Definitions

To say that P is closed under addition means that
the vector sum of any two points of P is in P.
 To say that P is closed under scalar
multiplication means that the scalar product of any
number (including zero) with any point of P is in P.
 To say that P is a subspace means that P satisfies
both properties:
 a subspace of a vector space is a subset that is
closed under both addition and scalar multiplication.

First we prove that
(a) P1={(x1, x2)^T| x1+x2=0} is a subspace of R^2
For any y=(y1, y2)^T, z=(z1,z2)^T in P1, we have
y1+y2=0,
z1+z2=0 Thus
(y1+z1)+(y2+z2)=0, Thus y+ z=(y1+z1, y2+z2) is in P1.
For any scalar alpha and any x=(x1, x2)^T in P1, we need to show that
alpha * x= (alpha *x1, alpha*x2) is in P1. We know that
alpha*x1+alpha*x2=alpha*(x1+x2)=0, since x1+x2=0. Thus
alpha * x is in P1.
Since x1, x2 are real number, P1 is a subset of R^2.
Thus according to the above definition. thus P1 is a subspace of R^2.
Now we prove that
( c) P3={(x1, x2)^T| x1=3*x2} is a subspace of R^2
For any y=(y1, y2)^T, z=(z1,z2)^T in P3, we have
y1=3*y2, z1=3*z2, thus
(y1+z1)=3*(y2+z2), i.e. y+z is in P3
For any scalar alpha and any x=(x1, x2)^T in P3, we need to show that
alpha * x is in P3. We know that
alpha*x1=alpha*(3*x2)=3* (alpha*x2), thus
alpha * x is in P3.
Also P3 is a subset of R^2, the proof is similar to that of P1. Thus P3 is
also a subspace of R^2.
The proofs for (b), (d), (e) are similar. I let you do so that you can learn
more.
Note:
T stands for transpose. i.e. (x1, x2)^T=
x1
x2
with big open ’ (‘ on the left and big close ‘) ‘on the right.