1. The problem statement, all variables and given/known data for the 2x2 matrix [a 12;12 b] is it a vector space 2. Relevant equations 1. If u and v are objects in V, then u+v is in V 2. u+v = v + u 3. u+(v+w) = (u+v)+w 4. There is an object 0 in V, called a zero vector for V, such that 0+u = u+0 = u for all u in V 5. For each u in V, there is an object -u in V, called a negative of u, such that u+(-u) = (-u)+u = 0 6. If k is any scalar and u is any object in V, then ku is in V 7. k(u+v) = ku+kv 8. (k+m)u = ku+mu 9. k(mu) = (km)(u) 10. 1u = u 3. The attempt at a solution axioms 1,4,5,6 fail to hold. axiom 1: for any vectors u = [u1,12; 12, u2], v = [v1,12 ; 12 v2]in the space, u + v must also be in the space this is obviously false, since u + v = [u1+v1,12 + 12; 12+12, u2 + v2] = [u1+v1,24; 24, u2 + v2] which is not in the space axiom 4: let u = [u1,12; 12, u2], we need to find a 0 vector such that u + 0 = 0 + u = u but 0 must have the form [z1,12; 12, z2] to be in the space and so u + 0 = [u1,12; 12, u2] + [z1,12; 12, z2] = [u1+z1,12+12 ; 12+12, u2+z2] = [u1+z1, 24 ; 24, u2+z2] which is not equal to vector u, regardless of z1 and z2 values. axiom 5: can't have an additive inverse, if there's no 0 vector axiom 6: let u = [u1,12; 12, u2] according to this axiom, for any scalar k, k*u is also in the space but this is false for any k =/= 1: k*u = k*[u1,12; 12, u2] =[k*u1,k*12; k*12, k*u2] which is not in space unless k = 1 however I am not sure if I did this correctly and need confirmation.