Vector spaces and subsets of Matrices 
Hi, I was wondering if anyone could help me with this linear algebra problem.
I have solved a) by using matrix multiplying by scalars and matrix addition/subtraction. And hence concluded that I should get the return of
Row 1: (3 3) Row 2: (-2 -5)
I have also solved b) by creating a matrix from the 2x2 matrices and used row reducing operations. From this I found each line to have a pivot variable and hence having a unique solution. Which thus means that they are linearly independent.
However I'm now stuck on parts c) and d). I know to prove that a subset is a subspace it must meet the three conditions:

*

*It is closed under addition


*It is closed under scalar multiplication


*The 0 vector is in the subset.
But I'm unsure how to prove that these do/do not hold true.
Any help is appreciated.
 A: Recall our basic definitions of matrices and arithmetic surrounding matrices...
Very loosely defined, a matrix is a rectangular array of elements with some number of rows and some number of columns (even possibly zero).  They can be pictured either with square brackets or with parentheses, both are common and mean the same thing.
We can add two matrices so long as they have the same number of rows and have the same number of columns.  To do so, we just perform the elementwise addition of each element.  For example, we have:
$$\begin{bmatrix}1&2\\3&4\end{bmatrix}+\begin{bmatrix}50&60\\70&80\end{bmatrix}=\begin{bmatrix}51&62\\73&84\end{bmatrix}$$
We can multiply a matrix by a scalar.  To do so, we just multiply each element of the matrix individually by the scalar.  For example, we have:
$$5\cdot \begin{bmatrix}1&2\\3&4\end{bmatrix}=\begin{bmatrix}5&10\\15&20\end{bmatrix}$$
This understanding of how to add matrices and multiply matrices by scalars is a critical first step in adequately understanding and defining the vector space in your question.  Make sure you are comfortable with matrices as objects.

Now, in your question you are interested very specifically at the set of $2\times 2$ matrices (matrices with two rows and two columns) with real entries whose top left, top right, and bottom right entries add up to be equal to zero.  That is to say, the set:
$$\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}~:~a+b+d=0\right\}$$
This includes for instance the matrices $\begin{bmatrix}1&2\\3&-3\end{bmatrix}$ since $1+2+(-3)=0$ as well as $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ since $1+0+(-1)=0$ and so on...
We ask whether or not this set (coupled with the scalar field of $\Bbb R$ and arithmetic operations as described earlier) satisfies the properties necessary to be called a vector space.
We should already know that the space of all $2\times 2$ real matrices is in fact a vector space over $\Bbb R$ and so we do not need to check here all of the mundane properties such as associativity of addition of vectors, distributivity of scalars over vectors, that the scalar field is actually a field and what not... we can instead focus all of our attention on proving that our set is closed under addition, is closed under scalar multiplication, and contains the "zero" vector (the additive identity of our set, in this case the $2\times 2$ matrix who contains only zeroes).
To check that the set is closed under addition, we need to check that any two matrices we take from our set such as $\begin{bmatrix}1&2\\3&-3\end{bmatrix}$ and $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, their sum will also wind up in the set.  Indeed, here we have $\begin{bmatrix}1&2\\3&-3\end{bmatrix}+\begin{bmatrix}1&0\\0&-1\end{bmatrix} = \begin{bmatrix}2&2\\3&-4\end{bmatrix}$ is also an element of the set since $2+2+(-4)=0$, the defining trait of what it means to be in our set.
Of course, there are infinitely many vectors in our set, we can't check every addition possible individually... so instead we choose to be general.
Suppose that $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and $\begin{bmatrix}a'&b'\\c'&d'\end{bmatrix}$ are two elements in our set of matrices.

 (the apostrophes here have no mathematical meaning and are just used in order to give the second set of variables similar but different names to tell them apart.  They could just as easily have been called $e,f,g,h$ or $\heartsuit,\clubsuit,\diamondsuit,\spadesuit$ or any other four symbols)

