ¼ða; bþ. We have v ¼ð5; Þ ¼u 8u ; and so ¼½; 8Š T. Method. Use, from Theorem 6.6, the fact that ¼ P ½vŠ E and the fact that ½vŠ E ¼½5; Š T : ¼ P 4 5 ½vŠ E ¼ ¼ 8

14 CHAPTER 6 Linear Mappings and Matrices 6.4. The vectors u ¼ð; ; 0Þ, u ¼ð; ; Þ, u ¼ð0; ; Þ form a basis S of R. Find (a) The change-of-basis matrix P from the usual basis E ¼ fe ; e ; e g to S. (b) The change-of-basis matrix Q from S back to E. 0 (a) Because E is the usual basis, simply write the basis vectors of S as columns: P ¼ (b) Method. Express each basis vector of E as a linear combination of the basis vectors of S by first finding the coordinates of an arbitrary vector v ¼ða; b; cþ relative to the basis S. We have a 0 x þ y ¼ a 4 b 5 ¼ x4 5 þ y4 5 þ z4 5 or x þ y þ z ¼ b c 0 y þ z ¼ c Solve for x; y; z to get x ¼ a b þ c, y ¼ 6a þ b c, z ¼ 4a b þ c. Thus, or v ¼ða; b; cþ ¼ða bþcþu þ ð 6a þ b cþu þð4a bþcþu ¼ ½ða; b; cþš S ¼½a bþc; 6a þ b c; 4a b þ cš T Using the above formula for and then writing the coordinates of the e i as columns yields e ¼ð; 0; 0Þ ¼ u 6u þ 4u e ¼ð0; ; 0Þ ¼ u þ u u and Q ¼ e ¼ð0; 0; Þ ¼ u u þ u 4 Method. Find P by row reducing M ¼½P; IŠ to the form ½I; P Š: M ¼ ¼½I; P Š Thus, Q ¼ P ¼ Suppose the x-axis and y-axis in the plane R are rotated counterclockwise 45 so that the new x 0 -axis and y 0 -axis are along the line y ¼ x and the line y ¼ x, respectively. (a) Find the change-of-basis matrix P. (b) Find the coordinates of the point Að5; 6Þ under the given rotation. (a) The unit vectors in the direction of the new x 0 - and y 0 -axes are u ¼ð p ffiffiffi ; pffiffiffi Þ and u ¼ ð p ffiffi ; pffiffi Þ (The unit vectors in the direction of the original x and y axes are the usual basis of R.) Thus, write the coordinates of u and u as columns to obtain " pffiffi pffiffiffi # P ¼ pffiffi pffiffiffi (b) Multiply the coordinates of the point by P : " pffiffiffi pffiffiffi # " pffiffiffi # 5 ffiffiffi pffiffiffi ¼ pffiffiffi 6 p (Because P is orthogonal, P is simply the transpose of P.)

15 CHAPTER 6 Linear Mappings and Matrices 6.6. The vectors u ¼ð; ; 0Þ, u ¼ð0; ; Þ, u ¼ð; ; Þ form a basis S of R. Find the coordinates of an arbitrary vector v ¼ða; b; cþ relative to the basis S. Method. this yields the system Express v as a linear combination of u ; u ; u using unknowns x; y; z. We have ða; b; cþ ¼xð; ; 0Þþyð0; ; Þþzð; ; Þ ¼ðx þ z; x þ y þ z; y þ zþ x þ z ¼ a x þ y þ z ¼ b y þ z ¼ c or x þ z ¼ a y þ z ¼ aþb y þ z ¼ c or x þ z ¼ a y þ z ¼ aþb z ¼ a b þ c Solving by back-substitution yields x ¼ b c, y ¼ a þ b c, z ¼ a b þ c. Thus, ¼½b c; a þ b c; a b þ cš T Method. Find P by row reducing M ¼½P; IŠ to the form ½I; P Š, where P is the change-of-basis matrix from the usual basis E to S or, in other words, the matrix whose columns are the basis vectors of S. We