$$\ U = \{(x,y,z,t) \in \mathbb R^4 \mid x-y+z = 0 \text{ and } x-y-2t=0 \} $$

I need to find a subspace $\ T \in \mathbb R^4$ so $\ T \oplus U = \mathbb R^4 $

So I know the bases of $U$ will be:

$\ u_1 = (1,1,0,0), u_2 = (2,0,-2,1) $ which are not linearly dep. So I need to find two vectors that will span $\ \mathbb R^4 $

Based on answer from similar question I understand that I can just pick two other base vectors $\ e_3, e_4 $ and just to make sure all four vectors are linearly independent(?). If they are linearly indep. does it mean that the only linear combination of the zero vector is $\ 0 = 0 \cdot u_1 + 0 \cdot u_2 + 0 \cdot e_3 + 0 \cdot e_4 $ and hence $\ U \cap T = \{0 \} $

Also I'm not sure I understood the answer here correctly. If I have two vectors in $\ \mathbb R^n$ does it mean I can just pick any $\ n -2 $ base vectors that I want and those two groups will be direct sum of $\ \mathbb R^n $?


Yes, no and sort of! (Maybe?)

Firstly, a subspace $T \subset \mathbb{R}^4$ is a subset, not an element, of $\mathbb{R}^4$.

Now given $u_1$ and $u_2$, yes, you need to find two other vectors $e_3$ and $e_4$ so that the four vector $u_1, u_2, e_3$ and $e_4$ span $\mathbb{R}^4$ (nitpick on what you may have meant above, but no two vectors will span $\mathbb{R}^4$) and yes, this means pick two other vectors that are linearly independent from $u_1$ and $u_2$ so that the four vectors are linearly independent and span $\mathbb{R}^4$.

One of the definitions for linear independence of a set of vectors is indeed that there is only one linear combination of the zero vector.

No, you cannot simply pick any two vectors to get a basis though. For example, in the above situation consider $e_3 = u_1$ and $e_4 = u_2$. Then with these four vectors you only get $U$ and not all of $\mathbb{R}^4$, even though $u_1$ and $u_2$ are base vectors.

I might be nitpicking here, but what do you mean by 'base vector'? Any non-zero vector is in some basis, so it's a little imprecise to call something a base vector without having the basis already. What you can do though is start with your two vectors, then find another that is linearly independent with the starting two, then find a fourth that is linear independent to all three, then find a fifth that is independent to the previous 4, etc... until you get $n$ linearly independent vectors of $\mathbb{R}^n$ that will be a basis. In this sense, yes, you can pick any vectors that work in this way.

To think about why, suppose we focus on your main example with $U$ and $T$. If you find $e_3$ linearly independent from $u_1$ and $u_2$, then find $e_4$ linearly independent from $u_1, u_2$ and $e_1$, but had $U \cap T \neq \{0\}$, then there are linear combinations of a non-zero vector in $U \cap T$ $$ au_1 + bu_2 = ct_1 + dt_2 $$ so that $au_1 + bu_2 - ct_1 - dt_2 = 0$, contradicting the fact that the four vectors are linearly independent! So $U \cap T = \{0\}$. The main point being linear independence, so that once you have $n$ (in this case 4) they must span $\mathbb{R}^n$ (in this case $\mathbb{R}^4$).

As a good thinking exercise about these ideas that may help solidify all these new concepts, why should any set of four vectors that are linearly independent automatically span $\mathbb{R}^4$? I.e., why is the spanning condition seemingly coming for free? (Idea: If we didn't span $\mathbb{R}^4$ with four linearly independent vectors, we would need to add another vector to the basis, and as with the $U\cap T \neq \{0\}$ situation this contradicts something.)

  • $\begingroup$ Is it because $\ \mathbb R^4 $ has exactly four elements as basis? Thanks a lot for your well explained answer. Just to make sure I understand you correctly, you say that n elements in $\ \mathbb R^n $ that are linearly independent will span $\ \mathbb R^n $ also, if we'll take that four vectors and split it to any two groups - U & T, so $\ U \cap T = 0 $ will always hold ?? $\endgroup$ – bm1125 Jun 4 '18 at 18:04
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    $\begingroup$ Sort of. Any basis of $\mathbb{R}^4$ has exactly four elements. You always have the obvious one, $(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)$ as a sanity check, and any linearly independent set of 5 vectors spans a five dimensional space, so with $\mathbb{R}^4$ being 4-dimensional, any set of 5 vectors in $\mathbb{R}^4$ must be linearly dependent. As per your last question, yes by the second last paragraph in the answer: if there were a non-zero vector in both $U$ and $T$, then your four vectors are linearly dependent. $\endgroup$ – Alex Jun 4 '18 at 18:38

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