Vector space $U\cap W$

Question

I need advice on this task, so if anyone had a similar dilemma it would help me.

The task is: Let $U$ be a subspace of space $\mathbb{R^4}$generated by vectors $u1=(1,2,0,-1), u2=(0,3,1,2), u3=(-1,1,1,3)$ and W a subspace generated by vectors $w1=(1,1,1,1), w2=(0,1,1,2), w3=(-1,0,0,1)$. Determine one base for vector spaces $U,W,U + W, U \cap W$.

I did it like this:

For $U$, I placed the vectors in the matrix, found the pivots, and determined that the base was made up of vectors $(1,0,0,0),(0,1,0,0)$.

I did the same for the vector space W.

$U + W$:

I placed the vectors in the matrix and found the pivots:

$$\left[\begin{matrix} 1 & 0 & -1 & 1 & 0 & -1\\ 2 & 3 & 1 & 1 & 1 & 0\\ 0 & 1 & 1 & 1 & 1 & 0\\ -1 & 2 & 3 & 1 & 2 & 1\\ \end{matrix}\right]\rightarrow \left[\begin{matrix} 1 & 0 & {\color{red}{-1}} & 1 & {\color{red}{0}} & {\color{red}{-1}}\\ 0 & 1 & {\color{red}{1}} & 1 & {\color{red}{1}} & {\color{red}{0}}\\ 0 & 0 & {\color{red}{0}} & -4 & {\color{red}{-2}} & {\color{red}{2}}\\ 0 & 0 & {\color{red}{0}} & 0 & {\color{red}{0}} & {\color{red}{2}}\\ \end{matrix}\right]$$

The basis of vector space $U + W$ are vectors: $ u1,u2,w1$.

My dilemma is whether the vectors are marked in red by the bases of the vector space $U \cap W$ ?

And you have two typos where you list the spanning set of $W$. — ancient mathematician, Sep 27 '21 at 15:19
Every linear combination of $(1, 0, 0, 0)$ and $(0,1,0,0)$ will have $0$ for its third and fourth coordinates. Each of the generators of $U$ has non-zero entries in the third and fourth coordinates. So how can $(1, 0, 0, 0)$ and $(0,1,0,0)$ form a basis for $U$? Their span doesn't even include the three known points of $U$. — Paul Sinclair, Sep 28 '21 at 03:49

score 1 · Answer 1 · answered Sep 28 '21 at 06:55

1

I don't understand what manipulations you are making or why. But here is a technique guaranteed to work when we want to find a basis of $U\cap W$ when we are given spanning sets of $U$ and $W$.

The vector $(a,b,c,d)$ lies in $U$ iff there exist scalars $x,y,z$ such that $x u_1+ y u_2 +z u_3= (a,b,c,d)$. Use Gauss elimination to find the conditions for the four equations to have a solution for $x,y,z$: the conditions are $2a-b+3c=0$,$a-2c+d=0$.

Do the same thing to see when $(a,b,c,d)$ lies in $W$: you'll get some more linear equations on $a,b,c,d$.

Now solve the whole set of equations in $a,b,c,d$ to find those $(a,b,c,d)$ in $U\cap W$. The Gauss elimination technique provides a basis for them.

answered Sep 28 '21 at 06:55

ancient mathematician

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I saw this method, so I was wondering if it could be like this?https://math.stackexchange.com/questions/1059700/find-a-basis-for-uw-and-u-cap-w – LogicNotFound Sep 28 '21 at 08:29
It's the same except in that case the intersection is trivial. – ancient mathematician Sep 28 '21 at 08:34
So I can sort the vectors into a matrix? When I find pivots, the vectors that determine pivots are the base of $U + V$? Those that do not specify pivots are the base $U\cap W $? – LogicNotFound Sep 28 '21 at 08:42
Sorry, I don't know what operations you are performing on your matrix. My answer is to the question in the first line of your post only. – ancient mathematician Sep 28 '21 at 08:51
Here is a direct question, when we cannot understand each other differently. Are vectors $u1, u2, w1$ base of $U+W$? – LogicNotFound Sep 28 '21 at 08:56
Yes they are. But of course the question is, does your process guarantee this in general? – ancient mathematician Sep 28 '21 at 10:19

Vector space $U\cap W$

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