have Thus; M ¼ ¼½I; P Š a b c P 6 ¼ 4 5 and ¼ P ½vŠ E ¼ 4 54 b 5 ¼ 4 a þ b c 5 c a b þ c 6.. Consider the following bases of R : S ¼ fu ; u g ¼ fð; Þ; ð; 4Þg and S 0 ¼ fv ; v g ¼ fð; Þ; ð; 8Þg (a) Find the coordinates of v ¼ða; bþ relative to the basis S. (b) Find the change-of-basis matrix P from S to S 0. (c) Find the coordinates of v ¼ða; bþ relative to the basis S 0. (d) Find the change-of-basis matrix Q from S 0 back to S. (e) Verify Q ¼ P. (f ) Show that, for any vector v ¼ða; bþ in R, P ¼ 0. (See Theorem 6.6.) (a) Let v ¼ xu þ yu for unknowns x and y; that is, a x þ y ¼ a ¼ x þ y or b 4 x 4y ¼ b Solve for x and y in terms of a and b to get x ¼ a b and y ¼ a þ b. Thus, or x þ y ¼ a y ¼ a þ b ða; bþ ¼ ð a Þu þða þ bþu or ½ða; bþš S ¼ ½ a b; a þ bšt (b) Use part (a) to write each of the basis vectors v and v of S 0 as a linear combination of the basis vectors u and u of S; that is, v ¼ð; Þ ¼ ð 9 Þu þð þ Þu ¼ u þ 5 u v ¼ð; 8Þ ¼ ð 6 Þu þð þ 4Þu ¼ 8u þ u

16 4 CHAPTER 6 Linear Mappings and Matrices Then P is the matrix whose columns are the coordinates of v and v relative to the basis S; that is, " P ¼ # 8 5 (c) Let v ¼ xv þ yv for unknown scalars x and y: a ¼ x þ y x þ y ¼ a or b 8 x þ 8y ¼ b Solve for x and y to get x ¼ 8a þ b and y ¼ a b. Thus, or x þ y ¼ a y ¼ b a ða; bþ ¼ ð 8a þ bþv þða bþv or ½ða; bþš S 0 ¼ ½ 8a þ b; a bš T (d) Use part (c) to express each of the basis vectors u and u of S as a linear combination of the basis vectors v and v of S 0 : u ¼ð; Þ ¼ ð 8 6Þv þð þ Þv ¼ 4v þ 5v u ¼ð; 4Þ ¼ ð 4 Þv þð9þ4þv ¼ 6v þ v Write the coordinates of u and u relative to S 0 as columns to obtain Q ¼ " # (e) QP ¼ ¼ 0 ¼ I (f ) Use parts (a), (c), and (d) to obtain " P 4 6 a ¼ Q ¼ b # 8a þ b 5 a þ b ¼ ¼½vŠ a b S Suppose P is the change-of-basis matrix from a basis fu i g to a basis fw i g, and suppose Q is the change-of-basis matrix from the basis fw i g back to fu i g. Prove that P is invertible and that Q ¼ P. Suppose, for i ¼ ; ;...; n, that w i ¼ a i u þ a i u þ...þ a in u n ¼ Pn a ij u j ðþ and, for j ¼ ; ;...; n, u j ¼ b j w þ b j w þþb jn w n ¼ Pn j¼ k¼ b jk w k Let A ¼½a ij Š and B ¼½b jk Š. Then P ¼ A T and Q ¼ B T. Substituting () into () yields w i ¼ Pn P n a ij b jk w k ¼ Pn P n a ij b jk w k j¼ j¼ k¼ Because fw i g is a basis, P a ij b jk ¼ d ik, where d ik is the Kronecker delta; that is, d ik ¼ if i ¼ k but d ik ¼ 0 if i 6¼ k. Suppose AB ¼½c ik Š. Then c ik ¼ d ik. Accordingly, AB ¼ I, and so Thus, Q ¼ P. k¼ QP ¼ B T A T ¼ðABÞ T ¼ I T ¼ I 6.9. Consider a finite sequence of vectors S ¼ fu ; u ;...; u n g. Let S 0 be the sequence of vectors obtained from S by one of the following elementary operations : () Interchange two vectors. () Multiply a vector by a nonzero scalar. () Add a multiple of one vector to another vector. Show that S and S 0 span the same subspace W. Also, show that S 0 is linearly independent if and only if S is linearly independent.. ðþ

