1)

Solución

2)

3)

UNIDAD VI

EJERCICIO 1

DETERMINE SI LATRANSFORMACIÓN DADA DE V W ES LINEAL O NO. I.- T: R2 R3 ; T (X, Y)= (X, Y) T [ (X1, Y1 ) + (X2 , Y2) ] = T [(X1, +X2 ) + (Y1 , Y2)] = [(Y1, +Y2 ) + (X1 , X2)]

(Y1, +Y2 ) = T(X1, Y1) Y (X1 , X2) = T(X2, Y2) T [ (X1, Y1 ) + (X2 , Y2) ] = T(X1, X2) + T (Y1, Y2) T [ K (X, Y) ] = T ( KX, KY) = KT(X, Y) = (Y, X)

SI ES UNA TRANSFORMACIÓN LINEAL EJERCICIO 2 2.- T: R3 R2 ; T (X, Y, Z) = (1, Z) T [ (X1, Y1 , Z1) + (X2 , Y2 , Z2) ] = T [(X1, +X2 ) + (Y1 + Y2) (Z1 + Z2)]

= [( 1 ) + (Z1 , Z2)] T [ (1, Z1) + (1, Z2)] = T(1, Z1) + T(1, Z2) T [ K (X, Y, Z) ] = T ( KX, KY, KZ) = KT(X, Y, Z) = K(1, Z) = (K, KZ)

NO ES UNA TRANSFORMACIÓN LINEAL. EJERCICIO 3 2.- T: R3 R2 ; T (X, Y, Z) = (X, Y) T [ (X1, Y1 , Z1) + (X2 , Y2 , Z2) ] = T [(X1, +X2 ) + (Y1 + Y2) (Z1 + Z2)] = [( X1, X2) + (Y1 , Y2)] T [ (X1, Y1) + (X1, Y2)] = T(X1, Y1) + T(X2, Y2) T [ K (X, Y, Z) ] = T ( KX, KY, KZ) = K(X, Y) = KT(X, Y)

SI ES UNA TRANSFORMACIÓN LINEAL. EJERCICIO 4 4.- T: R2 R2 ; T (X, Y) = (1, Y) T [ (X1, Y1 ) + (X2 , Y2 ) ] = T [(X1, +X2 ) + (Y1 + Y2)] = [( 1 ) + (Y1 , Y2)] T [ (1, Y1) + (1, Y2)] = T(1, Y1) + T(1, Y2) T [ K (X, Y) ] = T ( KX, KY,) = KT(X, Y) = K(1, Y) = (K, KY)

NO ES UNA TRANSFORMACIÓN LINEAL EJERCICIO 5 5.- T: R2 R2 ; T (X, Y) = (X + Y, X - Y, 3Y) T [ (X1, Y1 ) + (X2 , Y2 ) ] = T [(X1, +X2 ) + (Y1 + Y2)]

= [ (X1 + X2 +Y2 +Y2), (X1 + X2 -Y2 -Y2), (3Y1 +3Y2)] = [ (X1 + Y1 , X2 +Y2 , 3Y1 ) + ( X2 -Y2 , X2 -Y2 , 3Y2)] (X1 + Y1 , X2 +Y2 , 3Y1 ) = T (X1 , Y1) Y ( X2 -Y2 , X2 -Y2 , 3Y2) = T (X2 , Y2) T [ (X1, Y1 ) + (X2 , Y2 ) ] = T [(X1, +X2 ) + (Y1 + Y2)] T [ K (X, Y) ] = T ( KX, KY) = ( KX + K Y. KX – KY, 3KY) = k( X + Y, X – Y, 3Y) = KT (X, Y)

SI ES UNA TRANSFORMACIÓN LINEAL. OBTENER LA IMAGEN, NÚCLEO, RANGO Y NULIDAD DE LA TRANSFORMACIÓN DEFINIDA POR: EJEMPLO 1

T (X, Y, Z) = ( X + Y, Y + Z, 0 )

B = { (1, 0, 0), ( 0, 1, 0), ( 0, 0, 1)}

T = ( 1, 0, 0) = ( 1, 0, 0)

T = ( 0, 1, 0,) = ( 0, 1, 0)

T = ( 0, 0, 1) = ( 0, 1, 0) 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 0 0 α(1, 0, 0) + α( 0, 1, 0) = ( x, y , z) α = x α = y z = 0 II = { ( α, α, 0 ) / α, α, ∈ R } R = 2

( X + Y, Y + Z, 0 ) = ( 0, 0 ,0) X + Y = 0 X + Y = 0 0 = 0 X = -Y Y = - Z Z = Z SI Z = 1, y = -1, X = 1 BN { ( 1, -1, 1 )} α ( 1, -1. 1) = ( X. Y . Z) X = α Y = -α Z = α N = { (α, -α, α) / α ∈ R}

