Answer:
Yes, H is a subspace
Step-by-step explanation:
Recall that given a vector space V, a subset W of V is a subspace if and only if
- the 0 of V is in W
- given a, b in W, then a+b is in W
- given a real scalar r and a in W, then ra is in W.
In order to see if H is a subspace of [tex]M^{2\times 4}[/tex] we must check the three properties.
- It is clear that the matrix 0 in [tex]M^{2\times 4}[/tex] is in H since [tex]F0 = 0[/tex] (where the right hand 0 is the 0 vector in [tex]M^{3\times 4}[/tex].
- Let A,B in H. We want to check that A+B is in H. Since A,B in H we have that FA=0 and FB=0. We have that
[tex]F(A+B) = FA+FB = 0 +0 =0[/tex]
Then A+B is in H.
- given a real number r, and a matrix A in H, we want to check if rA is in H. Then
[tex] F(rA) = rFA = r0 = 0[/tex]
which shows that rA is in H.
Hence, H is a subspace of [tex] M^{2\times 4}[/tex]