17 CHAPTER 6 Linear Mappings and Matrices 5 Observe that, for each operation, the vectors S 0 are linear combinations of vectors in S. Also, because each operation has an inverse of the same type, each vector in S is a linear combination of vectors in S 0. Thus, S and S 0 span the same subspace W. Moreover, S 0 is linearly independent if and only if dim W ¼ n, and this is true if and only if S is linearly independent Let A ¼½a ij Š and B ¼½b ij Š be row equivalent m n matrices over a field K, and let v ; v ;...; v n be any vectors in a vector space V over K. For i ¼ ; ;...; m, let u i and w i be defined by u i ¼ a i v þ a i v þþa in v n and w i ¼ b i v þ b i v þþb in v n Show that fu i g and fw i g span the same subspace of V. Applying an elementary operation of Problem 6.9 to fu i g is equivalent to applying an elementary row operation to the matrix A. Because A and B are row equivalent, B can be obtained from A by a sequence of elementary row operations. Hence, fw i g can be obtained from fu i g by the corresponding sequence of operations. Accordingly, fu i g and fw i g span the same space. 6.. Suppose u ; u ;...; u n belong to a vector space V over a field K, and suppose P ¼½a ij Š is an n-square matrix over K. For i ¼ ; ;...; n, let v i ¼ a i u þ a i u þþa in u n. (a) Suppose P is invertible. Show that fu i g and fv i g span the same subspace of V. Hence, fu i g is linearly independent if and only if fv i g is linearly independent. (b) Suppose P is singular (not invertible). Show that fv i g is linearly dependent. (c) Suppose fv i g is linearly independent. Show that P is invertible. (a) Because P is invertible, it is row equivalent to the identity matrix I. Hence, by Problem 6.9, fv i g and fu i g span the same subspace of V. Thus, one is linearly independent if and only if the other is linearly independent. (b) Because P is not invertible, it is row equivalent to a matrix with a zero row. This means fv i g spans a substance that has a spanning set with less than n elements. Thus, fv i g is linearly dependent. (c) This is the contrapositive of the statement of part (b), and so it follows from part (b). 6.. Prove Theorem 6.6: Let P be the change-of-basis matrix from a basis S to a basis S 0 in a vector space V. Then, for any vector v V, we have P 0 ¼, and hence, P ¼ 0. Suppose S ¼ fu ;...; u n g and S 0 ¼ fw ;...; w n g, and suppose, for i ¼ ;...; n, w i ¼ a i u þ a i u þþa in u n ¼ Pn Then P is the n-square matrix whose jth row is ða j ; a j ;...; a nj Þ Also suppose v ¼ k w þ k w þþk n w n ¼ P n i¼ k iw i. Then Substituting for w i in the equation for v, we obtain v ¼ Pn i¼ ¼ Pn j¼ k i w i ¼ Pn