# Describe bases for the span of sets

• MHB
• mathmari

#### mathmari

Gold Member
MHB
Hey!

We have the subset $X_i$ of $\mathbb{R}^2$:
$$X_1 := \{(x,y) \in \mathbb{R}^2 : x + y = 0\}; \\ X_2 := \{(x,y) \in \mathbb{R}^2 : x + y = 1\} \\ X3 := \{(x,y) \in\mathbb{R}^2 : x^2 + y^2 = 0\}; \\ X4 := \{(x,y) \in \mathbb{R}^2 : x^2- y^2 = 0\}$$ We want to check which of these sets $X_i$ are linear subspace of the $\mathbb{R}$-vector space $\mathbb{R}^2$ and to describe bases for the span $\langle X_i\rangle$.

I have done the following:
1. We have that the set is nonempty, since $(0,0)\in X_1$.
2. Let $(x_1, y_1), (x_2, y_2)\in X_1$. Then we have that $x_1+y_1=0$ and $x_2+y_2=0$.

Then we get that $(x_1, y_1)+(x_2, y_2)=(x_1+ x_2, y_1+y_2)$: $$(x_1+x_2)+(y_1+y_2)=(x_1+y_1)+(x_2+y_2)=0+0=0$$ Therefore $(x_1, y_1)+(x_2, y_2)\in X_1$.
3. Let $(x_1, y_1)\in X_1$. Then we have that $x_1+y_1=0$.

Then we get for $\alpha \in \mathbb{R}$ that $\alpha\cdot (x_1, y_1)=(\alpha x_1, \alpha y_1)$: $$\alpha x_1+\alpha y_1=\alpha \cdot (x_1+y_1)=\alpha \cdot 0=0$$ Therefore $\alpha\cdot (x_1, y_1)\in X_1$.
So, $X_1$ is a linear subspace of $\mathbb{R}^2$.
1. We have that the set is nonempty, since $(1,0)\in X_2$.
2. Let $(x_1, y_1), (x_2, y_2)\in X_2$. Then we have that $x_1+y_1=1$ and $x_2+y_2=1$.

Then we get that $(x_1, y_1)+(x_2, y_2)=(x_1+ x_2, y_1+y_2)$: $$(x_1+x_2)+(y_1+y_2)=(x_1+y_1)+(x_2+y_2)=1+1=2\neq 1$$ Therefore $(x_1, y_1)+(x_2, y_2)\notin X_2$.
So, $X_2$ is not a linear subspace of $\mathbb{R}^2$.
• We have that $X_3=\{(x,y) \mid \mathbb{R}^2 : x^2 + y^2 = 0\}=\{(x,y) \mid \mathbb{R}^2 : x=y= 0\}=\{(0,0)\}$.
Therefore, we get the following:
1. We have that the set is nonempty, since $(0,0)\in X_3$.
2. W have for $\alpha \in \mathbb{R}$ that $\alpha\cdot (0,0)=(\alpha \cdot 0, \alpha \cdot 0)=(0,0)$. Therefore $\alpha\cdot (0,0)\in X_3$.
So, $X_3$ is a linear subspace of $\mathbb{R}^2$.
1. We have that the set is nonempty, since $(0,0)\in X_4$.
2. Let $(x_1, y_1), (x_2, y_2)\in X_2$. Then we have that $x_1=\pm y_1$ and $x_2=\pm y_2$.

Then we get that $(x_1, y_1)+(x_2, y_2)=(x_1+ x_2, y_1+y_2)$: $$(x_1+x_2)^2-(y_1+y_2)^2=x_1^2+2x_1x_2+x_2^2-y_1^2-2y_1y_2-y_2^2 \ \overset{x_1^2-y_1^2=x_2^2-y_2^2=0}{=} \ 2x_1x_2-2y_1y_2$$ If $x_1=-y_1$ and $x_2=y_2$ we get $-4y_1y_2$ which is not necessarily equal to $0$. Therefore $(x_1, y_1)+(x_2, y_2)\notin X_4$.
So, $X_4$ is not a linear subspace of $\mathbb{R}^2$.

Is everything correct? Could I improve something? (Wondering)

Could you give me a hint how we can describe bases for the span $\langle X_i\rangle$ ? (Wondering)

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Is everything correct? Could I improve something?

Hi mathmari!

It looks fine to me. (Nod)
Shouldn't we have closure for addition for $X_3$ though?

Could you give me a hint how we can describe bases for the span $\langle X_i\rangle$ ?

Can we find a non-zero vector in each of the linear subspaces? (Wondering)

It looks fine to me. (Nod)
Shouldn't we have closure for addition for $X_3$ though?

To show the closure of addition do we take twice the element $(0,0)$. i.e. do we do the following?
We have that $(0,0)+(0,0)=(0,0)\in X_3$.

(Wondering)

Can we find a non-zero vector in each of the linear subspaces? (Wondering)

Do we have to find a basis only for these sets that we have shown that they are a linear subspace of $\mathbb{R}^2$ ?

