Show that V is a subspace of M2x2 Matrices and Determine a basis A bit of information to start us off: Let V denote the set of all 2x2 matrices with equal column sums.
Show $V$ is a subspace of $M_{2\times 2}$ matrices:
and....
Determine a basis for $V$:
So for the first bit... $M$ is the set of all $2\times 2$ matrices, I used the definition of subspaces... A is a subset of B if for every element "a" in A, "a" is also in B. Am I able to then say, since $V$ is the set of all $2\times 2$ matrices with equal column sums, "$a$" being an element of $V$. Then "$a$" is a $2\times 2$ matrix. Since "$a$" is $2\times 2$ matrix, "$a$" is in $M_{2\times 2}$ as well. Therefore the set of all $2\times 2$ matrices with equal column sums is a subset of $M_{2\times2}$?
What I am really unsure of is how to determine a basis. I'm used to being given a spanning set and working from there. Thanks!
 A: Note that a $2\times2$ matrix A has equal column sums iff $e^T A (e_1-e_2) = 0$, where $e=(1,1)^T$. The operator $L(A) = e^T A (e_1-e_2)$ is a linear function of $A$, hence $V = \ker L$, and so is a subspace (The kernel of a linear operator is automatically a subspace).
Since $\dim \mathbb{R}^{2 \times 2} = 4 = \dim \ker A + \dim {\cal R} A$, we see that $\dim V = \dim \ker A = 4-1 =3$.
Intuitively, note that if you pick 3 elements of a $2 \times 2$ matrix, you can always choose the fourth to satisfy the equation.
So, let the $A_{22}$ element be the 'special' element. A standard basis for $\mathbb{R}^{2 \times 2}$ is $e_i e_j^T$ for $i,j = 1,2$. So, pick one of these and figure out what the $22$ element must be to satisfy $L(A) = 0$.
Working through the computation gives $L(e_i e_j^T-\alpha e_2 e_2^T) = \delta_{1j}-\delta_{2j} + \alpha$, hence $\alpha = \begin{cases} -1 & j=1 \\
+1 & j = 2\end{cases}$. This construction produces three non-zero matrices that lie in $V$ by construction. It is easy to show that these are linearly independent, hence the form a basis.
A: When generating a basis for a vector space, we need to first think of a spanning set, and then make this set linearly independent. I'll try to make this explanation well-motivated.
What is special about this space? Well, the columns have equal sums. Thus, let's start with the zero vector and try to generate some vectors in this space. 
\begin{bmatrix}
0 & 0 \\ 
0 & 0
\end{bmatrix}
The columns must be equal, so whenever we add some quantity to the left side, we also need to add it to the right. Thus, these are vectors in this space:
\begin{equation}
\begin{bmatrix}
1 & 1 \\ 
0 & 0
\end{bmatrix} ,
\begin{bmatrix}
0 & 0 \\ 
1 & 1
\end{bmatrix}
\end{equation}
But, we could have put that 1 on a different row. So here are two more vectors in this space:
\begin{equation}
\begin{bmatrix}
1 & 0 \\ 
0 & 1
\end{bmatrix} ,
\begin{bmatrix}
0 & 1 \\ 
1 & 0
\end{bmatrix}
\end{equation}
So, now with these four matrices, we can show that they span this space. I think the easiest way to show this is to simply start with an arbitrary matrix in this space.
\begin{bmatrix}
a & c \\ 
b & d
\end{bmatrix}
Here, $a + b = c + d$. Well, let's subtract some of our proposed basis vectors. 
$$\begin{bmatrix}
a & c \\ 
b & d
\end{bmatrix} - a\begin{bmatrix}
1 & 1 \\ 
0 & 0
\end{bmatrix} = \begin{bmatrix}
0 & c-a \\ 
b & d
\end{bmatrix}$$
$$\begin{bmatrix}
0 & c-a \\ 
b & d
\end{bmatrix} - d\begin{bmatrix}
0 & 0 \\ 
1 & 1
\end{bmatrix} = \begin{bmatrix}
0 & c-a \\ 
b-d & 0
\end{bmatrix}$$
Next, we can apply our relation that the columns are equal. Rearranging it, we find $c = a + b - d$. Let's substitute this into the top corner element.
$$\begin{bmatrix}
0 & c-a \\ 
b-d & 0
\end{bmatrix} = \begin{bmatrix}
0 & (a + b -d)-a \\ 
b-d & 0
\end{bmatrix} = \begin{bmatrix}
0 & b - d \\ 
b-d & 0
\end{bmatrix}$$
Since the diagonals are the same, we can subtract one of our diagonal matrices to get the zero matrix. 
$$\begin{bmatrix}
0 & b - d \\ 
b-d & 0
\end{bmatrix} = (b-d)\begin{bmatrix}
1 & 0 \\ 
0 & 1
\end{bmatrix} = \begin{bmatrix}
0 & 0 \\ 
0 & 0
\end{bmatrix}$$
Thus, we've shown that any arbitrary matrix is composed of our matrices. But, are our matrices linearly independent? Actually, they're not. We can add together the two row matrices, and subtract the two diagonal matrices, and get the zero matrix, thus finding a nontrivial linear combination that adds to zero.
$$\begin{bmatrix}
1 & 1 \\ 
0 & 0
\end{bmatrix} + \begin{bmatrix}
0 & 0 \\ 
1 & 1
\end{bmatrix} -\begin{bmatrix}
1 & 0 \\ 
0 & 1
\end{bmatrix} - \begin{bmatrix}
0 & 1 \\ 
1 & 0
\end{bmatrix}=\begin{bmatrix}
0 & 0 \\ 
0 & 0
\end{bmatrix}$$ 
Note that our construction process only required one of the diagonal matrices, however! We can simply remove the other matrix from our set. 
After that, we can prove the remaining three matrices are linearly independent by contradiction and brute force--let the set not be linearly independent. Then one can be removed. We observe that removing any one of the matrices would lead to one position in the remaining matrices both having a value of zero, so no matrices with a nonzero value for that position can be constructed. Thus, the set must be linearly independent.
A: More simplistically...
You must answer these three questions:


*

*is $0$ (in this case, this means the zero matrix) in $V$?

*if $A$ and $B$ are two matrices in V, is $A + B$ also in $V$?

*if $r \in \mathbb{R}$ (or whatever field it is referring to, probably $\mathbb{R}$) and $A \in V$, then is $rA$ also in $V$?


The answer to the first question is yes (what are the column sums of the zero matrix?)
If the column sum of $A$ is $\alpha$ and the column sum of $B$ called $\beta$, what will the column sum of $A + B$ be?
Similarly, what will the column sum of $rA$ be?
For these questions, the "show it is a subspace" part is the easier part. Once you've got that, maybe try looking at some examples in your note for the basis part and try to piece it together from the other answer.