j¼ a ij u j ðþ 0 ¼½k ; k ;...; k n Š T ðþ i¼ k i P n j¼ a ij u j ¼ Pn j¼ ða j k þ a j k þþa nj k n Þu j Accordingly, is the column vector whose jth entry is a j k þ a j k þþa nj k n P n a ij k i u j i¼ On the other hand, the jth entry of P 0 is obtained by multiplying the jth row of P by 0 that is, () by (). However, the product of () and () is (). Hence, P 0 and have the same entries. Thus, P 0 ¼ 0, as claimed. Furthermore, multiplying the above by P gives P ¼ P P 0 ¼ 0. ðþ

18 6 CHAPTER 6 Linear Mappings and Matrices Linear Operators and Change of Basis 6.. Consider the linear transformation F on R defined by Fðx; yþ ¼ð5x y; x þ yþ and the following bases of R : E ¼ fe ; e g ¼ fð; 0Þ; ð0; Þg and S ¼ fu ; u g ¼ fð; 4Þ; ð; Þg (a) Find the change-of-basis matrix P from E to S and the change-of-basis matrix Q from S back to E. (b) Find the matrix A that represents F in the basis E. (c) Find the matrix B that represents F in the basis S. (a) Because E is the usual basis, simply write the vectors in S as columns to obtain the change-of-basis matrix P. Recall, also, that Q ¼ P. Thus, P ¼ and Q ¼ P ¼ 4 4 (b) Write the coefficients of x and y in Fðx; yþ ¼ð5x y; x þ yþ as rows to get A ¼ 5 (c) Method. Find the coordinates of Fðu Þ and Fðu Þ relative to the basis S. This may be done by first finding the coordinates of an arbitrary vector ða; bþ in R relative to the basis S. We have ða; bþ ¼xð; 4Þþyð; Þ ¼ðx þ y; 4x þ yþ; and so Solve for x and y in terms of a and b to get x ¼ a þ b, y ¼ 4a b. Then ða; bþ ¼ ð a þ bþu þð4a bþu x þ y ¼ a 4x þ y ¼ b Now use the formula for ða; bþ to obtain Fðu Þ¼Fð; 4Þ ¼ð; 6Þ ¼ 5u u 5 and so B ¼ Fðu Þ¼Fð; Þ ¼ð; Þ ¼ u þ u Method. By Theorem 6., B ¼ P AP. Thus, B ¼ P AP ¼ 5 5 ¼ Let A ¼. Find the matrix B that represents the linear operator A relative to the basis 4 S ¼ fu ; u g ¼ f½; Š T ; ½; 5Š T g. [Recall A defines a linear operator A: R! R relative to the usual basis E of R ]. Method. Find the coordinates of Aðu Þ and Aðu Þ relative to the basis S by first finding the coordinates of an arbitrary vector ½a; bš T in R relative to the basis S. By Problem 6., ½a; bš T ¼ ð 5a þ bþu þða bþu Using the formula for ½a; bš T, we obtain Aðu Þ¼ ¼ ¼ 5u þ u 4 and Aðu Þ¼ ¼ 9 ¼ 89u þ 54u Thus; B ¼ 54 Method. Use B ¼ P AP, where P is the change-of-basis matrix from the usual basis E to S. Thus, simply write the vectors in S (as columns) to obtain the change-of-basis matrix P and then use the formula

