# Let $V$ the space formed by $(x,y,z,w) \in \mathbb{R}^4$ such that $x+z=y+w=0$ and …

Let $V$ the space formed by $(x,y,z,w) \in \mathbb{R}^4$ such that $x+z=y+w=0$ and $W=span((1,1,1,1),(1,-1,1,-1),(1,0,1,0)).$ Determine basis for $V+W$ and $V\cap W$.

First, I find a basis for $V,W$.
Basis for V: $B_1=((-1,0,1,0),(0,-1,0,1))$
Basis for W: $B_2=((1,-1,1,-1),(1,0,1,0))$

Now, i'm trying to find a basis for $V+W$.

By definition:
$V+W=\{v+w:v\in V, w\in W\}$

My first question: Is true $B_1\cup B_2=span\{V+W\}$?.
If $B_1\cup B_2=span\{V+W\}$, Then, $\{(-1,0,1,0),(0,-1,0,1),(1,-1,1,-1),(1,0,1,0)\}$ Generate $V+W$.
Then, i only need to show that vectors are linearly independent

Let $A= \left[ \begin {array}{cccc} -1&0&1&1\\ 0&-1&-1&0 \\ 1&0&1&1\\ 0&1&-1&0\end {array} \right]$ Then, $A'= \left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0\\ 0&0&0&1\end {array} \right]$

In consequence, $\{(-1,0,1,0),(0,-1,0,1),(1,-1,1,-1),(1,0,1,0)\}$ are linearly independent. Then, $B_1\cup B_2$ is basis for $V+W$.

Is correct my reasoning?

For, $V\cap W$ Let $z\in V\cap W$ then $z\in V$ and $z\in W$ That implies: $z=x_1(1,-1,1,-1)+x_2(1,0,1,0)$ with $x_1,x_2 \in \mathbb{R}$
$z=y_1(-1,0,1,0)+y_2(0,-1,0,1)$ with $x_1,x_2 \in \mathbb{R}$

Then:

$(0,0,0,0)=z-z=x_1(1,-1,1,-1)+x_2(1,0,1,0)-y_1(-1,0,1,0)-y_2(0,-1,0,1)=(x_1+x_2+y_1,-x_1+y_2,x_1+x_2-y_1,-x_1-y_2)=(0,0,0,0)$

Then:

$x_1+x_2+y_1=0$
$-x_1+y_2=0$
$x_1+x_2-y_1=0$
$-x_1-y_2=0$

Solving the system, we have: $x_1=x_2=y_1=y_2=0$.

Then, $V\cap W=\{0\}$ and his basis is the null vector.

Is correct?

• As Jose points out, span is a function applied to vectors and produces a VS, so you mean $\mathrm{span}(B_1 \cup B_2) = V+W$? Also your $B_1$ is false or your equations in line 1 are. Assuming that the basis is correct all else is. Can you think of a rigorous reasoning to why the equation holds? – mdave16 Oct 6 '17 at 22:43
• @amd i don't understand con you be more specific? – Bvss12 Oct 6 '17 at 23:05
• :O Yes, thanks, i see the error now. Thanks for all @amd – Bvss12 Oct 6 '17 at 23:11

No, it is not correct. The equality $B_1\cup B_2=\operatorname{span}\{V+W\}$ cannot be true, because $\operatorname{span}\{V+W\}$ is a vector space, whereas $B_1\cup B_2$ isn't.
• True, but $B_1\cup B_2$ can be a basis for $V+W$? I say that because that generate and his vectors are linearly independent. – Bvss12 Oct 6 '17 at 22:41
• @Bvss12 Yes. If $B_1$ is a basis of $V$ and $B_2$ is a basis of $W$, then $B_1\cup B_2$ generates $V+W$. If, furthermore, $B_1\cup B_2$ is linearly independent, then it is a basis of $V+W$. – José Carlos Santos Oct 6 '17 at 22:44
• @Bvss12 Almost. The vector space $\{0\}$ has no basis. – José Carlos Santos Oct 6 '17 at 22:47