# Vector space $U\cap W$

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 $$U \cap W$$ ?

• As $u_2=u_1+u_3$ you are wrong about a basis of $U$. Commented Sep 27, 2021 at 15:17
• And you have two typos where you list the spanning set of $W$. Commented Sep 27, 2021 at 15:19
• Yes, I accidentally transcribed incorrectly Commented Sep 27, 2021 at 15:39
• 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 $U$. Commented Sep 28, 2021 at 3:49

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.

• I saw this method, so I was wondering if it could be like this?math.stackexchange.com/questions/1059700/… Commented Sep 28, 2021 at 8:29
• It's the same except in that case the intersection is trivial. Commented Sep 28, 2021 at 8:34
• 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 $U\cap W$? Commented Sep 28, 2021 at 8:42
• 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. Commented Sep 28, 2021 at 8:51
• Here is a direct question, when we cannot understand each other differently. Are vectors $u1, u2, w1$ base of $U+W$? Commented Sep 28, 2021 at 8:56