19 CHAPTER 6 Linear Mappings and Matrices for P. This gives P ¼ and P ¼ 5 5 Then B ¼ P AP ¼ ¼ Let A ¼ : Find the matrix B that represents the linear operator A relative to the basis S ¼ fu ; u ; u g ¼ f½; ; 0Š T ; ½0; ; Š T ; ½; ; Š T g [Recall A that defines a linear operator A: R! R relative to the usual basis E of R.] Method. Find the coordinates of Aðu Þ, Aðu Þ, Aðu Þ relative to the basis S by first finding the coordinates of an arbitrary vector v ¼ða; b; cþ in R relative to the basis S. By Problem 6.6, Using this formula for ½a; b; cš T, we obtain ¼ðb cþu þ ð a þ b cþu þða b þ cþu Aðu Þ¼½4; ; Š T ¼ 8u þ u 5u ; Aðu Þ¼½4; ; 0Š T ¼ u 6u þ u Aðu Þ¼½9; 4; Š T ¼ u u þ 6u Writing the coefficients of u ; u ; u as columns yields 8 B ¼ Method. Use B ¼ P AP, where P is the change-of-basis matrix from the usual basis E to S. The matrix P (whose columns are simply the vectors in S) and P appear in Problem 6.6. Thus, 0 B ¼ P AP ¼ ¼ Prove Theorem 6.: Let P be the change-of-basis matrix from a basis S to a basis S 0 in a vector space V. Then, for any linear operator T on V, ½TŠ S 0 ¼ P ½TŠ S P. Let v be a vector in V. Then, by Theorem 6.6, P 0 ¼. Therefore, But ½TŠ S 0 0 ¼½TðvÞŠ S 0. Hence, P ½TŠ S P 0 ¼ P ½TŠ S ¼ P ½TðvÞŠ S ¼½TðvÞŠ S 0 Because the mapping v! 0 P ½TŠ S P ¼½TŠ S 0, as claimed. P ½TŠ S P 0 ¼½TŠ S 0 0 is onto K n, we have P ½TŠ S PX ¼½TŠ S 0X for every X K n. Thus, Similarity of Matrices 6.. Let A ¼ 4 6 and P ¼. 4 (a) Find B ¼ P AP. (b) Verify trðbþ ¼trðAÞ: (c) Verify detðbþ ¼detðAÞ: (a) First find P using the formula for the inverse of a matrix. We have " P ¼ #

20 8 CHAPTER 6 Linear Mappings and Matrices Then B ¼ P AP ¼ 4 ¼ (b) trðaþ ¼4 þ 6 ¼ 0 and trðbþ ¼5 5 ¼ 0. Hence, trðbþ ¼trðAÞ. (c) detðaþ ¼4 þ 6 ¼ 0 and detðbþ ¼ 5 þ 405 ¼ 0. Hence, detðbþ ¼detðAÞ Find the trace of each of the linear transformations F on R in Problem 6.4. Find the trace (sum of the diagonal elements) of any matrix representation of F such as the matrix representation ½FŠ ¼½FŠ E of F relative to the usual basis E given in Problem 6.4. (a) trðfþ ¼trð½FŠÞ ¼ 5 þ 9 ¼ 5. (b) trðfþ ¼trð½FŠÞ ¼ þ þ 5 ¼ 9. (c) trðfþ ¼trð½FŠÞ ¼ þ 4 þ ¼ Write A B if A is similar to B that is, if there exists an invertible matrix P such that A ¼ P BP. Prove that is an equivalence relation (on square matrices); that is, (a) A A, for every A. (b) If A B, then B A. (c) If A B and B C, then A C. (a) The identity matrix I is invertible, and I ¼ I. Because A ¼ I AI, we have A A. (b) Because A B, there exists an invertible matrix P such that A ¼ P BP. Hence, B ¼ PAP ¼ðP Þ AP and P is also invertible. Thus, B A. (c) Because A B, there exists an invertible matrix P such that A ¼ P BP, and as B C, there exists an invertible matrix Q such that B ¼ Q CQ. Thus, A ¼ P BP ¼ P ðq CQÞP ¼ðP Q ÞCðQPÞ ¼ðQPÞ CðQPÞ and QP is also invertible. Thus, A C Suppose B is similar to A, say B ¼ P AP. Prove (a) B n ¼ P A n P, and so B n is similar to A n. (b) f ðbþ ¼P f ðaþp, for any polynomial f ðxþ, and so f ðbþ is similar to f ðaþ: (c) B is a root of a polynomial gðxþ if and only if A is a root of gðxþ. (a) (b) The proof is by induction on n. The