If we consider all the sets, do we take the following non-zero vector?

$$X_1 \ : (x, y)=(x, -x)=x\cdot (1,-1); \\ X_2\ :\ (x,y)=(x,1-x)=(0,1)+x\cdot (1,-1) \\ X3 \ :\ (0,0);$$ Is everything correct? What about $X_4$ ? (Wondering)

To show the closure of addition do we take twice the element $(0,0)$. i.e. do we do the following?
We have that $(0,0)+(0,0)=(0,0)\in X_3$.

Yep. (Nod)

Do we have to find a basis only for these sets that we have shown that they are a linear subspace of $\mathbb{R}^2$ ?

If it's not a linear subspace no basis will do, so indeed, we only look at the linear subspaces. (Nerd)

If we consider all the sets, do we take the following non-zero vector?

$$X_1 \ : (x, y)=(x, -x)=x\cdot (1,-1); \\ X_2\ :\ (x,y)=(x,1-x)=(0,1)+x\cdot (1,-1) \\ X3 \ :\ (0,0);$$ Is everything correct? What about $X_4$ ? (Wondering)

$X_2$ and $X_4$ are not linear subspaces, so no basis can do the job. That is, if we try to span the subspace with the basis we will fail.
We found a non-zero vector in $X_1$. Good! Does it span the whole space?
The vector we have in $X_3$ is not a non-zero vector, so it's not part of a basis. Instead we have a subspace that contains only the zero vector. The corresponding basis is the empty basis. (Nerd)

If it's not a linear subspace no basis will do, so indeed, we only look at the linear subspaces. (Nerd)

Doesn't it hold that the span of every non-empty set is a subspace? (Wondering)

So, every $\langle X_i\rangle$ for $i=1,2,3,4$ is a linear subspace and so it has a basis, or not? (Wondering)

We have the following:
\begin{align*}X_1 :&= \{(x,y) \in \mathbb{R}^2 : x + y = 0\}=\{(x,-x)\mid x\in \mathbb{R}\}=\{x\cdot (1,-1)\mid x\in \mathbb{R}\} \\ X_2 :&= \{(x,y) \in \mathbb{R}^2 : x + y = 1\}=\{(x, 1-x)\mid x\in \mathbb{R}\}=\{(0,1)+x\cdot (1, -1)\mid x\in \mathbb{R}\} \\
X3 :&= \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 0\}=\{(0,0)\} \\ X4 :&= \{(x,y) \in \mathbb{R}^2 : x^2- y^2 = 0\}\end{align*}

But how are the spans of these sets look like? (Wondering)

Doesn't it hold that the span of every non-empty set is a subspace?

Yes. (Nod)

So, every $\langle X_i\rangle$ for $i=1,2,3,4$ is a linear subspace and so it has a basis, or not?

Yes.

We have the following:
\begin{align*}X_1 :&= \{(x,y) \in \mathbb{R}^2 : x + y = 0\}=\{(x,-x)\mid x\in \mathbb{R}\}=\{x\cdot (1,-1)\mid x\in \mathbb{R}\} \\ X_2 :&= \{(x,y) \in \mathbb{R}^2 : x + y = 1\}=\{(x, 1-x)\mid x\in \mathbb{R}\}=\{(0,1)+x\cdot (1, -1)\mid x\in \mathbb{R}\} \\
X3 :&= \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 = 0\}=\{(0,0)\} \\ X4 :&= \{(x,y) \in \mathbb{R}^2 : x^2- y^2 = 0\}\end{align*}

But how are the spans of these sets look like? (Wondering)

$X_1$ is a linear subspace with basis $\{(1,-1)\}$.
The span is the set of all linear combinations of its vectors, which in this case is just $X_1$.

$X_2$ has (0,1) and (1,0) in it. The set of all linear combinations of those 2 vectors is $\mathbb R^2$.
Btw, $X_2$ is a so called affine space.

$X_3$ is a trivial subspace as it contains only the zero vector.

$X_4$ has (1,1) and (1,-1) in it. So just like $X_2$ its span is $\mathbb R^2$. (Thinking)

$X_1$ is a linear subspace with basis $\{(1,-1)\}$.
The span is the set of all linear combinations of its vectors, which in this case is just $X_1$.

$X_2$ has (0,1) and (1,0) in it. The set of all linear combinations of those 2 vectors is $\mathbb R^2$.
Btw, $X_2$ is a so called affine space.

$X_3$ is a trivial subspace as it contains only the zero vector.

$X_4$ has (1,1) and (1,-1) in it. So just like $X_2$ its span is $\mathbb R^2$. (Thinking)

So, do we have the following?

A basis of $\langle X_1\rangle$ is $\{(1,-1)\}$.

A basis of $\langle X_2\rangle$ is $\{(1,0), (0,1)\}$.

The basis of $\langle X_3\rangle$ is the empty set.

A basis of $\langle X_4\rangle$ is $\{(1,0), (0,1)\}$.

(Wondering)

Yep. (Nod)

Yep. (Nod)

Ok! Thank you! (Yes)
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