Linear Algebra: Vector Subspaces problem I am currently stuck on number 16:

So theorem 4.5 says:
If $W$ is a non empty subset of a vector space $V$, then $W$ is a subspace of $V$ if and only if the following closure conditions hold:


*

*If $\mathbf{u}$ and $\mathbf{v}$ are in (subspace) $W$ then $\mathbf{u} + \mathbf{v}$ is in $W$.

*If $\mathbf{u}$ is in $W$ and $c$ is any scalar, then $c\mathbf{u}$ is in $W$.


Questions:
1) What is defined as a closure? 
2) I tried doing 16 by saying that A is the 2x2 matrix $\left(\begin{array}{cc}1&0\\0&0\end{array}\right)$, and that B is the 2x2 matrix $\left(\begin{array}{cc}0&0\\0&1\end{array}\right)$. Both of these are singular. So if you add the two together you get $\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$ which is nonsingular. Thus we can conclude that it is not closed under addition. So no.1 is not satisfied.
3) How do I know if condition 2 is satisfied? If I multiply some scalar, such as 2, by matrix A from 2), I would get $\left(\begin{array}{cc}2&0\\0&0\end{array}\right)$. Is this considered to be part of the subspace?
 A: *

*"Closure" in this context means that if you take things inside the set, and you perform the required operation (adding them, or scalar multiplication in this case), then the result will not fall "outside" the set, it will stay "inside"; it all happens inside a "closed environment", so to speak.  It's not a something, but it's a property that the set may (or may fail to) have.

*I think you mean 15, not 16. That is an example to show that in the case of $n=2$, the set in question is not "closed under vector sums" (your first "closure condition"), so that shows 15 is not a subspace. You may want to modify it slightly to show this works for any $n\gt 1$, not just $n=2$; (and for bonus points, figure out why, in the case of $n=1$, the set of singular matrices is a subspace.

*If your set consists of all singular $2\times 2$ matrices, then the matrix
$$\left(\begin{array}{cc}
2 & 0\\
0 & 0
\end{array}\right)$$
is in the set if and only if it is a singular $2\times 2$ matrix. Is it a singular $2\times 2$ matrix? It's not whether it is or it is not "considered part of the subspace", it's whether it is or is not in the set. That is an objective condition (remember, the bouncer at the door only lets people in the set if they are members, and it's not a matter of "consideration"; either the object is in the set, or not). 
(As it happens, the set of singular $n\times n$ matrices is closed under scalar multiplication; that is, the second of your "closure conditions" does hold; if you know that a square matrix is singular if and only if its determinant is $0$, you can prove this using some simple properties of the determinant).
You say you are stuck on 16, the set of all $n\times n$ matrices $A$ such that $A^2=A$. 
A very easy matrix with this property is the identity matrix. What happens if you multiply the identity  matrix by a scalar? Will it still have the property? Why or why not? If not, give a specific example. What if you add the identity m atrix to itself; will the resulting matrix will be equal to its own square?
