linear algebra how do i calculate the matrix transformation from matrix space to another matrix space suppose i have a linear trasformation $$T(X)=\begin{pmatrix}1 & 1\\
1 & 1
\end{pmatrix}X,T:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})$$
and i want to find the eigenvalues, i started by calculating the transformation of the basis matrices.
$$T\left(\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right)=\left(\begin{array}{cc}
1 & 1\\
1 & 1
\end{array}\right)\left(\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right)=\left(\begin{array}{cc}
1 & 0\\
1 & 0
\end{array}\right)$$
and did it for the rest of the basis vectors.
now, when i build the transformation matrix does it matter if i organize them by row or by column?
i mean is there a difference between 
$$\left[T\right]_{E}=\left(\begin{array}{cccc}
1 & 0 & 1 & 0\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
0 & 1 & 0 & 1
\end{array}\right)%%
%%
\left[T\right]_{E}=\left(\begin{array}{cccc}
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1\\
1 & 0 & 1 & 0\\
0 & 1 & 0 & 1
\end{array}\right)$$
english is not my main language so i had some hard time figuring out what to google in order to find a solution.
edit: i have the matrices
$$\left(\begin{array}{cc}
1 & 0\\
1 & 0
\end{array}\right)
\left(\begin{array}{cc}
0 & 1\\
0 & 1
\end{array}\right)
\left(\begin{array}{cc}
1 & 0\\
1 & 0
\end{array}\right)
\left(\begin{array}{cc}
0 & 1\\
0 & 1
\end{array}\right)
$$
and the problem is how to order their entries: as rows or columns?
thank you!
 A: If I understand, you’re considering whether to order your basis for $M_2$ as
$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 &1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, $$
or
$$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 &0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} .$$
If so, then this won’t matter. It just amounts to a change of basis, which won’t alter eigenvalues.
EDIT: If we have a linear transformation $T : V \to V$, where $V$ is a $d$-dimensional vector space, we can encode it as a square matrix as follows. Let $\{v_1, \ldots, v_d \}$ be a basis for $V$. Define a $d \times d$ matrix $A = [a_{i, j}]_{i, j = 1}^d$ by looking at $T v_j$ and writing $T v_j = a_{1, j} v_1 + \cdots + a_{d, j} v_d = \sum_{i = 1}^d a_{i, j} v_j$. The expression is unique because $\{ v_1, \ldots, v_d \}$ is a basis for $V$. This means that the $j$th column of the matrix $A$ depicts how $T$ acts on $v_j$. It is, however, vital that we keep the order of the basis consistent when calculating eigenvalues, i.e. we can't swap them around.
So if we were to take our basis to be
\begin{align*}
v_1 & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} , & v_2 & = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \\
v_3 & = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, & v_4 & = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} ,
\end{align*}
then we'd compute our matrix as follows.
\begin{align*}
T v_1 & = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \\
& = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} \\
& = v_1 + v_3 , \\
T v_2 & = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \\
& = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} \\
& = v_2 + v_4, \\
T v_3 & = v_1 + v_3 , \\
T v_4 & = v_2 + v_4 .
\end{align*}
The matrix for $T$ under this basis would then be
$$A = \begin{bmatrix} 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \end{bmatrix} .$$
Now for fun, let's reorder the basis, and consider the basis
\begin{align*}
w_1 & = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, & w_2 & = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} , \\
w_3 & = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, & w_4 & = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}
\end{align*}
This is how I interpreted your question about "rows" and "columns", I'm sorry if I misunderstood you.
If we compute the operator on this basis, we'll get
\begin{align*}
T w_1 & = w_1 + w_2, & T w_2 & = w_1 + w_2, \\
T w_3 & = w_3 + w_4, & T w_4 & = w_3 + w_4 ,
\end{align*}
yielding a matrix
$$\begin{bmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 1 
\end{bmatrix} .$$
As far as computing eigenvalues, these matrices are both just as good as each other. They're what we call equivalent, meaning that there exists an invertible $d \times d$ matrix $S$ such that $A = S B S^{-1}$. In general, if we take some linear transformation $T : V \to V$, and we take any two bases of $V$, say $\{ v_1, \ldots, v_d \}, \{ w_1, \ldots, w_d\}$, and we write the matrix $A$ for $T$ based on $\{ v_1, \ldots, v_d\}$, and the matrix $B$ for $T$ based on $\{ w_1, \ldots, w_d \}$, then $A$ and $B$ will be equivalent. Any two equivalent matrices will have the same characteristic polynomials, and thus the same eigenvalues.
I also wanna comment that this is specifically the way you build these square matrices when $T$ maps from a vector space $V$ to itself. In general, we'll have linear transformations $T : V \to W$, where we obviously can't use one basis to compute $T$. Instead we'll need a basis for $V$ and a basis for $W$. For that, you might look at Section 1.4 of these notes.
