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Let $U_1,U_2$ be vector subspaces from $\in \mathbb R^5$.

$$\begin{align*}U_1 &= [(1,0,1,-2,0),(1,-2,0,0,-2),(0,2,1,2,2)]\\ U_2&=[(0,1,1,1,0),(1,2,1,2,1),(1,0,1,-1,0)] \end{align*}$$ (where [] = [linear span])

Calculate a basis from $U_1+U_2$ and a vector subspace $W \in \mathbb R^5$ so that $U_1+U_2=(U_1 \cap U_2) \oplus W$. ($\oplus$ is the direct sum).

I have the following so far. I calculated a basis from $U_1 \cap U_2$ in the previous exercise and got the following result: $(1,0,1,-1,0)$. I've also calculated a basis from $U_1+U_2$ and got that the standard basis from $\mathbb R^5$ is a basis.

So I suppose now I should solve the following:

standard basis from $\mathbb R^5$ = $(1,0,1,-1,0)\oplus W$

I thought I should get 4 additional vectors and they should also respect the direct sum criterion, that their intersection $= \{0\}$, however my colleagues have this:

$W = \{(w_1,w_2,0,w_3,w_4) | w_1,w_2,w_3,w_4 \in \mathbb R\}$. Where did I go wrong?

Many many many thanks in advance!

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    $\begingroup$ Note that your colleagues' $W$ has 4 arbitrary parameters and that $(1,0,1,-1,0)$ is not in $W$. (In fact, $W$ is the span of $e_1, e_2, e_4, e_5$ where $e_i$ is the usual $i$th unit vector). $\endgroup$ Commented Dec 18, 2011 at 10:24

2 Answers 2

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It seems that you're fine. The $W$ given by your colleagues' has as a basis $\{e_1,e_2, e_4,e_5 \}$ where $e_i$ is the standard $i^{\rm th}$ unit vector in $\Bbb R^5$. Moreover, their $W$ does not contain $ (1,0,1,-1,0)$ (any vector in $W$ has 0 in its third coordinate); thus, this vector together with $e_1$, $e_2$, $e_4$, $e_5$ will give a basis for $\Bbb R^5$. So, then, $\mathbb R^5$ = $(1,0,1,-1,0)\oplus W$.

Your approach would be more or less the same. I imagine your colleagues' interpreted the question as "exhibit the subspace $W$", rather than "find a basis of the subspace $W$".

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  • $\begingroup$ Hi David, thanks for the fast reply! Question: would the basis(es?) $\{e_2, e_3, e_4, e _5\},\{e_1, e_2, e_3, e _5\}$ also be valid? Using your reasoning, there's no way to get $(1,0,1,-1,0)$ (always a 0 in one of the coordinates that should be 1). Can I just simply write that (any vector in $W$ has 0 in its third coordinate) or do I have the proof something more? many thanks again! $\endgroup$
    – Clash
    Commented Dec 18, 2011 at 11:05
  • $\begingroup$ My reasoning to reply the question would be the following. We have the vector $(1,0,1,-1,0)$ and we want to expand it to basis $\mathbb R^5$. Let's try using the vectors $\{e_2, e_3, e_4, e _5\}$. I then perform gauß and see that I get the standard basis. Then I finalize by writing that W can not be written as a linear combination from $\{e_2, e_3, e_4, e _5\}$. Would this be ok? $\endgroup$
    – Clash
    Commented Dec 18, 2011 at 11:08
  • $\begingroup$ @clash The two bases you give will work. You could also say "$W$ is the set of all vectors with 0 third coordinate". $\endgroup$ Commented Dec 18, 2011 at 11:10
  • $\begingroup$ @clash for your second comment, you want to say "(1,0,1,-1,0) can not be written as a linear combination from $\{e_2,e_3,e_4,e_5\}$" (not $W$, as you wrote). $\endgroup$ Commented Dec 18, 2011 at 11:13
  • $\begingroup$ oops, thanks david! $\endgroup$
    – Clash
    Commented Dec 18, 2011 at 11:22
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You could also find your $W$ like this: all you have to do is complete the matrix

$$ \begin{pmatrix} 1 & * & * & * & * \\ 0 & * & * & * & * \\ 1 & * & * & * & * \\ -1 & * & * & * & * \\ 0 & * & * & * & * \\ \end{pmatrix} $$

in such a way that it has rank $5$. You can do this almost as you want, but maybe an easy strategy is like this: sure enough these firs two columns are linearly independent, aren't they?

$$ \begin{pmatrix} 1 & 1 & * & * & * \\ 0 & 0 & * & * & * \\ 1 & 0 & * & * & * \\ -1 & 0 & * & * & * \\ 0 & 0 & * & * & * \\ \end{pmatrix} $$

So, you keep going the same way. These first three columns are linearly independent too:

$$ \begin{pmatrix} 1 & 1 & 0 & * & * \\ 0 & 0 & 1 & * & * \\ 1 & 0 & 0 & * & * \\ -1 & 0 & 0 & * & * \\ 0 & 0 & 0 & * & * \\ \end{pmatrix} $$

So, let's try again with

$$ \begin{pmatrix} 1 & 1 & 0 & 0 & * \\ 0 & 0 & 1 & 0 & * \\ 1 & 0 & 0 & 1 & * \\ -1 & 0 & 0 & 0 & * \\ 0 & 0 & 0 & 0 & * \\ \end{pmatrix} $$

Still four linearly independent columns. Now, the only risk is with the last one:

$$ \begin{pmatrix} 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \\ \end{pmatrix} $$

If you keep on just putting the next vector of the standard basis of $\mathbb{R}^5$, this matrix has rank four. Never mind: just replace your last column with this one:

$$ \begin{pmatrix} 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ \end{pmatrix} $$

This matrix has rank $5$. Hence, your vector $(1,0,1,-1,0)$ and $e_1, e_2, e_3, e_5$ are linearly independent vectors. Hence the sum $[(1,0,1,-1,0)] + [e_1, e_2, e_3, e_5]$ is a direct sum and is equal to $\mathbb{R}^5$. So, you can take $W = [e_1, e_2, e_3, e_5]$. (The solution of this problem is far away from being unique.)

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  • $\begingroup$ thanks agusti, this is indeed an easy to way to see it! $\endgroup$
    – Clash
    Commented Dec 18, 2011 at 12:10

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