If z is neither real nor pure imaginary show that z is a ba

If z is neither real nor pure imaginary, show that {z, } is a basis of C. (b) If z is n s. In each case use Theorem 4 to decide if S is a basis of V (a) r = M22; 1 11 10 11 [0 01 [0 0 1 1, 11 1], 11 1, 101

Solution

5) In order for S to be basis for M22, all the matrices M22 must be independent

aM1 + bM2 + cM3 + dM4 = [0 0;0 0]

a[1 1; 1 1] + b[0 1;1 1] + c[0 0;1 1] + d[0 0;0 1] = [0 0; 0 0]

a = 0

a + b = 0

Hence b must be zero

a + b + c = 0 and (a+b+c+d) = 0

Therefore, the only solution will be (a=b=c=d=0)

Hence the set of matrices form basis for M22