Can you convert a non "normal" complex square matrix into a "normal" one? I've read the definition that "a complex square matrix ${\bf A}$ is normal if it commutes with its conjugate transpose".
I've also read that "A is a normal matrix iff there exists a unitary matrix U such that UAU^(-1) is a diagonal matrix.".
Given the matrix
$A = \begin{bmatrix}a+ib&-cd\\-ce&a+ib\end{bmatrix}$
I have found that the complex matrix is not normal: $AA^* \neq A^*A$
I know I can do a change of basis from the standard one to one that utilizes the eigenvectors ("eigen basis"), which inevitably transforms $A$ into a normal matrix with the eigenvalues on the diagonal. Aside from this "trivial" solution, can a complex square matrix that is not normal, be converted to one that is normal? Is there a given methodology/algorithm for this? If the method involves a change of basis, can an I retain an orthonormal basis? 
Here is what I tried so far:
I've found the eigenvectors and eigenvalues of $A$. The eigenvectors are
$ V_1 = \left[\frac{(ed)^{1/2}}{e},1 \right]^T , V_2 = \left[-\frac{(ed)^{1/2}}{e},1 \right]^T$
The eigenvectors are not orthogonal, so simply normalizing them does not provide an orthonormal basis (part of my objective). I then normalized both eigenvectors and chose one to be fixed as the first basis vector. I then used the Gram-Schmidt procedure to find the orthonormal basis vector. After that I checked if the orthonormal basis vectors satisfies $U \times U^* = U^* \times U = I$, which means its normal, and $U^* = U^{-1}$. (NOTE: * means conjugate transpose)
I then applied the change of basis to $A$ to get
$ \begin{bmatrix} a + ib + c(ed)^{1/2} & -c(e - d) \\ 0 & a + ib - c(ed)^{1/2} \end{bmatrix}$
Clearly the only way to make it diagonal is to set $d=e$. If that were the case, then the resulting diagonal is the eigenvalues, which result from utilizing an "eigen basis". So it would seem that $A$ is only a normal matrix when $d = e$ (which is only true since the diagonals of $A$ are the same).
In my attempt, where I "chose one to be fixed as the first basis vector", would choosing a different orthonormal basis ever transform $A$ into a normal matrix? At the start, $A$ uses a standard basis (orthonormal) but is not a normal matrix. I tried to make a change of basis using an orthonormal "eigen basis", but still the transform matrix is not normal. 
This led be to believe there is a requirement that $A$ be normal to begin with. After doing some more digging, this actual makes sense and it should have been clear from the start: $ A = CDC^{-1}$ where $C$ is my change of basis and $D$ is the transformation matrix with respect to the new basis. What I tried was to force $C$ to be unitary, which showed that $D$ is not diagonal, because $A$ was not normal to begin with.
From a different perspective, if I force $D$ to be diagonal with the eigenvalues (result from $C$ being the "eigen basis"), then $A$ is only normal if $C$ is unitary. But as I said prior, the eigen basis is not orthogonal, hence $A$ is not normal.
Call it one big proof (to my self) by contradiction that any change of basis will not influence whether the complex transformation matrix is normal.
So, if basis changes don't help, are there other alternatives? What about changes in the coordinates ("This is, I believe, as good as you can get")? I'm open to any suggestions really. Maybe I'm just stuck with non normal matrix?
 A: Let's take another viewpoint on the matter: $A$ corresponds to a linear map $\Phi\colon \mathbb C^2 \to \mathbb C^2, \quad x\mapsto Ax$. 
Conjugating $A$ with some unitary matrix corresponds to a change of basis from the standard basis to another orthogonal basis of $\mathbb C^2$.
Being normal can be expressed in terms of $\Phi$, it means that $\Phi^*$ and $\Phi$ commute, where $\Phi^* \colon \mathbb C^2 \to \mathbb C^2$ is the unique linear map  which satisfies $\langle \Phi(v),w\rangle = \langle v,\Phi^*(w)\rangle$ (this map exists, since our vector space is $2$-dimensional). Whether $\Phi$ and $\Phi^*$ commute does not depend on the choice of the orthonormal basis. If there is an orthogonal basis of $\mathbb C^2$, such that the transformation matrix of $\Phi$ has diagonal form, then $\Phi$ will automatically commute with $\Phi^*$ (since the transformation matrix of $\Phi^*$ wrt to this basis is just the conjugate-transpose). So if $A$ is diagonalizable over an  orthogonal basis, then it is already normal. 
In terms of matrices, if you conjugate $A$ to $U^*AU$, then the new matrix is normal if and only if the old one is: $(U^*AU)^* = U^* A^* U$, which commutes with $U^* A U $ iff $A$ commutes with $A^*$.
What you can do: conjugate with a non-unitarian matrix. This corresponds to chosing a possibly non-orthogonal basis. In your case, just take $S=(V_1|V_2)$, then $S^{-1}AS=D$, where $D=\text{diag}(a+ib+c\sqrt{ed}~,~a+ib-c\sqrt{ed})$. Then $D$ is a normal matrix, since it is in diagonal form. This is what you called the "trivial way". Of course, you can find many normal matrices which are similar to $D$ and therefore similar to $A$. Just take any unitary matrix $U$ and consider
$$B=U^* S^{-1} A S U = U^* D U$$
$B$ is still similar to $A$ (we only did a change of basis), but since we did a unitarian change of basis from $D$, the matrix $B$ is normal (since $D$ was normal).
Other coordinate changes then "change of basis" are not an appropriate tool to deal with when considering linear maps. They break the meaning of the transformation matrix.
