$M_{2\times 2}(R)$ linear combination $\begin{pmatrix} 1&0\\0&1 \end{pmatrix}$, $\begin{pmatrix} 1&1\\0&1 \end{pmatrix}$, and $\begin{pmatrix} 1&0\\1&1 \end{pmatrix}$ do not generate $M_{2\times 2}(R)$ because each of these matrices has equal diagonal entries. So any linear combination of these matrices has equal diagonal entries.  
I understand why they can not generate $M_{2\times 2}(R)$, but what does the sentence "any linear combination of these matrices has equal diagonal entries" means? 
 A: It means for any scalars $a,b,c$, the two diagonal entries of the matrix $a\begin{pmatrix} 1&0\\0&1 \end{pmatrix}+b\begin{pmatrix} 1&1\\0&1 \end{pmatrix}+c\begin{pmatrix} 1&0\\1&1 \end{pmatrix}$ are equal to each other.
Remark: while the given explanation that the three matrices do not generate $M_{2,2}(\mathbb{R})$ is a correct one, I think it's quite unnatural. Actually, you cannot span $M_{2,2}(\mathbb{R})$ by three matrices because the matrix space, regarded as a vector space, is four-dimensional.
A: Presumably, you mean that these three matrices, $M_1, M_2 ,M_3 $, do not generate $\mathbb M_{2\times 2}(\mathbb R)$ as a vector space over $\mathbb R$. Any scalar multiple $r\cdot M_i$ is a matrix where the two diagonal entries are equal. Thus, any linear combination of the form $r_1\cdot M_1 + r_2\cdot M_2 + r_3 \cdot M_3$ will also be a matrix with equal diagonal entries. Thus, the given matrices generate a subspace of $M_{2\times 2}(\mathbb R)$ of matrices with equal diagonal entries. Since not all $2\times 2$ matrices have equal diagonal matrices, the given matrices do not generate all of $M_{2\times 2}(\mathbb R)$.
A: Just my 2 cents. Consider the two matrices:
$A=\bigl(\begin{smallmatrix}
1&0\\ 0&1
\end{smallmatrix} \bigr)$ and $B= \bigl(\begin{smallmatrix}
1&1\\ 0&1
\end{smallmatrix} \bigr)$
Then any linear combination between these two matrices $c_1 A +c_2 B$  will necessarily have diagonal entries $c_1+c_2$. 
I guess this is what you mean by linear combination of them have equal diagonal entries.