result holds for n ¼ by hypothesis. Suppose n > and the result holds for n. Then B n ¼ BB n ¼ðP APÞðP A n PÞ¼P A n P Suppose f ðxþ ¼a n x n þþa x þ a 0. Using the left and right distributive laws and part (a), we have P f ðaþp ¼ P ða n A n þþa A þ a 0 IÞP ¼ P ða n A n ÞP þþp ða AÞP þ P ða 0 IÞP ¼ a n ðp A n PÞþþa ðp APÞþa 0 ðp IPÞ ¼ a n B n þþa B þ a 0 I ¼ f ðbþ (c) By part (b), gðbþ ¼0 if and only if P gðaþp ¼ 0 if and only if gðaþ ¼P0P ¼ 0. Matrix Representations of General Linear Mappings 6.. Let F: R! R be the linear map defined by Fðx; y; zþ ¼ðx þ y 4z; x 5y þ zþ. (a) Find the matrix of F in the following bases of R and R : S ¼ fw ; w ; w g ¼ fð; ; Þ; ð; ; 0Þ; ð; 0; 0Þg and S 0 ¼ fu ; u g ¼ fð; Þ; ð; 5Þg

21 CHAPTER 6 Linear Mappings and Matrices 9 (b) Verify Theorem 6.0: The action of F is preserved by its matrix representation; that is, for any v in R, we have ½FŠ S;S 0 ¼½FðvÞŠ S 0. (a) From Problem 6., ða; bþ ¼ ð 5a þ bþu þða bþu. Thus, Fðw Þ¼Fð; ; Þ ¼ð; Þ ¼ u þ 4u Fðw Þ¼Fð; ; 0Þ ¼ð5; 4Þ ¼ u þ 9u Fðw Þ¼Fð; 0; 0Þ ¼ð; Þ ¼ u þ 8u Write the coordinates of Fðw Þ, Fðw Þ; Fðw Þ as columns to get ½FŠ S;S 0 ¼ (b) If v ¼ðx; y; zþ, then, by Problem 6.5, v ¼ zw þðy zþw þðx yþw. Also, FðvÞ ¼ðxþy 4z; x 5y þ zþ ¼ ð x 0y þ 6zÞu þð8xþy 5zÞu Hence; ¼ðz; y z; x yþ T x 0y þ 6z and ½FðvÞŠ S 0 ¼ 8x þ y 5z z Thus, ½FŠ S;S 0 ¼ 4 y x 5 x 0y þ 6z ¼ ¼½FðvÞŠ x þ y 5z S 0 x y 6.. Let F: R n! R m be the linear mapping defined as follows: Fðx ; x ;...; x n Þ¼ða x þþa n x n, a x þþa n x n ;...; a m x þþa mn x n Þ (a) Show that the rows of the matrix ½FŠ representing F relative to the usual bases of R n and R m are the coefficients of the x i in the components of Fðx ;...; x n Þ. (b) Find the matrix representation of each of the following linear mappings relative to the usual basis of R n : (i) F: R! R defined by Fðx; yþ ¼ðx y; x þ 4y; 5x 6yÞ. (ii) F: R 4! R defined by Fðx; y; s; tþ ¼ðx 4y þ s 5t; 5x þ y s tþ. (iii) F: R! R 4 defined by Fðx; y; zþ ¼ðx þ y 8z; x þ y þ z; 4x 5z; 6yÞ. (a) We have Fð; 0;...; 0Þ ¼ða ; a ;...; a m Þ Fð0; ;...; 0Þ ¼ða ; a ;...; a m Þ ::::::::::::::::::::::::::::::::::::::::::::::::::::: Fð0; 0;...; Þ ¼ða n ; a n ;...; a mn Þ and thus; a a... a n a ½FŠ ¼ a... a n 6 4::::::::::::::::::::::::::::::::: 5 a m a m... a mn (b) By part (a), we need only look at the coefficients of the unknown x; y;... in Fðx; y;...þ. Thus, 8 ðiþ ½FŠ ¼ ; ðiiþ ½FŠ ¼ ; ðiiiþ ½FŠ ¼ Let A ¼ 5. Recall that A determines a mapping F: R! R defined by FðvÞ ¼Av, 4 where vectors are written as columns. Find the matrix ½FŠ that represents the mapping relative to the following bases of R and R : (a) The usual bases of R and of R. (b) S ¼ fw ; w ; w g ¼ fð; ; Þ; ð; ; 0Þ; ð; 0; 0Þg and S 0 ¼ fu ; u g ¼ fð; Þ; ð; 5Þg. (a) Relative to the usual bases, ½FŠ is the matrix A.

