Linear Algebra

190 Chapter Three. Maps Between Spaces(these are easy to check). Then, as described in the preamble, for any member⃗v of the domain, we can express the image h(⃗v) in terms of the h( ⃗ β)’s.( ( 2 1h(⃗v) = h(c 1 · + c0)2 · )4)( (2 1= c 1 · h( ) + c0)2 · h( )4)⎛ ⎞ ⎛ ⎞⎛⎞⎛⎞⎛⎞⎛⎞= c 1 · (0 ⎝ 1 0⎠− 1 ⎝ 0 −2⎠+ 1 ⎝ 1 0⎠) + c 2 · (1 ⎝ 1 0⎠− 1 ⎝ 0 −2⎠+ 0 ⎝ 1 0⎠)20 0 10 0 1⎛ ⎞⎛1= (0c 1 + 1c 2 ) · ⎝0⎠ + (− 1 2 c 1 − 1c 2 ) · ⎝ 0 ⎞⎛−2⎠ + (1c 1 + 0c 2 ) · ⎝ 1 ⎞0⎠001Thus,For instance,with Rep B (⃗v) =⎛)then Repc D ( h(⃗v) ) = ⎝ 0c ⎞1 + 1c 2−(1/2)c 1 − 1c 2⎠.21c 1 + 0c 2(c1⎛( ( (4 1 4with Rep B ( ) = then Rep8)2)D ( h( ) ) = ⎝8) 2⎞−5/2⎠.B1We will express computations like the one above with a matrix notation.⎛⎝ 0 −1/2 1⎞−1⎠1 0B,D(c1c 2)B⎛= ⎝ 0c ⎞1 + 1c 2(−1/2)c 1 − 1c 2⎠1c 1 + 0c 2In the middle is the argument ⃗v to the map, represented with respect to thedomain’s basis B by a column vector with components c 1 and c 2 . On the rightis the value h(⃗v) of the map on that argument, represented with respect to thecodomain’s basis D by a column vector with components 0c 1 + 1c 2 , etc. Thematrix on the left is the new thing. It consists of the coefficients from the vectoron the right, 0 and 1 from the first row, −1/2 and −1 from the second row, and1 and 0 from the third row.This notation simply breaks the parts from the right, the coefficients and thec’s, out separately on the left, into a vector that represents the map’s argumentand a matrix that we will take to represent the map itself.D