Question 3698
 1. Sol:   
         About W1, since a1 – a3 – a4 = 0, or a1 = a3 + a4.
         We see that {(s+t,r,s,t,u)|r,s,t,u in F} is a general
         form of vectors in W1.

         Choose a1=1, a3=-a4 = 1 and then a4=0, a1=a3=1
         we see that (1,a2,1,-1,a5), (1,a2,1,0,a5) are linear independent
         of W1 for all scalars a2,a5 in F.
         Thus, {(1,1,1,-1,0), (1,0,1,-1,1),(1,1,1,0,0),(1,0,1,0,1)}
         is a set of 4 independent vectors.
         Also, note W1 is a proper subspace of F5 and hence dimW1 <=4.
        We claim that {(1,1,1,-1,0), (1,0,1,-1,1),(1,1,1,0,0),(1,0,1,0,1)}      
        a basis of W4 and hence Dim W1 = 4.
 
         For W2, since a2 = a3, a3 = a4 and a1 + a5 = 0.
         We see that {(s,t,t,t,-s)|s,t in F} is a general
         form of vectors in W2. 
         Clearly, dim W2 = 2 from this general form.
         By choosing a2=a3 = 1, a3=a4=1, and a1=a5 = 0.
         And, choose a2=a3=a4 =0, and a1= -a5=1.
         We obtain two independent vectors (1,1,1,0,0) and
         (1,0,0,0,-1) in W2. 
         Thus, {(1,1,1,0,0),(1,0,0,0,-1)} is a basis of W2 and
         dim W2 = 2.

 2. Note the dimension of M n x n (F) isn nxn = n^2.
    There are (1+2+3+..+n) = n(n+1)/2 free entries in
    an upper upper triangular of M n x n (F).
    Hence, dim W = n(n+1)/2.

         [Discussion of question 1:
         First of all , try to find the dimensions of W1 and W2.
         Since the restraint of W1 is the linear equation 
         a1 – a3 – a4 = 0, so dim W1 = 5 -1 = 4.
         For W2, there are three linear restraints, namely,  
         a2 = a3, a3 = a4 and a1 + a5 = 0
         if they are independent,then dim W2 = 5 -3 = 2. 
         Since, clearly,two of the 3 equations are independent,
         we see that dim W2 = 2 or 3. ]

 Kenny