I'm having trouble understanding the following proof.
If are vectors in a vector space , then:
(a) The set of all linear combinations of is a subspace of .
(b) is the smallest subspace of that contains in the sense that every other subspace of that contains must contain .
I understand how to prove (a).
Now I have questions regarding proof (b):
The proof of (b) is given like this:
Each vector is a linear combination of since we can write
Therefore, the subspace contains each of the vectors .
Let be any other subspace that contains
Since is closed under addition and scalar multiplication, it must contain all linear combinations of . Thus contains each vector of
( My Question ) How does the author deduce that contains all linear combinations of in the last sentence of above proof?
How does the statement that " is closed under addition and scalar multiplication" make the statement that " contains all linear combinations of '' true?
a non-empty subset W of V is a subspace if and only if for all w1,w2 in W, and a in F, w1+w2 and aw1 are in W.
these properties are called "closure of vector addition and scalar multiplication" and are enough to ensure that W is a
full-fledged vector space in its own right (using the same vector addition and scalar multiplication as in V).
so if {v1,v2,...,vr} are all in W, then so are the scalar multiples a1v1, a2v2,....arvr and thus also the sum:
a1v1 + a2v2 +....+ arvr, because of the closure properties listed above (properties any subspace of V has.
think about this: if W is to be a subspace, then it has to satisfy all of the vector space axioms, or else it
wouldn't be a subspace, just a subset).
Sorry for reopening this thread. Something is not clear to me.
According to definition of subspace is a subspace of if and only if:
a) If and are vectors in then is in .
b) If is any scalar and is any vector in , then is in .
So in the above proof is in :
so
which is in .
Deveno did you mean this that is in when you said "so if {v1,v2,...,vr} are all in W, then so are the scalar multiples a1v1, a2v2,....arvr and thus also the sum: a1v1 + a2v2 +....+ arvr, because of the closure properties listed above" ?
you just use the properties of closure repeatedly.
v1 and v2 are in W, so so are a1v1 and a2v2 (a1 and a2 are both scalars, and v1,v2 are both vectors in W). this is property (b) used twice.
since we have established that both a1v1 and a2v2 are in W, then a1v1 + a2v2 is in W, by property (a).
again, by property (b), since v3 is in W, a3v3 is in W. so then, by property (a) again:
a1v1 + a2v2 + a3v3 = (a1v1 + a2v2) + a3v3 is in W.
we can keep doing this, adding each term ajvj to:
a1v1 + a2v2 +...+ a(j-1)v(j-1) until we reach arvr.