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## Homework Statement

for the 2x2 matrix [a 12;12 b] is it a vector space

## Homework 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

## 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.

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