By the fact that they are elements of our set, that means by definition of our set that they satisfy $a+b+d=0$ and $a'+b'+d'=0$ respectively.
We ask whether their sum also belongs to the set, that is whether $\begin{bmatrix}(a+a')&(b+b')\\(c+c')&(d+d')\end{bmatrix}$ satisfies the property that the top left, top right, and bottom right entries add up to equal zero... that is whether or not $(a+a')+(b+b')+(d+d')=0$
Indeed, by a bit of algebraic rearranging, we have $$(a+a')+(b+b')+(d+d')=(a+b+d)+(a'+b'+d')=0+0=0$$
Here, we used what we knew about the original pair of matrices, that they were in the set as well and so individually had their entries satisfying the property.  This shows that regardless what two matrices we picked, so long as they were from our set their sum will also be in our set.
As for scalar multiplication, this follows similarly.  We let $k$ be any element from our scalar field and $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ be an element from our set of matrices.  We ask if $k\cdot \begin{bmatrix}a&b\\c&d\end{bmatrix}$ is also a matrix in our set.
Indeed, $k\cdot \begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}ka&kb\\kc&kd\end{bmatrix}$ and we have the sum of the top left, top right, and bottom right entries happens to be:
$$ka+kb+kd = k(a+b+d)=k(0)=0$$
Thus, our set is indeed closed under scalar multiplication.
We move on to asking about "if zero is an element of our set."  It is important here to understand what we mean by "zero."  There are many different mathematical objects, all of whom might be given the name "zero."  In this context, we mean very specifically that element from our proposed vector space who acts as an additive identity, an element who when added to any other will not change the result.  In our specific problem, that "zero" is very specifically the $2\times 2$ matrix whose entries are all zero, $\begin{bmatrix}0&0\\0&0\end{bmatrix}$.
So... we ask... is $\begin{bmatrix}0&0\\0&0\end{bmatrix}$ a matrix in our set?  Well, the sum of the top left, top right, and bottom right entries are $0+0+0=0$ so yes, it is an element of our set.

 It is worth pointing out that this condition can be replaced with the condition "Our set is non-empty" and the set of definitions would have been equivalent.  After all, a set containing "zero" is nonempty since it contains at least zero, and a set who is nonempty and closed under scalar multiplication would include some $V$ and also contain $0\cdot V$, the result of the scalar zero multiplied by the vector $V$ which results in the vector "zero."

By showing these three properties to hold true, we have accomplished what we wanted in showing the set to indeed be a vector subspace of $\mathcal M_{2\times 2}(\Bbb R)$ (what is written in your screenshot simply as $M$)

Now, on to the question of finding a basis and a dimension for this, the common technique is to try to think in terms of "free-variables."  If we can change a variable to a "free-variable" and allow it to freely change without affecting its ability to stay in our set, we do so, but keeping in mind that some variables are dependent on one another so not all can be simultaneously turned into free-variables.  Think of it like the process of parametrizing a curve.
In your case, $c$ is a completely free variable, not relying on anything else.  We could set $c$ as anything and it would never have any impact on whether or not it is an element of our set.  Let us replace $c$ by a free variable, what I will label as $t$.
Next, we can set some (but not all) of the others to free variables.  Letting $a$ be replaced by the free variable $r$ and letting $b$ be replaced by the free variable $s$ we see that $d$ is then forced to be equal to $-r-s$ in order to satisfy the defining property of our set that $a+b+d=0$.
That is to say, every element in our space can be written in the form:
$$\left\{\begin{bmatrix}r&s\\t&-r-s\end{bmatrix}~:~r,s,t\in\Bbb R\right\}$$
We can split this up and factor this as well, factoring the free variables outside, giving us:
$$\left\{r\cdot \begin{bmatrix}1&0\\0&-1\end{bmatrix} + s\cdot \begin{bmatrix}0&1\\0&-1\end{bmatrix}+t\cdot\begin{bmatrix}0&0\\1&0\end{bmatrix}~:~r,s,t\in\Bbb R\right\}$$
Well... this is precisely what is meant when we talk about the span of the vectors $\begin{bmatrix}1&0\\0&-1\end{bmatrix},\begin{bmatrix}0&1\\0&-1\end{bmatrix},\begin{bmatrix}0&0\\1&0\end{bmatrix}$.
These vectors, which can be trivially shown to be linearly independent and span our space, then satisfy the definition of what it means to be a basis of our space.
Finally, we note that there are three vectors in our basis for our space and so the dimension of the space is simply the number of vectors in the basis and is then equal to $3$.
It should be emphasized that this is just one among many possible choices of basis.  It is a well known theorem however that a vector space will always have the same number of basis elements regardless which basis is used to describe it and so the dimension is unambiguous and does not rely on any specific basis.  For us, we could have had as a basis $\begin{bmatrix}1&2\\3&-3\end{bmatrix},\begin{bmatrix}2&3\\4&-5\end{bmatrix},\begin{bmatrix}0&0\\42\pi&0\end{bmatrix}$ for instance and been just as correct.  It would not have necessarily been what was written in a solutions manual or what the majority of people would have come up with (like the earlier choice I gave would be) but would still have been just as mathematically correct.
