Intersection of two Matrices I have two spaces of matrices which are given below 

$$W_1 = \begin{cases} \\  \begin{pmatrix}
 a&-a \\
 c& d\end{pmatrix}\end{cases} \text{where a,c,d $\in$ $\mathbb{R}$}  $$
$$W_2=\begin{cases} \\  \begin{pmatrix}
 a&-b \\
 -a& d\end{pmatrix}\end{cases} \text{where a,b,d $\in$ $\mathbb{R}$}$$ 

now i have to find the $W_1 \cap W_2$
i don't know how to solve this question systematically .I know that answer is $\begin{pmatrix}
a & -a\\ 
-a & d
\end{pmatrix}$ but i got this answer by some random logic .Please provide me the proper solution of this 
Thankyou. 
 A: We have to find all matrices $A=\begin{bmatrix} w & x \\ y & z \end{bmatrix}$ that are in both $W_1$ and $W_2$, in other words $A \in W_1 \cap W_2$.  So the gameplan is this:

Given any matrix $A \in W_1$, find any restrictions needed for it to also be in $W_2$.  Similarly, given any matrix $A \in W_2$, find any restrictions needed for it to also be in $W_1$

Let's follow the gameplan:


*

*Given matrix $A=\begin{bmatrix} w & x \\ y & z \end{bmatrix} \in W_1$, we must have $x=-w$ by the definition of $W_1$, so
$$A=\begin{bmatrix} w & -w \\ y & z \end{bmatrix}$$
But for $A$ to be in $W_2$, we also need that $y=-w$. So
$$A=\begin{bmatrix} w & -w \\ -w & z \end{bmatrix}$$
which is exactly the form me wanted.

*Similarly, the same reasoning can be used starting with $A=\begin{bmatrix} w & x \\ y & z \end{bmatrix} \in W_2$, to show that it is in $W_1$ only if the matrix has the desired form (you should try this direction yourself).  
So we conclude that 
$$ W_1 \cap W_2 = \left\{ A\in\mathbb{R}^{2\times 2}:A= \begin{bmatrix} w & -w \\ -w & z \end{bmatrix} \right\} $$
A: Consider a matrix $\begin{pmatrix}
 r&s \\
 t&u\end{pmatrix}$ which is in both $W_1$ and  $W_2$.
You must have $s=-r$ to be in $W_1$ and $t=-r$ to be in $W_2$
The matrix therefore has the form $\begin{pmatrix}
 r& -r \\
 -r&u\end{pmatrix}$
A: The negative sign in front of the $b$ in $W_2$ is a red herring.
A $2\times2$ real matrix is in $W_1$ iff the top right entry is the negative of the top left entry. A $2\times2$ real matrix is in $W_2$ iff the bottom left entry is the negative of the top left entry.
Therefore a matrix is in both sets iff the top right entry is the negative of the top left entry and the bottom left entry is the negative of the top left entry (this is what a set intersection means: the elements in both sets).
