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Let $U$ be a proper subspace of a finite-dimensional vector space $V$ . Find a basis for $V$ containing no element of $U$.

We have no answer key for this question which i find annoying since that is the case for many questions in uni.

But if we look at a $R^2$ with basis vectors {$e_1, e_2$} and for example $U$ spans the vector $(1, 1)$. I get that this is a much simpler case but shouldn't it be the same for all other vector spaces of dimensions $n$? I struggle a lot with formally proving/showing that something is true and I'm not sure how to think when looking at arbitrary vector spaces.

Also I saw somewhere an explanation that said that: Let {$u_1, u_2, ... , u_k$} be a basis for $U$. Because $U$ is a subspace of $V$ {$u_1, u_2, ... , u_k$} are also in $V$. By adding linearly independent vectors from $V$ to the set of basis vectors of $U$ we eventually get a basis for $V$ {$u_1, u_2, ... , u_k, v_{k+1}, ... , v_n$}. And by removing all vectors that are in $U$ we get a basis {$v_{k+1}, ... , v_n$} for $V$ that contains no elements of $U$.

This I don't get however. Because how can they just remove the basis vectors of the subspace $U$? Wouldn't there be too few vectors to form the basis for $V$ after removing those? If we go back to the simpler case using the above strategy $U$ is spanned by {$u_1$} and $V$ = $R^2$. Adding one linearly independent vector from $V$ to the set gets us a basis for $V$ {$u_1, v_2$} and after removing $u_1$ we only have {$v_2$} left which is not a basis for $V$.

Am I wrong thinking this way? Can someone please explain.

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    $\begingroup$ That explanation looks like something ChatGPT would dream up. Is that where you got it? Regardless, yes, that argument is very wrong. $\endgroup$
    – JonathanZ
    Feb 22 at 22:40
  • $\begingroup$ In the future, ideally right now, identify exactly where you found the material you are attempting to quote, so that (i) we can verify that you are reporting its content accurately; and (ii) if it is incorrect, we all know that is not a reliable source of information. $\endgroup$ Feb 22 at 22:52
  • $\begingroup$ I'm sorry this was my first post here so I'm not quite accustomed to exactly what to include and what not to. But thank you for correcting me I will definitely do that next time. The explanation in the post someone gave me on Discord and I don't know how I would link this. I didin't identify it as a ChatGPT explanation and this is probably due to my inexperience in this field of math, but since the source isn't really something I would call reliable it definitely could be. I just brought it up since maybe it was correct and I missed something. Again I'm sorry for not making everything clear. $\endgroup$ Feb 22 at 23:03
  • $\begingroup$ Well, as you can tell, a lot of us here have been burned by ChatGPT stuff, and are all kinds of tired of it. But yeah, I don't think a link to a Discord would have been worth including. Although if they told you they got it from ChatGPT, you should be aware that ChatGPT stuff isn't permitted here. (We're still in the process of adding a banner that makes this clear to new users.) $\endgroup$
    – JonathanZ
    Feb 23 at 1:15
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    $\begingroup$ I'm not sure if it was ChatGPT since I wasn't explicitly told so, but it might have been and I understand if that is obvious and a nuisance for answerers so I will make a bigger effort to verify such claims in the future. But I have used it in the past myself to try and get questions answered, so thank you again I will stay far away from it from now on. I did not really understand what kind of place Math.StackExchange was but I just read up on guidelines regarding how to ask questions and some other things on the meta page and will take those into consideration when asking questions. $\endgroup$ Feb 23 at 1:30

2 Answers 2

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What you "saw somewhere" is of course incorrect, as $v_{k+1},\ldots,v_n$ cannot be a basis for $V$ if you had a basis for $V$ with $n$ elements; this one just has $n-k$ elements.

