- #1

mathmari

Gold Member

MHB

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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:

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)

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:

- We have that the set is nonempty, since $(0,0)\in X_1$.
- 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 $(x_1, y_1)+(x_2, y_2)\in X_1$. - 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 $\alpha\cdot (x_1, y_1)\in X_1$.

- We have that the set is nonempty, since $(1,0)\in X_2$.
- 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 $(x_1, y_1)+(x_2, y_2)\notin X_2$.

- 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:- We have that the set is nonempty, since $(0,0)\in X_3$.
- W have for $\alpha \in \mathbb{R}$ that $\alpha\cdot (0,0)=(\alpha \cdot 0, \alpha \cdot 0)=(0,0)$. Therefore $\alpha\cdot (0,0)\in X_3$.

- We have that the set is nonempty, since $(0,0)\in X_4$.
- 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 $(x_1, y_1)+(x_2, y_2)\notin X_4$.

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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