N = 1 EJEMPLO 2

2.- T ( X, Y , Z) = ( 2Y + Z, 3X + 1, 1+ Y) B = { (1, 0, 0), ( 0, 1, 0), ( 0, 0, 1)}

T = ( 1, 0, 0) = ( 0, 4, 1)

T = ( 0, 1, 0,) = ( 2, 1, 1)

T = ( 0, 0, 1) = ( 1, 1, 1) 0 4 1 1 1 1 0 1 1 1 1 1 2 1 2 2 1 2 0 -1 -1 0 -1 -1 1 1 1 0 4 1 0 4 1 0 0 7 α(1, 1, 1) + α( 0, -1, -1) +α (0, 0, -7)= ( x, y , z) α = x α - α = y

α - α - 7α = Z II = { ( α, α - α, α - α - 7α, 0 ) / α, α, α ∈ R } R = 3 ( 2Y + Z, 3X + 1, 1+ Y) = ( 0, 0, 0) 2Y + Z = 0 3X + 1= 0 1+ Y = 0 X = - 1/3 Y = - 1 Z = 2 BN { ( - 1/3, -1, 2 )} α (- 1/3, -1, 2 ) = ( X. Y . Z) X = -1/3α Y = -α Z = 2α N = { ( -1/3α, -α, 2α) / α ∈ R}

N = 1

EJEMPLO 3 3.- T ( X, Y , Z) = (3X + Y, Z + 1, 3Y + 1) B = { (1, 0, 0), ( 0, 1, 0), ( 0, 0, 1)} T = ( 1, 0, 0) = ( 3, 1, 1) T = ( 0, 1, 0,) = ( 1, 0 ,4) T = ( 0, 0, 1) = ( 0, 2, 1) 3 1 1 1 1 4 1 1 1 1 1 4 1 1 4 3 1 1 0 -2 -11 0 -2 -11 0 2 1 0 2 1 0 2 1 0 0 -10 α(1, 1, 4) + α( 0, -2, -11) +α (0, 0, -10)= ( x, y , z) α = x α - α = y

4α -11 α - 10α = Z II = { ( α, α - α,4 α - 11α - 10α, ) / α, α, α ∈ R } R = 3 ( 3X + Y, Z + 1, 3Y + 1) = ( 0, 0, 0) 3X + Y= 0 Z + 1= 0 3Y + 1 = 0 X = 1/9 Y = - 1/3 Z =-1 BN { ( 1/9, -1/3, -1 )} α ( 1/9, -1/3, -1 ) = ( X. Y . Z) X = -1/9α Y = -1/3α Z = -1α N = { ( 1/9α, -1/3α, -1α) / α ∈ R}

N = 1

EJEMPLO 4 4.- T ( X, Y , Z) = ( X + Z, Y, X + Z) B = { (1, 0, 0), ( 0, 1, 0), ( 0, 0, 1)} T = ( 1, 0, 0) = ( 1, 0, 1) T = ( 0, 1, 0,) = ( 0, 1, 0) T = ( 0, 0, 1) = ( 1, 0, 1) 1 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 0 0 α(1, 0, 1) + α( 0, 1, 0) = ( x, y , z) α = x α = y α= Z II = { ( α, α) / α, α ∈ R }

R = 2 ( X + Z, Y, X + Z) = ( 0, 0, 0) X + Z= 0 Y= 0 X + Z = 0 X = -Z Y = 0 Z = Z SI Z = 1, X = -1, Y = 0 BN { ( -1, 0, 1 )} α ( -1, 0, 1 ) = ( X. Y . Z) X = -α Y = 0 Z = α N = { ( -α, 0, α) / α ∈ R}

N = 1

EJEMPLO 5 5.- T ( X, Y , Z) = ( X + Z, 0) B = { (1, 0, 0), ( 0, 1, 0), ( 0, 0, 1)} T = ( 1, 0, 0) = ( 1, 0, 0) T = ( 0, 1, 0,) = ( 0, 1, 0) T = ( 0, 0, 1) = ( 1, 0, 0) 1 0 0 1 0 0 0 1 0 0 1 0 1 0 0 0 0 0 α(1, 0, 0) + α( 0, 1, 0) = ( x, y , z) α = x α = y 0= Z II = { ( α, α, 0) / α, α ∈ R } R = 2 ( X + Z, Y, 0) = ( 0, 0, 0) X + Z= 0 Y= 0

0 = 0 X = -Z Y = 0 Z = Z

SI Z = 1, X = -1, Y = 0 BN { ( -1, 0, 1 )} α ( -1, 0, 1 ) = ( X. Y . Z) X = -α Y = 0 Z = α N = { ( -α, 0, α) / α ∈ R}

N = 1