Can you help me check whether what I did is right or wrong?
Here are the questions:
Which of the following sets are subspaces of $\mathbb{R}^n$?
(a) The plane of $\mathbb{R}^3$ that passes through (7,-3,2) and is
perpendicular to to the vector <1,1,-1>(b) The plane in $\mathbb{R}^3$ that passes through (2,1,3) and
perpendicular to the vector <1,1,-1>(c) The set of all vectors with in $\mathbb{R}^2$ with their tail at the
origin and their tip at the point on the curve $y = x^2$.
Here is my attempt:
(a) The equation of the plane is $1(x-7)+1(y+3)-1(z-2)= 0 \implies x+y-z-2 = 0 $.
The coordinates of the 0 vector are <0,0,0>. Therefore since 0+0-0-2 = 0 is impossible, (a) is not a subspace
(b) The equation of the plane is $1(x-2)+1(y-1)-(z-3)=0 \implies x+y-z = 0$.
Let the vector $u_1 = <1,-1,0>$ and $u_2 = <3,-3,0>$.
The coordinates of the zero vector <0,0,0> satisfy the equation of the plane
$u_1 + u_2 = <(1+3),(-1-3), (0+0)> = 0$.
$t(u_1) = t<1,-1,0> = 0$ and therefore (b) is a subspace.
Can you guys explain (c) to me and check whether (a) and (b) are correct?
Best Answer
To show that a set of vectors $V$ is a subspace of a vector space, it is sufficient and necessary to show that:
1. $0 \in V$
2. $V$ is closed under addition -- i.e. if $u,v \in V$, then $u+v\in V$.
3. $V$ is closed under scalar multiplication -- if $v \in V$ and $a$ is a scalar, then $av \in V$.
Your proof for (a) is correct, since you showed that 1. does not hold.
For (b) you showed that 1. holds. To show that 2. and 3. hold for (b), find two vectors $u,v$ in the plane which are linearly independent, i.e. such that every point in the plane can be written as $t_1 u + t_2 v$ for some scalars $t_1, t_2$. As it stands, you have not given two such vectors yet.
For (c) find an example where either 2. or 3. fails to hold.