Finding linear transformation for vector space of matrices Good evening everyone,
I understand how to find a determinant. What does it mean to have a linear transformation from the space $V$ of $2\times 2$ upper triangular matrices to $V$.
Also, how did they arrive to the solution? What are the questions should I ask myself when working with problems like this?


Thank you.
 A: Let $T: V \to V$ be a linear transformation that transforms every $2\times 2$ upper triangular matrix $M$ to a new upper triangular matrix. 
Firstly, we need to determine a basis of the vector space $$V = \{X: X \text{ is a $2\times 2$ upper triangular matrix}\}.$$
We can "easily" observe that a "standard" basis of $V$ is the set of matrices:
$$B = \left\{ \underbrace{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}}_{E_1},\underbrace{\begin{bmatrix} 0 & 1\\ 0& 0 \end{bmatrix}}_{E_2}, \underbrace{\begin{bmatrix} 0 & 0 \\ 0 & 1\end{bmatrix}}_{E_3}  \right\}.$$
All these matrices in $B$ are in $V$, they are linearly independent and they span $V$.

Now, we need to apply our transformation to every matrix in $B$ and express the transformed matrix in terms of $E_1, E_2, E_3$. 
Thus, we have:
$$T(E_1) = \begin{bmatrix} 2 & 3 \\ 0 & 4\end{bmatrix} \cdot E_1 = \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = 2E_1 + 0E_2 + 0E_3$$
The above coefficients are going to form the first column of our transformation matrix $A$.
Following the same procedure for $E_2$ we have:
$$T(E_2) =\begin{bmatrix} 2 & 3\\ 0 & 4 \end{bmatrix} \cdot E_2 = \begin{bmatrix} 0 & 2 \\ 0 & 0 \end{bmatrix}= 0E_1 + 2E_2 + 0E_3.$$
The above coefficients are going to form the second column of our transformation matrix $A$.
Accordingly, we have $$T(E_3) = \begin{bmatrix} 0 & 3\\ 0 & 4\end{bmatrix} = 0E_1+3E_2+4E_3.$$
Hence, we have:
$$A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 3 \\ 0 & 0 & 4\end{bmatrix}.$$

Practically, what does this transformation matrix $A$ mean?
Assume we have a row vector $[E_1 \quad E_2 \quad E_3]$. If we apply the transformation matrix to the previous vector i.e.
$$[E_1 \quad E_2 \quad E_3]\cdot \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 3\\ 0 & 0 & 4\end{bmatrix}=[2E_1 \quad 2E_2 \quad 3E_2 +4E_3]= [T(E_1)\quad T(E_2)\quad T(E_3)],$$
we get a row vector with the transformed matrices.

Addition to your comment:
As said before, the transformation is $T: V\to V$. We have the basis of $V: \beta =  (E_1, E_2, E_3).$ 
According to this, the matrix representation of a linear transformation $T:V\to V$ in the ordered base $\beta$  is the $3\times 3$ matrix $A$ defined as:
$$A = [T]_\beta = \Big[ [T(E_1)]_\beta\quad [T(E_2)]_\beta \quad [T(E_3)]_\beta\Big]$$

Now, we can treat the problem as following. Assume that we have the transformation $V\ni \begin{pmatrix} x & y \\ 0 & z\end{pmatrix} \mapsto \begin{pmatrix} x' & y' \\ 0 & z'\end{pmatrix}\in V.$  Essentialy, it's like transforming the vector $[x\quad y \quad z]$ into the vector $[x' \quad y' \quad z']$.
Thus, we can consider the transformation as
$$T(\mathbf{x}) = \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 & 3 \\ 0 & 0 & 4\end{bmatrix}\mathbf{x},$$
where $\mathbf{x} = [x \quad y \quad z]^T$.
