Thread: Proving a group is a linear subspace

1. Proving a group is a linear subspace

$W = \{(\alpha_1,...,\alpha_n) \in R^n | \alpha_1 = 0\}$

Prove W is a linear subspace of $R^n$

I know that W is a subspace. I just don't understand the answer my book gives.

The first step is to prove that W is not empty - For example, the zero vector is in W, so W is not empty.

The second step is to prove that for every two vectors in W, their sums are also in W. Here's where I'm a bit confused by the books answer - here's what it says -
if a = $(\alpha_1,...,\alpha_n)$and b = $(\beta_1,...,\beta_n)$ are in W, so $\alpha_1 = \beta_1 = 0$.
a + b = $(\alpha_1 + \beta_1,...,\alpha_n +\beta_n)$
however, $\alpha_1 + \beta_1 = 0 + 0 = 0$
therefore a + b $\in$ W

And they did the same type of thing for the scalar step...

Why was it enough that they only checked the first elements of the group of vectors. How do they know, for example, that $\alpha_n + \beta_n \in W$???

2. First, I should probably point out that your title does not make sense. You want to show that a set is a linear subspace.

Your strategy is correct: you need to show that the set is closed under vector addition and scalar multiplication. That is, if $a,b\in W$, you want to show that $a+b\in W$, and etc... But what does it mean to be in $W$? It means that the first component is zero. Can you check that $a+b$ has zero as its first component?

3. Originally Posted by jayshizwiz
$W = \{(\alpha_1,...,\alpha_n) \in R^n | \alpha_1 = 0\}$

Prove W is a linear subspace of $R^n$

I know that W is a subspace. I just don't understand the answer my book gives.

The first step is to prove that W is not empty - For example, the zero vector is in W, so W is not empty.

The second step is to prove that for every two vectors in W, their sums are also in W. Here's where I'm a bit confused by the books answer - here's what it says -
if a = $(\alpha_1,...,\alpha_n)$and b = $(\beta_1,...,\beta_n)$ are in W, so $\alpha_1 = \beta_1 = 0$.
a + b = $(\alpha_1 + \beta_1,...,\alpha_n +\beta_n)$
however, $\alpha_1 + \beta_1 = 0 + 0 = 0$
therefore a + b $\in$ W

And they did the same type of thing for the scalar step...

Why was it enough that they only checked the first elements of the group of vectors. How do they know, for example, that $\alpha_n + \beta_n \in W$???
They don't say that and it doesn't even make sense. For all n, $\alpha_n$ and $\beta_n$ are NUMBERS, not vectors. Their sum is a number, not a member of W.

Because the condition for a vector to be in W is that the first number in the "ordered n-tuples" is a 0:
if v is in W, then v is of the form $(0, a_1, a_2, \cdot\cdot\cdot, a_n)$. The sum of two such things is of the form $(0, a_1, a_2,\cdot\cdot\cdot, a_n)+ (0, b_1, b_2, \cdot\cdot\cdot, b_n)= (0, a_1+ b_1, a_2+ b_2, \cdot\cdot\cdot, a_n+ b_n)$, again an "ordered n-tuple" in which the first number is 0 and so a member of W.

4. Thanks again. I always get confused whether we're dealing with numbers or vectors for some reason.

First, I should probably point out that your title does not make sense. You want to show that a set is a linear subspace.
Sorry, I'm trying my best to explain myself. I don't study in English so I don't know the exact phrases sometimes. Well, as long as you understand me...