Then the diagram V T B W C V T C B W commutes ie T C B v B Tv C for all v V

# Then the diagram v t b w c v t c b w commutes ie t c

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Then the diagram V T / [ · ] B W [ · ] C V [ T ] C B / W commutes, i.e. [ T ] C B [ v ] B = [ Tv ] C for all v V . Proof. To show both matrices are equal, we must show that their entries are equal. We know ([ T ] C B ) i,j = e * i [ Tv j ] C . Hence ([ T ] C B [ v ] B ) i, 1 = n X j =1 ([ T ] C B ) i,j ([ v ] B ) j, 1 = n X j =1 e * i [ Tv j ] C ([ v ] B ) j, 1 = e * i n X j =1 ([ v ] B ) j, 1 [ Tv j ] C = e * i " n X j =1 ([ v ] B ) j, 1 Tv j # C = e * i " T n X j =1 ([ v ] B ) j, 1 v j # C = e * i [ Tv ] C = ([ Tv ] C ) i, 1 as [ · ] C , e * i , and T are linear transformations, and v = n X j =1 ([ v ] B ) j 1 v j . Remarks 3.4.7 . When we proved 2.4.3 and 2.5.6 (1), we used the coordinate map implicitly. The algorithm of Gaussian elimination proves the hard result of 2.4.2, which in turn, proves these technical theorems after we identify the vector space with F n for some n . Proposition 3.4.8. Suppose S, T L ( V, W ) and λ F , and suppose that B = { v 1 , . . . , v n } is a basis for V and C = { w 1 , . . . , w m } is a basis for W . Then [ S + λT ] C B = [ S ] C B + λ [ T ] C B . Thus [ · ] C B : L ( V, W ) M m × n ( F ) is a linear transformation. 49
Proof. First, we note that this follows immediately from 3.4.2 and 3.4.6 as [ S + λT ] C B x = [( S + λT )[ x ] - 1 B ] C = [ S [ x ] - 1 B + λT [ x ] - 1 B ] C = [ S [ x ] - 1 B ] C + λ [ T [ x ] - 1 B ] C = [ S ] C B x + λ [ T ] C B x for all x F n . We give a direct proof as well. We have that Sv j = m X i =1 ([ S ] C B ) i,j w i and Tv j = m X i =1 ([ T ] C B ) i,j w i for all j [ n ]. Thus ( S + λT ) v j = Sv j + λTv j = m X i =1 ([ S ] C B ) i,j w i + λ m X i =1 ([ T ] C B ) i,j w i = m X i =1 (([ S ] C B ) i,j + λ ([ T ] C B ) i,j ) w i . for all j [ n ], and ([ S + λT ] C B ) i,j = ([ S ] C B ) i,j + λ ([ T ] C B ) i,j , so [ S + λT ] C B = [ S ] C B + λ [ T ] C B . Proposition 3.4.9. Suppose T L ( U, V ) and S L ( V, W ) , and suppose that A = { u 1 , . . . , u m } is a basis for U , B = { v 1 , . . . , v n } is a basis for V , and C = { w 1 , . . . , w p } is a basis for W . Then [ ST ] C A = [ S ] C B [ T ] B A . Proof. We have that STu j = S n X i =1 ([ T ] B A ) i,j v i = n X i =1 ([ T ] B A ) i,j ( Sv i ) = n X i =1 ([ T ] B A ) i,j p X k =1 ([ S ] C B ) k,i w k ! = p X k =1 n X i =1 ([ S ] C B ) k,i ([ T ] B A ) i,j ! w k = p X k =1 ( [ S ] C B [ T ] B A ) k,j w k for all j [ m ], so [ ST ] C A = [ S ] C B [ T ] B A . Exercises V, W will denote vector spaces. Let T L ( V, W ). Exercise 3.4.10. Exercise 3.4.11. Exercise 3.4.12. Suppose dim( V ) = n < and dim( W ) = m < , and let B, C be bases for V, W respectively. (1) Show that L ( V, W ) = M m × n ( F ). (2) Show T is invertible if and only if [ T ] C B is invertible. Exercise 3.4.13. Suppose V is finite dimensional and B, C are two bases of V . Show [ T ] B [ T ] C . 50
Chapter 4 Polynomials In this section, we discuss the background material on polynomials needed for linear algebra. The two main results of this section are the Euclidean Algorithm and the Fundamental Theorem of Algebra. The first is a result on factoring polynomials, and the second says that every polynomial in C [ z ] has a root. The latter result relies on a result from complex analysis which is stated but not proved. For this section, F is a field. 4.1 The Algebra of Polynomials Definition 4.1.1. A polynomial p over F is a sequence p = ( a i ) i Z 0 where a i F for all i Z 0 such that there is an n N such that a i = 0 for all i > n . The minimal n Z 0 such that a i = 0 for all i > n (if it exists) is called the degree of p , denoted deg( p ), and we define the degree of the zero polynomial, the sequence of all zeroes, denoted 0, to be -∞ .

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