By Gabriela Jeronimo, Juan Sabia y Susana Tesauri
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Ws } es una base de S, resulta que αi = 0 para todo 1 ≤ i ≤ r y βj = 0 para todo r + 1 ≤ j ≤ s. Luego dim(S + T ) = r + (s − r) + (t − r) = s + t − r = dim S + dim T − dim(S ∩ T ). 2 Suma directa Un caso de especial importancia de suma de subespacios se presenta cuando S ∩ T = {0}. 44 Sea V un K-espacio vectorial, y sean S y T subespacios de V . Se dice que V es suma directa de S y T , y se nota V = S ⊕ T , si: 1. V = S + T , 2. S ∩ T = {0}. Ejemplo. Sean S = {x ∈ R3 : x1 + x2 + x3 = 0} y T = < (1, 1, 1) >.
B+C =B C . B = B iii) V = R>0 , K = Q. b √ n ⊗ : m am n ⊗a= Ejercicio 3. Sea V un espacio vectorial sobre K, k ∈ K, v ∈ V . v = 0 ⇒ k = 0 ´o v = 0 ii) iv) −0 = 0 −(−v) = v Ejercicio 4. i) Sea v ∈ R2 un vector fijo. Se define la funci´on fv : R2 → R2 de la siguiente forma: fv (x, y) = (x, y) + v Interpretar geom´etricamente el efecto de fv sobre el plano (fv se llama la traslaci´ on en v). (x − 2, y − 1) + (2, 1) (Este espacio se notar´a R2(2,1) para distinguirlo de R2 con la suma y el producto usual.
Se tiene que dim S = 2, dim T = 1 y S ∩ T = {0}. Entonces dim(S + T ) = 3, de donde S + T = R3 . Luego, R3 = S ⊕ T . 45 Sea V un K-espacio vectorial. Sean S y T subespacios de V tales que V = S ⊕ T . Entonces, para cada v ∈ V , existen u ´nicos x ∈ S e y ∈ T tales que v = x + y. Demostraci´ on. Existencia: Como V = S + T , para cada v ∈ V existen x ∈ S, y ∈ T tales que v = x + y. Unicidad: Supongamos que v = x + y y v = x + y con x, x ∈ S, y, y ∈ T . Entonces x − x = y − y y x − x ∈ S, y − y ∈ T , luego x − x ∈ S ∩ T = {0}.
Algebra Lineal by Gabriela Jeronimo, Juan Sabia y Susana Tesauri
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