Proof of homogenous

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mahjk17

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How should I prove this problem:

"Prove that the set of all solutions x of the linear systems Ax = b forms a subspace if and only if the system is homogenous."

I proved it directly, will this attempt suffice:

Suppose x1 and x2 are solutions; we need to show that c1x1 + c2x2 is also a solution. Because x1 is a solution, Ax1 = 0. Similarly, Ax2 = 0. Then for any scalars c1; c2, A(c1x1 + c2x2) = c1Ax1 + c2Ax2 = c10 + c20 = 0: So c1x1 + c2x2 is also a solution.
 
Your proof is fine. Of course a subspace must be non-empty, which you did not argue (but is trivial to show--why?).

Note though, that is is an if and only if. The above shows that if b=0, then {x| Ax=b} is a vector subspace (this is sometimes called the "kernel", "nullspace", or the "0-eigenspace" for A - a very important subspace). Now you must show that if {x|Ax=b} is a subspace, then b=0.
 
Daon2, I am not familiar with the nonempty set and for the second part of your statement, that is what I was having trouble with; trying to show that if Ax=b is a subspace then the only solution of b is b=0, but I don't know where to start from that.
 
It is part of the definition of being a subspace! Simply show that the zero vector belongs to the set.

Second, if S = {x|Ax=b} is a subspace, then x belongs to S implies cx belongs to S for all scalars c. In particular, pick any non-zero c. Then Ax = b, and A(cx) = b. But what is A(cx)?
 
b = A(cx) = cAx = cb. Pick c=-1 for instance. This says b = -b, or 2b = 0 => b=0.

An easier way which I overlooked: Subspaces must contain the zero vector. So 0 belongs to S => 0 = A0 = b.
 
Some texts, instead of saying "non-empty, closed under scalar multiplication, and closed under addition" as the definition of a subspace, say "contains the 0 vector, closed under scalar multiplication, and closed under addition".

Of course, those are the same. Obviously, if it contains the 0 vector it is non-empty. If a set is non-empty, it contains some vector, v. Since it is closed under scalar multiplication, it contains -v and, then, since it is closed under addition, it contains v+(-v)= 0.
 
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