I have this problem.

Let $X\subseteq R^3$ That defined by : $$X=\{(x_1,x_2,x_3)|x_1^2+x_2^2-x_3^2=1\}$$

Find $Sp(X)$, dim(X), and find basis for $X$

My solution :

Let $x_1,x_2 \in R$

$$x_1^2+x_2^2-x_3^2=1 \implies x_1^2+x_2^2-1=x_3^2$$

$$[B]_u= (x_1^2,x_2^2,x_1^2+x_2^2-1)$$ $$dim(X)=2$$ $$X = Sp\{(x_1^2,x_2^2,x_1^2+x_2^2-1)\} = Sp\{x_1^2(1,0,1-1)+x_2^2(0,1,1-1)\}=Sp(\{(1,0,0),(0,1,0)\}$$

But $Sp(\{(1,0,0),(0,1,0)\}$ is incorrect since (for exmaple) $(2,0,0) \notin X$

I pretty sure the rest is correct, Any idea how to get the span of X?

Thank you!


The set $X$ that you've described is not a subspace of $\mathbb R^3$, so it doesn't have a basis. For instance, $(1, 0, 0) \in X$, but $(2, 0, 0) \notin X$, showing that it's not closed under scalar multiplication.

On the other hand, we have $(1,0,0) \in X$, $(0, 1, 0) \in X$ and $(1, 1, 1) \in X$, from which we can conclude that $(0,0,1)$ is in the span of $X$, hence all of $\mathbb R^3$ is in $span(X)$, so $span(X) = \mathbb R^3$.

Because $X$ is not a subspace, it does not have a dimension (as a subspace). As a manifold, it has dimension 2. But I doubt that this is relevant to your question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.