Check my proof for quality (Inner Product Space / Subspaces)

[RJLiberator]

Homework Statement

S = a non-empty set of vecotrs in V
S' = set of all vectors in V which are orthogonal to every vector in S
Show S' = subspace of V

Homework Equations

Subspace requirements.
1. 0 vector is there
2. Closure under addition
3. Closure under scalar multiplication

The Attempt at a Solution

1. 0 is held within S' by definition here
2. Let v and v' exist within S' and let s exist within S.
<v+v',s>=<v,s>+<v',s> = 0+0 =0 which is what we want since
<v,s>=0 and <v',s>=0.
3. if x is a scalar, then <xv,s> = x<v,s> =x*0 = 0

therefore we have satisfied all the necessary requirements and have shown that S' is a subspace of V.

Question: In my notes, the instructor lists a fourth requirement, namely If v exists within S, S a subspace, then -v exists within s.
Must I prove this as well?

 

[Mark44]

Homework Statement

S = a non-empty set of vecotrs in V
S' = set of all vectors in V which are orthogonal to every vector in S
Show S' = subspace of V

Homework Equations

Subspace requirements.
1. 0 vector is there
2. Closure under addition
3. Closure under scalar multiplication

The Attempt at a Solution

1. 0 is held within S' by definition here

I wouldn't say "by definition". The 0 vector is orthogonal to every vector in S, so $0 \in S'$.
Minor quibble -- "is an element of S'" ($\in S'$) is used more commonly than "is held within".

2. Let v and v' exist within S' and let s exist within S.
<v+v',s>=<v,s>+<v',s> = 0+0 =0 which is what we want since
<v,s>=0 and <v',s>=0.

The conclusion here is that if $v, v' \in S'$, then $v + v' \in S'$. This shows closure of addition.

3. if x is a scalar, then <xv,s> = x<v,s> =x*0 = 0

The above shows that if $v \in S'$ and x is a scalar, the $xv \in S'$, showing closure of scalar multiplication.

therefore we have satisfied all the necessary requirements and have shown that S' is a subspace of V.

Question: In my notes, the instructor lists a fourth requirement, namely If v exists within S, S a subspace, then -v exists within s.
Must I prove this as well?

-v = (-1) * v, right? If you have already shown that the set is closed under scalar multiplication, then that also includes the case where the scalar is -1.

 

[RJLiberator]

Excellent. It looks good with a few minor touch ups.

Thank you for your check and analysis of my writing. It is always helpful.

I see the importance in saying "this shows closure under ____" as that is what we set out to prove.

Thank you.