Matrice di una trasformazione lineare in una data base
ciao a tutti,
dovrei calcolare la matrice nella base
$G:={(2/3,-1/3,1,0),(0,-1,0,1),(0,1,0,1),(1,0,1,0)}$
della trasformazione lineare
$y'=2x^1+x^2-x^3+x^4$
$y''=x^1-x^2-x^3-x^4$
$y'''=3x^2+x^3+3x^4$
a me viene
$((-9,3/2,-1/2,8),(-6,1,-2,5),(3,-1/2,7/2,-2))$
ho fatto bene? grazie mille
dovrei calcolare la matrice nella base
$G:={(2/3,-1/3,1,0),(0,-1,0,1),(0,1,0,1),(1,0,1,0)}$
della trasformazione lineare
$y'=2x^1+x^2-x^3+x^4$
$y''=x^1-x^2-x^3-x^4$
$y'''=3x^2+x^3+3x^4$
a me viene
$((-9,3/2,-1/2,8),(-6,1,-2,5),(3,-1/2,7/2,-2))$
ho fatto bene? grazie mille
Risposte
per trovarla ho risolto il sistema:
$(2,1,-1,1) = a'(2/3,-1/3,1,0) + b'(0,-1,0,1) + c'(0,1,0,1) + d'(1,0,1,0)$
$(1,-1,-1,-1) = a''(2/3,-1/3,1,0) + b''(0,-1,0,1) + c''(0,1,0,1) + d''(1,0,1,0)$
$(0,3,1,3) = a'''(2/3,-1/3,1,0) + b'''(0,-1,0,1) + c'''(0,1,0,1) + d'''(1,0,1,0)$
trovando:
$ a'=-9, b'=3/2, c'=-1/2, d'=8$
$ a''=-6, b''=1, c''=-2, d''=5$
$ a'''=3, b'''=-1/2, c'''=7/2, d'''=-2$
$(2,1,-1,1) = a'(2/3,-1/3,1,0) + b'(0,-1,0,1) + c'(0,1,0,1) + d'(1,0,1,0)$
$(1,-1,-1,-1) = a''(2/3,-1/3,1,0) + b''(0,-1,0,1) + c''(0,1,0,1) + d''(1,0,1,0)$
$(0,3,1,3) = a'''(2/3,-1/3,1,0) + b'''(0,-1,0,1) + c'''(0,1,0,1) + d'''(1,0,1,0)$
trovando:
$ a'=-9, b'=3/2, c'=-1/2, d'=8$
$ a''=-6, b''=1, c''=-2, d''=5$
$ a'''=3, b'''=-1/2, c'''=7/2, d'''=-2$
up!