Instead, let us consider the following: if $v_1,\ldots,v_n$ form a basis for $V$, and $1\lt i\leq n$, then $$v_1+v_i, v_2+v_i,\ldots, v_{i-1}+v_i, v_i, v_{i+1},\ldots,v_n$$ is still a basis. Why? Well, clearly it spans! If $\mathbf{x}\in V$, then we can write $$\mathbf{x} = \alpha_1v_1+\cdots +\alpha_nv_n.$$ Then $$\begin{align*} \mathbf{x} &= \alpha_1v_1+\cdots + \alpha_n v_n\\ &= \alpha_1(v_1+v_i) - \alpha_1v_i + \alpha_2(v_2+v_i) - \alpha_2v_i + \cdots \\&\qquad \mathop{+}\alpha_{i-1}(v_{i-1}+v_i) - \alpha_{i-1}v_i + \alpha_iv_i + \cdots + \alpha_n v_n\\ &= \alpha_1(v_1+v_i) + \alpha_2(v_2+v_i) + \cdots + \alpha_{i-1}(v_{i-1}+v_i)\\ &\qquad \mathop{+} (\alpha_i-(\alpha_1+\cdots+\alpha_{i-1}))v_i + \alpha_{i+1}v_{i+1} + \cdots + \alpha_n v_n. \end{align*}$$ Since it is a spanning set of vectors with exactly $n$ vectors, it is a basis.

Now, let $u_1,\ldots,u_k$ be a basis for $U$. Extend to a basis $u_1,\ldots,u_k,v_{k+1},\ldots,v_n$ of $V$. Note that $v_{k+1}\notin U$, and that $k+1\leq n$. Consider $$u_1+v_{k+1}, u_2+v_{k+1},\ldots, u_k+v_{k+1},v_{k+1},\ldots,v_n.$$ This is a basis. We know for sure that $v_{k+1},\ldots,v_n$ are not in $U$. Show that $u_i+v_{k+1}$, with $i=1,\ldots,k$, is also not in $U$.

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  • $\begingroup$ Okay! So since {$u_1, ... , u_k$} is a basis for $U$ meaning that they are the maximum amount of vectors that are linearly independent in the subspace $U$, adding $v_{k+1}$ to $u_i$, $i = 1, ... , k$, changes all of those basis vectors to no longer be in $U$, since $v_{k+1}$ is linearly independent from them. $\endgroup$ Feb 22 at 23:42
  • $\begingroup$ Or there are no linear combinations of {$u_1, ... ,u_k$} that result in any of the vectors in {$u_1+v_{k+1}, ... ,u_k+v_{k+1}$}. That makes a lot of sense thank you so much! I don't know if what I just wrote suffices to show $u_i+v_{k+1}$, with $i=1,...,k$, is not in U but I understand. $\endgroup$ Feb 22 at 23:48
  • $\begingroup$ @JonathanChang I don't think any of that is correct. In fact, there are certainly linear combinations of those vector that are in $U$. Instead, note that for each $j$, $u_j$ is in $U$, and $v_{k+1}$ is not in $U$. Can the sum of a fector that is in $U$ and one that is not in $U$ result in a vector i. $U$. $\endgroup$ Feb 23 at 1:50
  • $\begingroup$ Oh you're right about the linear combinations. I think what I was trying to say was that none of {$u_1+v_{k+1}, ... , u_k+v_{k+1}$} lie on any linear combination of {$u_1, ... , u_k$}. Hence, it is a basis for $V$ containing no element of $U$. Could you use the argument that it is because we are adding a linearly independent vector to all of the basis vectors of $U$ or adding another dimension to it which in turn would make the new set {$u_1+v_{k+1}, ... , u_k+v_{k+1}$} not in $U$? $\endgroup$ Feb 23 at 14:53
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    $\begingroup$ @JonathanChang Again, I don't understand what you are trying to argue. This is really simple: if $u\in U$ and $v\notin U$, then $u+v\notin U$. Because if $u+v\in U$, then because $U$ is a subspace, $(u+v)-u=v\in U$, but $v\notin U$. $\endgroup$ Feb 23 at 15:03
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By removing elements of $U$ you of course no länger have a basis for the whole space, you reduce the dimension.

You can extend the basis in the way you described by vectors $v_j \in V\setminus U$ and just add $v_{k+1}$ to every basis vector $u_i$ of $U$. This is still a maximal linearly independent set and does not contain elements of $U$. This is exactly what happens in your example aswell.

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  • $\begingroup$ Thank you for the explanation I understand it now. But what exactly do you mean with that it is the same thing that happens in my example? $\endgroup$ Feb 22 at 23:52
  • $\begingroup$ Sorry I should have clarified: Take $\mathbb{R}^2$, Basis $\{(1,0),(0,1)\}$. The first vector spans a subspace $U$, and $\{(1,1)=(1,0)+(0,1), (0,1)\}$ is a basis in the style I tried to describe. $\endgroup$
    – Nuffie
    Feb 23 at 9:50

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