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Is the following set a subspace of $\Bbb R^{n}$? If it is, find the base and dimension. $A=\{(x_1,x_2,...,x_n): x_i\in\Bbb Z, \forall i\in \{1,...n\}\}$

This is what I did, I'm not sure it's right:

To check whether some set is a subspace of another set, we should check that the closure property holds. So $\forall a,b \in A$ it must be true that $\alpha a+\beta b \in A$.

Let $a=(x_1,x_2)$ and $b=(x_1,x_2,x_3)$.

$\alpha a+\beta b\in A\iff \alpha(x_1,x_2)+\beta(x_1,x_2,x_3) \in A\iff (\alpha x_2,\alpha x_2)+(\beta x_1, \beta x_2, \beta x_3) \in A\iff (\alpha x_1+\beta x_1, \alpha x_2+\beta x_2, \beta x_3) \in A$

How do I now show that this new vector I got is or isn't in $A$? And what does $\Bbb R^{n}$ represent?

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  • $\begingroup$ Is $e_1 / 2$ in $A$? $\endgroup$ – user251257 Oct 31 '16 at 12:18
  • $\begingroup$ What does that represent? $\endgroup$ – lmc Oct 31 '16 at 12:23
  • $\begingroup$ $e_1 = (1,0,0,\dotsc,0)\in A$. $\endgroup$ – user251257 Oct 31 '16 at 12:24
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To be subspace of $\mathbb{R^n}$, you need to show that $ax-y\in A$ for $x,y\in A$ and $a\in \mathbb{R}$ (NOT in $A$)

Take $(1,0,\dots,0)\in A$, $0.5\in \mathbb{R}$.
Clearly, $0.5(1,0,\dots,0)=(0.5,0,\dots,0)\not\in A$ since the first entry of the element is not in $\mathbb{Z}$.

Hence $A$ is not a subspace of $\mathbb{R}^n$.

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$\mathbb R^n$ is the set of all n-tuples $(x_1,...,x_n)$ with $x_i \in \mathbb R.$

We have $x:=(1,1,...,1) \in A$. Is $\frac{1}{2}x \in A$ ?

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  • $\begingroup$ It's not? What I did above makes no sense then? $\endgroup$ – lmc Oct 31 '16 at 12:43
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Although you have quantified $a,b$ in your expression $\alpha a + \beta b$, you have failed to quantify $\alpha,\beta$.

What you must show is that $\forall a,b \in A$ and $\forall \alpha,\beta \in \mathbb{R}$ it must be true that $\alpha a + \beta b \in A$.

And if you attempt to do that, you will easily find counterexamples by careful choice of $\alpha,\beta$ (as in the other answers).

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