Question:Every other subspace of V that contains v1,v2,...,vr must contain W?How?

**x3bnm** · May 16th 2011, 01:43 PM

I'm having trouble understanding the following proof.

If $v_1 , v_2 , ..., v_r$ are vectors in a vector space $V$ , then:
(a) The set $W$ of all linear combinations of $v_1 , v_2 , ..., v_r$ is a subspace of $V$ .
(b) $W$ is the smallest subspace of $V$ that contains $v_1 , v_2 , ..., v_r$ in the sense that every other subspace of $V$ that contains $v_1 , v_2 , ..., v_r$ must contain $W$ .

I understand how to prove (a).
Now I have questions regarding proof (b):

The proof of (b) is given like this:
Each vector $v_i$ is a linear combination of $v_1 , v_2 , ..., v_r$ since we can write
$v_i = 0v_1 + 0v_2+... + 1v_i+...+0v_r$

Therefore, the subspace $W$ contains each of the vectors $v_1 , v_2 , ..., v_r$ .

Let $W'$ be any other subspace that contains $v_1 , v_2 , ..., v_r$

Since $W'$ is closed under addition and scalar multiplication, it must contain all linear combinations of $v_1 , v_2 , ..., v_r$ . Thus $W'$ contains each vector of $W$

( My Question ) How does the author deduce that $W'$ contains all linear combinations of $v_1 , v_2 , ..., v_r$ in the last sentence of above proof?

How does the statement that " $W'$ is closed under addition and scalar multiplication" make the statement that " $W'$ contains all linear combinations of $v_1 , v_2 , ..., v_r$ '' true?

**Plato** · May 16th 2011, 01:57 PM

Originally Posted by x3bnm

How does the statement that " $W'$ is closed under addition and scalar multiplication" make the statement that " $W'$ contains all linear combinations of $v_1 , v_2 , ..., v_r$ '' true?

Well we know that $W'$ is a subspace.
It contains each $v_k$ so it contains any linear combination of those.

**x3bnm** · May 16th 2011, 02:04 PM

Originally Posted by Plato

Well we know that $W'$ is a subspace.
It contains each $v_k$ so it contains any linear combination of those.

First of all thanks for replying.

But still I'm not understanding. Sorry about that.

How does being a subspace that contains $v_k$ make the subspace to contain linear combination of those? I don't understand. Can you explain?

**Deveno** · May 16th 2011, 02:22 PM

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).

**x3bnm** · May 16th 2011, 02:33 PM

Originally Posted by Deveno

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).

Thank you Deveno for the explanation. I understand now.

**x3bnm** · May 16th 2011, 03:13 PM

Originally Posted by Deveno

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 $W$ is a subspace of $V$ if and only if:

a) If $u$ and $v$ are vectors in $W$ then $u + v$ is in $W$ .
b) If $k$ is any scalar and $u$ is any vector in $W$ , then $ku$ is in $W$ .

So in the above proof $v_1, v_2,...,v_3$ is in $W'$ :

$u = (a_1 * v_1) + (a_2 * v_2) + .... + (a_r * v_r)$
$v = (b_1 * v_1) + (b_2 * v_2) + .... + (b_r * v_r)$

so $u + v = (a_1 + b_1)v_1 + (a_2 + b_2)v_2 +...+(a_r + b_r)v_r$
which is in $W'$ .

Deveno did you mean this that $u + v$ is in $W'$ 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" ?

**Deveno** · May 16th 2011, 05:51 PM

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.

**x3bnm** · May 17th 2011, 01:31 PM

Originally Posted by Deveno

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.

Thank you Deveno.

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