Representing linear transformations as matrices. What benefit, if any, is there to *not* expressing them as diagonal matrices? This question concerns something quite basic and I've done a questionable job of explaining my confusions, because I'm confused. I apologize.
I was playing with linear transformations earlier today, graphing the points as I calculated the matrices so that I could properly understand what I was doing. I encountered something that surprised me.
Operating the linear transform
$\left(\begin{matrix}
1&2\\3&4
\end{matrix}\right)$ on the vector $\left(\begin{matrix}1\\2\end{matrix}\right)$,
$\begin{align*}
\left(\begin{matrix}1&2\\3&4\end{matrix}\right) \left(\begin{matrix}1\\2\end{matrix}\right) &= \left(\begin{matrix}1i + 4j \\ 3i + 8j\end{matrix}\right)\\
&= \left(\begin{matrix}5\\11\end{matrix}\right)
\end{align*}
$
($1i + 4j$ and $3i + 8j$ are graphed alongside $(1,2)$ and $(5,11)$ below.)
So, the transformed $x$ value, for example, depends on both the $x$ and $y$ values of the original vector? This was something that I'd always accepted, but graphing it makes it seem strange. What benefit is there to doing this for linear transformations? Why would a diagonal matrix not always be used/chosen for a transform?
(Since, for example, in this case,
$\left(\begin{matrix}5&0\\0&\frac{11}{2}\end{matrix}\right)$
gives the same output vector and makes more sense intuitively.)


In the image above, the x component of the transformed blue vector is given by the addition of the purple components. This is how I assumed the transformation would be graphed, before realizing it was folly, as the axes of the final vector (first case, second case) are not the same as the axes of the original ($x$,$y$) or intermediate two (contributions), as regraphed below.


Interpreting the matrices as compact notation for linear equations, this makes sense to a degree. The total price for a number of item $x$ and a number of item $y$ is calculated $n_1x + n_2y$, so contributions from both axes makes sense. But then, the total price would be a scalar, and occupy one component of the transformed vector. The second component would be the total price of another seemingly unrelated combination of items $x$ and $y$. What is the benefit of having multiple axes here? What is useful about being able to see/communicate the total price of one combination of items at the same time as a different combination? (Does this have any applications, or are the two total prices only represented as one vector for the sake of compactness?)
 A: Here are three brief reasons (this list is deliberately not exhaustive), beyond the issue with producing $D$ pointed out in the comments, why one might use a basis in which a given transformation is not diagonal:


*

*Often one already wants to use a certain basis particularly well-suited to the problem at hand. Sometimes one is happy to work with any basis with a certain property (say, orthonormality w.r.t. a particular inner product), but not all linear transformations admit a diagonalizing basis which has that property.

*There are some linear transformations of finite-dimensional vector spaces that are not diagonal w.r.t. any basis. Consider, e.g., the transformation $\Bbb R^2 \to \Bbb R^2$ defined by left multiplication by the matrix
$$\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}.$$

*Sometimes one is interested in working with more than one linear transformation at a time. Typically it is easiest to work in a fixed basis, but we can readily find two (or more) linear transformations (even diagonalizable ones) which are not simultaneously diagonalizable, that is, for which there is no basis w.r.t. which all the matrices are diagonalizable.
