sir please All choose not subspace of a subset of the R3 and

sir, please.

All choose, not subspace of a subset of the R^3. and explain the reason why. (a) { (0,y,z)|y,z, R} (b) { (1,y,z)|y,z, R} (c) {(0;y;z)|y,z, R} {(x,0,z)|x,z, R} (d) {(0,0,0)} (e) {a(1,1,0) + b(2,0,1)|a, b, R} (f) { (x, y, z) I z - y + 3x =0}

Solution

An arbitrary element of R³ is of the form ( a ,b,c) where a,b, c R. Let the subset of R³ be called S in all the cases.

(a) An arbitrary element of this sunset S is of the form ( 0, x, y ) where x , y R.. Let ( 0, x₁ , y₁ ) and ( 0, x₂ , y₂ ) be two arbitrary elements of S. Then ( 0, y₁ , z₁ ) + ( 0, y₂ , z₂ ) = ( 0, y₁ + y₂ , z₁ + z₂ ) also S. so that S is closed under addition. Further, if be an arbitrary scalar, then ( 0, y, z ) = ( 0, y, z) also belongs to S as x and y R. Thus S is closed under scalar multiplication. Further , the zero vector S as   can be 0 also. Thus S is a subspace of R³.

(b) An arbitrary element of S is of the form ( 1, y, z). Let ( 1, y₁ , z₁ ) and (1, y₂ , z₂ ) be two arbitrary elements of S. Then ( 1, y₁ , z₁ ) + ( 1, x₂ , y₂ ) = ( 2, y₁ + y₂ , z₁ + z₂ )   which does not belong to S. Thus S is not closed under addition and , therefore, S is not a subspace of R³.

(c) An arbitrary element of S is of the form ( 0, y₁ , z₁ ) + ( x₁ , 0 , z₂ ) = ( x_1, y_1, z₁ + z₂  ) = ( x_1, y_1, w₁), ( say). It is similar to an arbitrary element of R³ . Thus, S = R³ and therefore S is a subspace of R³ ( Every vector space is a subset, and therefore, a subspace of itself).

(d) An arbitrary element of S is of the form( 0, 0, 0).Since( 0, 0, 0)+( 0, 0 ,0) = (0, 0, 0) and ( 0, 0, 0)= ( 0, 0, 0) ,   S is closed under addition and scalar multiplication and is, therefore, a subspace of R³.

(e) An arbitrary element of S is of the form a (1, 1, 0) + b ( 2, 0, 1) = ( a + 2b, a, b). Let ( a₁ + 2b₁, a₁, b₁). and   ( a₂ + 2b₂, a₂ , b₂ ).be two arbitrary elements of S. Then, ( a₁ + 2b₁, a₁, b₁) + ( a₂ + 2b₂, a₂ , b₂ ) = ( a₁ + a₂ + 2b₁ + 2b₂ , a₁ + a₂ , b₁ + b₂ ) S so that S is closed under addition. Further, ( a + 2b, a, b) = (a + 2b , a , b  ) S so that S is closed under scalar multiplication. Therefore S is a subspace of R^3.

^{( f)} An arbitrary element of S is of the form ( x, y, z) where z - y + 3x = 0. Let ( x₁ , y₁ , z₁ ) and ( x₂, y₂ , z₂ ) be two arbitrary elements of S. Then, z₁ - y₁ + 3x₁ = 0 and z₂ - y₂ + 3x₂ = 0 . Further, ( x₁ , y₁ , z₁ ) + ( x₂, y₂ , z₂ ) = ( x₁ + x₂ , y₁ + y₂ , z₁ + z₂ ). Also z₁ + z₂ - ( y₁ + y₂ ) + 3 (x₁ + x₂) = z₁ - y₁ + 3x₁ + z₂ - y₂ + 3x₂ = 0 + 0 = 0. Thus S is closed under addition. Further, ( x, y, z) = ( x , y , z ) and z - y + 3 x = (z - y + 3x) = *0 = 0. Thus S is closed under scalar multiplication. Therefore, S is a subspace of R³.