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Find $U + W$ and $U \cap W$ for $U =\{\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} \in \mathbb{R}^5 | \left\{\begin{array}{l}x_2=2x_1-x_3\\x_4=3x_5\end{array}\right. \}$, $W=\{\begin{pmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{pmatrix} \in \mathbb{R}^5 | x_3+x_4=0 \}.$

So I found out that:

$U = Sp\{ \begin{pmatrix}1\\2\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\-1\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\0\\3\\1\end{pmatrix} \}$ and $W=Sp\{ \begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\0\\0\end{pmatrix},\begin{pmatrix}0\\0\\1\\-1\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\0\\1\end{pmatrix} \}$

Now... how do I calculate $U + W$ and $U \cap W$?

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  • $\begingroup$ For a systematic, mechanical way to do this, there is the Zassenhaus algorithm. $\endgroup$ – amd Oct 19 '18 at 21:05
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First recall the formula $dim(U+W)+dim(U\cap W)= dim W + dim U$. From the definitions and your computations it it clear that $dim U=3$ and $dim W=4$. Thus if we find a vector in $U$ but not in $W$ it is clear that $U+W= \mathbb{R}^5$: $(0,-1,1,0,0)$ which you presented does the trick. Now the formula gives us $dim (U \cap W)=3+4-5=2$: if we can find two linearly independent vectors in which are in $U$ and $W$ we have found basis for the intersection subspace. Notice $(1,2,0,0,0)$ and $(0,1,-1,1, 1/3)$ satisfy the requirements.

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  • $\begingroup$ How have you picked $(0,-1,1,0,0)$? $\endgroup$ – C. Cristi Oct 19 '18 at 12:59
  • $\begingroup$ Can you be more explicit on what you just did? $\endgroup$ – C. Cristi Oct 19 '18 at 13:09
  • $\begingroup$ I picked $(0,-1,1,0,0)$ by looking at the base of $U$ you presented and choosing the first vector which is not in $W$. If a subspace of some finite dimensional space $V$ has codimension $1$ then it is immediate that adding one vector you get the whole space. As for giving a more explicit proof I would not know how to do that: the explanation I gave you is quite direct and concise. The basic idea is to use the formula correlating the dimensions to understand how many vectors you have to pick for the bases of $U +W$ and $U \cap W$. What is that you find unclear? $\endgroup$ – N.B. Oct 19 '18 at 15:17
  • $\begingroup$ I find more a lot more different vectors other than $(0,-1,1,0,0)$... say $(1,2,0,0,0)$ why cannot this work? $\endgroup$ – C. Cristi Oct 21 '18 at 8:08
  • $\begingroup$ Because $(1,2,0,0,0)$ is already in $W$: since $U \cap W$ is not trivial there exists non-zero vectors which are in both $U$ and $W$. When understanding $U +W$ you want find vectors in $U$ but not in $W$ so that you are extending $W$ by the span of these vectors. $\endgroup$ – N.B. Oct 22 '18 at 9:50
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Observe that all vectors in U are of the form $\begin{pmatrix}x_1 \\ 2x_1- x_3 \\ x_3 \\ 3x_5 \\ x_5\end{pmatrix}= x_1 \begin{pmatrix}1 \\ 2 \\ 0 \\ 0 \\ 0 \end{pmatrix}+ x_3\begin{pmatrix}0 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}+ x_5\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$ so U is a three dimensional subspace spanned by $\begin{pmatrix}1 \\ 2 \\ 0 \\ 0 \\ 0 \end{pmatrix}$, $\begin{pmatrix}0 \\ -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}$, and $\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$.

And vectors in W are of the form $\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ -x_3 \\ x_5\end{pmatrix}=$$ x_1\begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}+$$ x_2\begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}+$$ x_3\begin{pmatrix}0 \\ 0 \\ 1 \\ -1 \\ 0\end{pmatrix}+$$ x_5\begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 1\end{pmatrix}$. So W is a four dimensional subspace spanned by those four vectors.

The question, really, is 'how may of the vectors spanning U are in W'. If all of them are U+ W= W. If there 1 that is not (and there cannot be more than 1) then U+ W= $R^5$. $U\cap W$ is the subspace spanned by those basis vectors of U that are in W.

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