Show that origin in R^n is the n-tuple 0=(0,0, .,0)

[woundedtiger4]

Dear all,

This definition/property was given as a remark in my lecture notes. And I have attempted to prove it (I have attached the picture of proof). Can you please tell me if I am correct or wrong (& possibly where am I wrong?)

Thanks in advance.

 

[HallsofIvy]

I'm not sure I understand what you are attempting to prove. You say "show that the origin in Rⁿ is ..." What exactly do you mean by "the orgin"? I would have defined that as "(0, 0, 0, ..., 0)"! But it looks like what you are really trying to prove is that "the additive identity" of the vector space Rⁿ." The defining property of the additive identity, 0, is that v+ 0= v for any vector v. Since a vector in Rⁿ is of the form (x_1, x_2, ..., x_n) and addition is defined "coordinate-wise", yes, you only need observe that (x_1, x_2, ..., x_n)+ (0, 0, ..., 0)= (x_1+ 0, x_2+ 0, ..., x_n+ 0)= (x_1, x_2, ..., x_n).

 

[Fredrik]

First of all, I second everything that HallsofIvy said.

Second, that photo is not a good enough way to show us your work. Some of your handwriting is impossible to understand. I will give it a try though.

"Suppose that $x_i\in\mathbb R^n$, $\forall i=1,\dots,n$". It wasn't until I read the next few lines several times that I understood that the comma between the x and the $\in$ is actually an $i$. You don't need n n-tuples. You just need one.

Then you wrote "Let $p,{}^\circ\! V\in i$ such that $p<{}^\circ\! V$". I'm guessing that you meant something like "let $j,k\in\{1,\dots,n\}$ be such that j<k". It's OK to use the symbol p as an index, but I don't understand why you would use a notation that looks like ${}^\circ\! V$.

"If $x_p+0_p=x_p$ (...something...) then $x_{{}^\circ\! V}+0_{{}^\circ\! V}=x_{{}^\circ\! V}$". So if the pth n-tuple has a certain property, then the ${}^\circ\! V$th n-tuple must have a similar property? Why?

"$0_p\in 0_i$" Are you really saying that the pth component of the zero n-tuple is an element of the ith component of the zero n-tuple?

OK, I'm going to stop here. I think it's a very good idea to post attempted proofs to get input on how you write them. But you need to put in a much bigger effort when you do so. You need to type the problem statement and your solution. Handwritten notes are not OK, especially not when they're this hard to read.

 

[Mark44]

I second Fredrik's comment about posting photos of your work. Please enter your work as text or use LaTeX to format it.

 

[WWGD]

I am not sure I get the question either, but you can see at (0,0,..,0) as the only 0-dimensional subspace. Show any 0-dimensional subspace must coincide with (0,0,..,0).

 

[Fredrik]

I am not sure I get the question either, but you can see at (0,0,..,0) as the only 0-dimensional subspace. Show any 0-dimensional subspace must coincide with (0,0,..,0).

I assume that what you mean by that is that any 0-dimensional subspace must be the singleton set {(0,0,...,0)}. For this to be a good approach, you need a definition of "0-dimensional subspace" other than "contains (0,0,...,0) and nothing else". I would suggest "contains the additive identity and doesn't contain any linearly independent subsets". This takes us back to doing what Halls suggested, and also proving that every singleton subset of $\mathbb R^n$ other than {(0,0,...,0)} is linearly independent. This stuff about linear independence looks like an unnecessary complication. I would just do what Halls suggested.

 

[woundedtiger4]

I assume that what you mean by that is that any 0-dimensional subspace must be the singleton set {(0,0,...,0)}. For this to be a good approach, you need a definition of "0-dimensional subspace" other than "contains (0,0,...,0) and nothing else". I would suggest "contains the additive identity and doesn't contain any linearly independent subsets". This takes us back to doing what Halls suggested, and also proving that every singleton subset of $\mathbb R^n$ other than {(0,0,...,0)} is linearly independent. This stuff about linear independence looks like an unnecessary complication. I would just do what Halls suggested.

Thanks a lot