22 0 CHAPTER 6 Linear Mappings and Matrices (b) From Problem 9., ða; bþ ¼ ð 5a þ bþu þða bþu. Thus, Fðw Þ¼ ¼ 4 ¼ u þ 8u 4 4 Fðw Þ¼ ¼ ¼ 4u þ 4u 4 0 Fðw Þ¼ ¼ ¼ 8u þ 5u Writing the coefficients of Fðw Þ, Fðw Þ, Fðw Þ as columns yields ½FŠ ¼ Consider the linear transformation T on R defined by Tðx; yþ ¼ðx y; x þ 4yÞ and the following bases of R : E ¼ fe ; e g ¼ fð; 0Þ; ð0; Þg and S ¼ fu ; u g ¼ fð; Þ; ð; 5Þg (a) Find the matrix A representing T relative to the bases E and S. (b) Find the matrix B representing T relative to the bases S and E. (We can view T as a linear mapping from one space into another, each having its own basis.) (a) From Problem 6., ða; bþ ¼ ð 5a þ bþu þða bþu. Hence, Tðe Þ¼Tð; 0Þ ¼ð; Þ ¼ 8u þ 5u 8 and so A ¼ Tðe Þ¼Tð0; Þ ¼ ð ; 4Þ ¼ u u 5 (b) We have Tðu Þ¼Tð; Þ ¼ ð ; Þ ¼ e þ e and so B ¼ Tðu Þ¼Tð; 5Þ ¼ ð ; Þ ¼ e þ e 6.5. How are the matrices A and B in Problem 6.4 related? By Theorem 6., the matrices A and B are equivalent to each other; that is, there exist nonsingular matrices P and Q such that B ¼ Q AP, where P is the change-of-basis matrix from S to E, and Q is the change-of-basis matrix from E to S. Thus, P ¼ ; Q ¼ 5 ; Q ¼ 5 5 and Q AP ¼ 8 ¼ ¼ B Prove Theorem 6.4: Let F: V! U be linear and, say, rankðfþ ¼r. Then there exist bases V and of U such that the matrix representation of F has the following form, where I r is the r-square identity matrix: A ¼ I r Suppose dim V ¼ m and dim U ¼ n. Let W be the kernel of F and U 0 the image of F. We are given that rank ðfþ ¼r. Hence, the dimension of the kernel of F is m r. Let fw ;...; w m r g be a basis of the kernel of F and extend this to a basis of V: Set fv ;...; v r ; w ;...; w m r g u ¼ Fðv Þ; u ¼ Fðv Þ;...; u r ¼ Fðv r Þ

23 CHAPTER 6 Linear Mappings and Matrices Then fu ;...; u r g is a basis of U 0, the image of F. Extend this to a basis of U, say fu ;...; u r ; u rþ ;...; u n g Observe that Fðv Þ ¼ u ¼ u þ 0u þþ0u r þ 0u rþ þþ0u n Fðv Þ ¼ u ¼ 0u þ u þþ0u r þ 0u rþ þþ0u n :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Fðv r Þ ¼ u r ¼ 0u þ 0u þþu r þ 0u rþ þþ0u n Fðw Þ ¼ 0 ¼ 0u þ 0u þþ0u r þ 0u rþ þþ0u n :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: Fðw m r Þ¼0 ¼ 0u þ 0u þþ0u r þ 0u rþ þþ0u n Thus, the matrix of F in the above bases has the required form. SUPPLEMENTARY PROBLEMS Matrices and Linear Operators 6.. Let F: R! R be defined by Fðx; yþ ¼ð4x þ 5y; x yþ. (a) Find the matrix A representing F in the usual basis E. (b) (c) Find the matrix B representing F in the basis S ¼ fu ; u g ¼ fð; 4Þ; ð; 9Þg. Find P such that B ¼ P AP. (d) For v ¼ða; bþ, find and ½FðvÞŠ S. Verify that ½FŠ S ¼½FðvÞŠ S Let A: R! R be defined by the matrix A ¼ 5. 4 (a) Find the matrix B representing A relative to the basis S ¼ fu ; u g ¼ fð; Þ; ð; 8Þg. (Recall that A represents the mapping A relative to the usual basis E.) (b) For v ¼ða; bþ, find and ½AðvÞŠ S For each linear transformation L on R, find the matrix A representing L (relative to the usual basis of R ): (a) L is the rotation in R counterclockwise by 45. (b) L is the reflection in R about the line y ¼ x. (c) (d) L is defined by Lð; 0Þ ¼ð; 5Þ and Lð0; Þ ¼ð; Þ. L is defined by Lð; Þ ¼ð; Þ and Lð; Þ ¼ð5; 4Þ Find the matrix representing each linear transformation T on R relative to the usual basis of R : (a) Tðx; y; zþ ¼ðx; y; 0Þ. (b) Tðx; y; zþ ¼ðz; y þ z; x þ y þ zþ. (c) Tðx; y; zþ ¼ðx y 4z; x þ y þ 4z; 6x 8y þ zþ Repeat Problem 6.40 using the basis S ¼ fu ; u ; u g ¼ fð; ; 0Þ; ð; ; Þ; ð; ; 5Þg Let L be the linear transformation on R defined by Lð; 0; 0Þ ¼ð; ; Þ; Lð0; ; 0Þ ¼ð; ; 5Þ; Lð0; 0; Þ ¼ð; ; Þ (a) Find the matrix A representing L relative to the usual basis of R. (b) Find the matrix B representing L relative to the basis S in Problem Let D denote the differential operator; that is, Dð f ðtþþ ¼ df =dt. Each of the following sets is a basis of a vector space V of functions. Find the matrix representing D in each basis: (a) fe t ; e t ; te t g. (b) f; t; sin t; cos tg. (c) fe 5t ; te 5t ; t e 5t g.

