Q:

Showing that a subset of a given space of matrices is a subspace

I've been given the following

$V$: The set of $2 \times 2$ matrices with real entries
$H := \{ A \in V : A \; \text{is symmetric} \}$

And I need to prove that $H$ is a subspace of $V$. So far, I have no idea where to start.  I'm supposing that I have a matrix $b$ and $c$, both belonging to the set $H$. 
I'll appreciate any help.  Thanks

A:

To prove $H$ is a subspace of $V$, recall what you have to prove: closure under scalar multiplication and closure under addition. 
Symbolically, let $A,B\in H$ and let $k$ be a scalar.  Then you want to show that

$A+B \in H$
$kA \in H$

Since $V$ is the space of $2\times 2$ real matrices, and $H$ is the subset of those matrices which are symmetric, it will help you to write the $A,B \in H$ explicitly:
$$A = \begin{bmatrix}
a & b\\ 
b & a
\end{bmatrix} \;\;\; B = \begin{bmatrix}
c & d\\ 
d & c
\end{bmatrix}$$
You can check that these are indeed the general form for symmetric $2\times 2$ real matrices (provided $a,b,c,d$ are real numbers), and thus $A = A^T, B= B^T$.
From there, you want to find $A+B$, and also find, for arbitrary constant $k$, the matrix $kA$ (or $kB$, it won't matter).  Show that both matrices are in $H$ - that is, show $A+B=(A+B)^T$ and $kA = (kA)^T$.
This will give closure under scalar multiplication and addition, and thus conclude the proof that $H$ is a subspace of $V$.

Addendum:
Sometimes when you come across questions of this nature, they won't be a subspace.  I'm not guaranteeing either for the $H,V$ in question.  But if you want to show something is not a subspace, you can show that one of the closure properties does not hold.  A simple counterexample using specific matrices/scalar $k,A,B$ also is valid (you don't have to use the general form - if any one example violates the closure axioms, then it's simply not a subspace).