Linear Algebra Question

Question

H={[a,b,c,d]: a+3b=c, b+c+a=d}

 

Show the subset H is a subspace or give a counter example

 

I am very confused can someone help

Answer 1

(1) It must contain the zero vector

(2) It must be closed under addition

(3) It must be closed under scalar multiplication

 

condition 1: It must contain the zero vector

 [0,0,0,0]

            a + 3b = 0 + 3(0) = 0 + 0 = 0 = c

         b+c+a = 0 +0+0=0 = d

 

 Yes, the zero vector is in H.

 

condition 2: It must be closed under addition.

 

   Let U and V be elements of H , where U = (u1,u2,u3,u4) and V = (v1,v2,v3,v4)

 

   Then u1 + 3*u2 = u3 and u2 + u3+u1 = u4 AND

           v1  + 3*v2 = v3 and v2 + v3 + v1 = v4 <--- since U and V are in H, these properties hold. 

                                                                       <--- please note and label the bold face equations ALPHA and

                                                                       <--- the underlined,italic equations BETA, respectively

 

Let Z = U + V = ( u1 + v1,  u2 + v2,  u3 + v3, u4 + v4) = (z1,z2,z3,z4)

 

  (u1 + v1) + 3( u2 + v2) = u1 + v1 + 3*u2 + 3*v2    <--- distributive property

                                      = (u1 + 3*u2) + (v1 + 3*v2)  <--- commutative and associative properties

                                      = u3 + v3  <---- by substitution of the ALPHA equations above

                                      = z3 <--- substitution

 

So far so good...

 

   (u2 + v2) + (u3+v3) + (u1+v1) = 

     (u2 + u3 + u1) + (v2 + v3 + v1) =   <---- associative and commutative properties

     u4 + v4  =              <--- by substitution of the BETA equations above

     z4    <---- substitution

 

Closure under addition is proven.

 

Condition 3: It must be closed under scalar multiplication

 

Let vector X=(x1,x2,x3,x4) be an element of H and k be a fixed scalar constant.

 

Then (x1 + 3*x2 = x3) and (x2+x3+x1 = x4) since vector X is in H.  <--- call these equations DELTA and EPSILON

                                                                                                                  respectively

 

Let Z = kX = (k*x1, k*x2, k*x3, k*x4) = (z1, z2, z3, z4)

 

k*x1 + 3*k*x2 = k * (x1 + 3*x2) <--- distributive

                     = k * x3 <--- substitute equation DELTA

                    = z3  <--- substitution

 

so far, so good....

 

k*x2 + k*x3 + k*x1 = k*(x2+x3+x1)   <--- distributive

                               = k*x4 <--- substitution of equation EPSILON

                               = z4

 

Closure under scalar multiplication is proven.

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