24 CHAPTER 6 Linear Mappings and Matrices Let D denote the differential operator on the vector space V of functions with basis S ¼ fsin y, cos yg. (a) Find the matrix A ¼½DŠ S. (b) Use A to show that D is a zero of f ðtþ ¼t þ Let V be the vector space of matrices. Consider the following matrix M and usual basis E of V: M ¼ a b c d and E ¼ ; 0 ; ; Find the matrix representing each of the following linear operators T on V relative to E: (a) TðAÞ ¼MA. (b) TðAÞ ¼AM. (c) TðAÞ ¼MA AM Let V and 0 V denote the identity and zero operators, respectively, on a vector space V. Show that, for any basis S of V, (a) ½ V Š S ¼ I, the identity matrix. (b) ½0 V Š S ¼ 0, the zero matrix. Change of Basis 6.4. Find the change-of-basis matrix P from the usual basis E of R to a basis S, the change-of-basis matrix Q from S back to E, and the coordinates of v ¼ða; bþ relative to S, for the following bases S: (a) S ¼ fð; Þ; ð; 5Þg. (c) S ¼ fð; 5Þ; ð; Þg. (b) S ¼ fð; Þ; ð; 8Þg. (d) S ¼ fð; Þ; ð4; 5Þg Consider the bases S ¼ fð; Þ; ð; Þg and S 0 ¼ fð; Þ; ð; 4Þg of R. Find the change-of-basis matrix: (a) P from S to S 0. (b) Q from S 0 back to S Suppose that the x-axis and y-axis in the plane R are rotated counterclockwise 0 to yield new x 0 -axis and y 0 -axis for the plane. Find (a) (b) (c) The unit vectors in the direction of the new x 0 -axis and y 0 -axis. The change-of-basis matrix P for the new coordinate system. The new coordinates of the points Að; Þ, Bð; 5Þ, Cða; bþ Find the change-of-basis matrix P from the usual basis E of R to a basis S, the change-of-basis matrix Q from S back to E, and the coordinates of v ¼ða; b; cþ relative to S, where S consists of the vectors: (a) u ¼ð; ; 0Þ; u ¼ð0; ; Þ; u ¼ð0; ; Þ. (b) u ¼ð; 0; Þ; u ¼ð; ; Þ; u ¼ð; ; 4Þ. (c) u ¼ð; ; Þ; u ¼ð; ; 4Þ; u ¼ð; 5; 6Þ Suppose S ; S ; S are bases of V. Let P and Q be the change-of-basis matrices, respectively, from S to S and from S to S. Prove that PQ is the change-of-basis matrix from S to S. Linear Operators and Change of Basis 6.5. Consider the linear operator F on R defined by Fðx; yþ ¼ð5x þ y; x yþ and the following bases of R : S ¼ fð; Þ; ð; Þg and S 0 ¼ fð; Þ; ð; 4Þg (a) Find the matrix A representing F relative to the basis S. (b) Find the matrix B representing F relative to the basis S 0. (c) Find the change-of-basis matrix P from S to S 0. (d) How are A and B related? 6.5. Let A: R! R be defined by the matrix A ¼. Find the matrix B that represents the linear operator A relative to each of the following bases: (a) S ¼ fð; Þ T ; ð; 5Þ T g. (b) S ¼ fð; Þ T ; ð; 4Þ T g.