I was asked this:
Let $W_1,W_2 \subseteq \mathbb{R^4}$ subspaces that have defined like so:
$W_1 = \{ (x_1,x_2,x_3,x_4) \in \mathbb{R}^4 | x_1 + x_3 = x_2 + x_4, x_1 - x_3 = x_2 - x_4 \}$.
$W_2 = \{ (x_1,x_2,x_3,x_4) \in \mathbb{R}^4 | x_1 + x_3 = -x_2 - x_4, x_1 - x_3 = -x_2 + x_4 \}$.
Let $W=W_1+W_2$, Is $W=W_1 \oplus W_2$? Is $W=\mathbb{R}^4$?
I was trying to prove it like so:
Since, We're given that $W_1+W_2 =W$ so we only need to check if $W_1 \cap W_2 = \{ (0,0,0,0) \}$. Let $w=(x_1,x_2,x_3,x_4)\in W_1 \cap W_2$, since it defined $x_1+x_3$ as $x_2+x_4$ and $-x_2-x_4$, but $x_2+x_4 \neq-x_2-x_4$ except for $0$, therefore $\pm 0\pm 0=\pm 0 \pm 0$ is the only possible solution, which means $w=(0,0,0,0)$ and ${w}=W$.
We need to check now if $W=\mathbb{R}^4$ so we can do it like so: $dim(W_1) + dim(W_2)= dim(W) = dim( \{ 0 \} ) = 0 \Rightarrow 0 \neq dim(\mathbb{R})$, therefore $W \neq \mathbb{R}^4$.
But I feel like I got lost somewhere in the proofs, especially in proving $W=\mathbb{R}^4$. Thank you.
The vector space $W_1 \cap W_2$ corresponds to the solutions of the system
$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ -1 & -1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ -1 & 1 & 1 & -1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} =0 $.
The matrix below has determinant $-8$ and hence the dimension of the solutions space is $0$. This proves that $0 = W_1 \cap W_2$. In particular $W_1 + W_2 = W_1 \oplus W_2$.
Now, $dim(W_1 \oplus W_2)= dim(W_1)+ dim(W_2)$ minus the dimension of the intersection. Both spaces have dimension $2$, the intersection has dimension 0, hence the sum has dimension $4$. Here is the second affirmative answer.