Prove that is a subspace We have the set $V$ of square $3x3$ matrices that sattisfy these two conditions:   
1) the sum of the first row and the third row is the vector $(0, 0, 0)^T$,
2) the sum of the elements on the main diagonal is $0$.
I should prove the set is subspace of $R^{3x3}$ and then find some five-element set of generators of this subspace.
To prove this set is a  subspace, I need to show the set is non empty, is closed under vector addition and scalar multiplication, that I can do. What I am not clear about is the second part. How do I find the set of generators here? And what is the purpose of the two conditions here? 
Thank you.
 A: As I understand you have proved that it is a subspace. To find a system of generator of such subspace you may proceed as foloows: 


*

*Take any matrix in $V$, let us say  $$ \begin{pmatrix}  a &b & c  \\ d& e&f  \\  g& h& i \end{pmatrix}$$

*What we have as a given: sum first and last row is the zero vector then  $ g=-a$ and  $ h=-b$ and $ i=-c$, so our arbitrary matrix can be written as $$ \begin{pmatrix}  a &b & c  \\ d& e&f  \\  -a & -b & -c  \end{pmatrix}$$  

*We have also that the sum of the diagola is  zero, $ a+e-c=0$,  so  $c=a+e $ thus your matrix can be written as
$$ \begin{pmatrix}  a &b & a+e  \\ d& e&f  \\  -a & -b & -a-e  \end{pmatrix}$$
Hence you arbitrary matrix can be written as
$$ \begin{pmatrix}  a & b & a+e  \\ d& e&f  \\  -a & -b & -a-e  \end{pmatrix} = a\begin{pmatrix}  1 & 0 & 1  \\ 0& 0&0  \\  -1 & 0 & -1  \end{pmatrix} +b\begin{pmatrix}  0 & 1 & 0  \\ 0& 0&0  \\  0 & -1 &0 \end{pmatrix}   +e \cdots$$
Continue, you will find the five desired  elements that generate the subspace  $V$.

A: Let $A \colon= [a_{ij}]_{3 \times 3}$ be an element of $V$. Then we must have 
$$a_{31} = -a_{11}, \ a_{32}=-a_{12}, \ a_{33} = -a_{13}, $$
and also 
$$a_{11} + a_{22}+ a_{33} = 0.$$
The last condition then becomes
$$a_{11}+ a_{22} -a_{13} = 0.$$
Or, 
$$a_{22} = a_{13} - a_{11}.$$
Thus, we can choose $a_{11}$, $a_{12}$, $a_{13}$, $a_{21}$, and $a_{23}$ independently, and the rest of the $a_{ij}$ are given by the formulas
$$a_{31} = -a_{11}, \ a_{32}=-a_{12}, \ a_{33} = -a_{13} \ \mbox{ and } \ a_{22} = a_{13} - a_{11}.$$
Therefore, the basis elements (or, what you call, generators) of this subspace can be found by taking one of the entries $a_{11}$, $a_{12}$, $a_{13}$, $a_{21}$, and $a_{23}$ to be $1$ turn by turn, while leaving all the others to be $0$, and of course using the above formulas to find the remaining entries of the matrix. 
Hope you can continue from here. 
