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Let $W$ be the set of all linear combinations of $v_1,...,v_n$. Then $W$ is a subspace of $V$.

Proof. Let $y_1,...,y_n$ be numbers. Then

$(x_1 v_1+...+x_n v_n)+(y_1 v_1+...+y_n v_n)=(x_1+y_1)v_1+...+(x_n+y_n)v_n$.

Thus the sum of two elements of $W$ is again an element of $W$. Furthermore, if $c$ is a number, then

$c(x_1 v_1+...+x_n v_n)=cx_1 v_1+...+cx_n v_n$

is a linear combination of $v_1,...,v_n$, and hence is an element of $W$.

Finally,

$\mathbb{0}=0v_1+...+0v_n$

is an element of $W$. This proves that $W$ is a subspace of $V$.

My question is: How could writer say that'the sum of two elements of $W$ is again an element of $W$'? Also, How could writer say that $v$ is closed under scaalar multiplication. Can you explain clearly?

3 Answers 3

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Note that the proof is carelessly formulated. It starts "Let $y_1, \dots, y_n$ be numbers. Then $(x_1 v_1 + \dots + x_n v_n) + (y_1v_1 + \dots + y_n v_n)=(x_1+y_1) v_1+ \dots +(x_n+y_n)v_n$." What are the $x$'s? Also, it is prudent to point out that every element of $W$ is of the form $x_1 v_1 + \dots + x_n v_n$ for certain $x_1, \dots, x_n \in {\mathbb R}$.

But, let's start with what $W$ actually is; from your comments on the other answers, I guess this may be what you're actually struggling with. It is the subset of $V$ given by $$W = \{\alpha_1 v_1 + \dots \alpha_n v_n : \alpha_1, \dots, \alpha_n \in {\mathbb R}\}.$$

Now, to prove that the sum of two elements of $W$ is again an element of $W$, take two elements $u, v \in W$. This means, by definition of $W$, that there are $x_1, \dots, x_n \in {\mathbb R}$ and $y_1, \dots, y_n \in {\mathbb R}$ such that $$u = x_1 v_1 + \dots + x_n v_n \text{ and } v = y_1 v_1 + \dots + y_n v_n.$$ Then, as in your proof, $$u + v = (x_1 + y_1) v_1 + \dots + (x_n + y_n) v_n.$$ (So, at the risk of being too verbose, there do indeed exist $\alpha_1, \dots, \alpha_n \in {\mathbb R}$ such that $u + v = \alpha_1 v_1 + \dots + \alpha_n v_n$. Namely, $\alpha_1 = x_1 + y_1, \dots, \alpha_n = x_n + y_n$ will do.) Therefore, by definition of $W$, $u + v \in W$.

Now, can you formulate the proof that $W$ is closed under scalar multiplication yourself?

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    Yes I can. But, I still have a question: Assume $u,v$ in $W$. We wiil show that $u+v$ in $W$. And, how do you know $u+v$ in $W$. You said that it is by definition, what do the definition say? – 2017-02-26
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    It say if $u,v$ in $W$ then $u+v$ in $W$, i.e., I think, initialy we should assume $u,v$ in $W$. Later, we must show $u+v$ in $W$. But, I think we didn't show this. Can you convince me $u+v$ in $W$? – 2017-02-26
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    @Kahler By definition, $W$ is the subset of $V$ consisting of all linear combinations of $v_1, \dots v_n$. This is another way of saying that $W$ is defined to be $\{\alpha_1 v_1 + \dots + \alpha_n v_n : \alpha_1, \dots, \alpha_n \in {\mathbb R}\}$. – 2017-02-26
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    @Kahler All it should need to convince you that $u+v\in W$ is to echibit numbers $\alpha_1,\ldots, \alpha_n$ with $u+v=\alpha_1v_1+\ldots+\alpha_nv_n$. Apparently $\alpha_1:=x_1+y_1,\ldots, \alpha_n:=x_n+y_n$ does it. – 2017-02-26
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HINT: $W$ is the subset of elements of $V$ that are linear combinations of $v_1, v_2 \cdots v_n$. Now aren't $(x_1 + y_1)v_1 + (x_2 + y_2)v_2 + \cdots + (x_n + y_n)v_n$ and $(cx_1)v_1 + (cx_2)v_2 + \cdots (cx_n)v_n)$ linear combinations of $v_1, v_2 \cdots v_n$?

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    These are linear comb. of $v_1,...,v_n$ but how do you know $(x_1+y_1)v_1+...+(x_n+y_n)v_n$ is in $W$ and $cx_1 v_1+...+cx_n v_n$ is in $W$? – 2017-02-26
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    @Kahler W is the subset of ALL linear combination of the vectors and obviously $(x_1 + y_1)v_1 + (x_2 + y_2)v_2 + \cdots + (x_n + y_n)v_n$ and $(cx_1)v_1 + (cx_2)v_2 + \cdots (cx_n)v_n)$ are linear combinations of the vectors. – 2017-02-26
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    @Kahler If you're more familiar with fields you can see that V is a vectors space over a field $F$. Then we have that $W = \{ \sum c_iv_i | c_i \in F\}$. Obviously the closure of the fields under addition and multiplication gives us the wanted result. – 2017-02-26
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    Okey. So, what is the ALL linear comb. mean? What is the trick of ALL here? – 2017-02-26
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    @Kahler We have that if an element is of the form $c_1v_1 + c_2v_2 + \cdots c_nv_n$ for some $c_1,c_2,\cdots c_n$ are elements of the field $F$ then the vector is in W. That's the meaning of ALL linear combination, – 2017-02-26
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$W$ is the set of all linear combination of $v_1, v_2....v_n$.
Now $(x_1+y_1)v_1 + (x_2+y_2)v_2+......(x_n+y_n)v_n$ is also a linear combination of $v_1, v_2....v_n$.
$(cx_1)v_1+(cx_2)v_2+....(cx_n)v_n$ is also a linear combination of $v_1, v_2....v_n$.

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    Okey these are linear comb. of $v_1,...,v_n$ but how do you know $(x_1+y_1)v_1+...+(x_n+y_n)v_n$ is in $W$ and $cx_1 v_1+...+cx_n v_n$ is in $W$? – 2017-02-26
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    Because $W$ contains all linear combinations of $v_1,v_2.....v_n$ – 2017-02-26