Submanifold of $2\times 2$ Complex Matrices Using Transversality? Let $M$ be the set of all $2\times 2$ complex matrices with $a_{21}=\bar a_{12}$. $M$ is a smooth manifold diffeomorphic to $\mathbb{R}^6$. Let $W$ be the subset consisting of all matrices in $M$ with two equal eigenvalues.
Problem: What is the dimension of $W$? Is $W$ a submanifold of $M$? If not, what is the smallest closed subset $C$ of $W$ that one can remove so that $W-C$ is a subamnifold of $M$?
Attempt:
The characteristic polynomial is $p_A(t)=(t-a_{11})(t-a_{22})-|a_{12}|^2$, and we want this to have to equal roots:
i.e. $p_A(t)=(t-a_{11})(t-a_{22})-|a_{12}|^2=(t-\lambda)^2$
So we get the two equations:
$\lambda=(a_{11} +a_{22})/2$ and $\lambda^2= a_{11}a_{22}-|a_{12}|^2$, which combine to give 
$(a_{11} - a_{22})^2/2 +|a_{12}|^2=0$.  ($W$ consists of the matrices satisfying this )
Now if we define $f=(a_{11} - a_{22})^2/2 +|a_{12}|^2$
, then $f$ is a smooth function from our manifold into $\mathbb{C}\cong \mathbb{R}^2$. So if we can show that $0$ is a regular value (possibly w/ some restriction of the domain), we will have that that space is a submanifold of dimension $6-2=4$ (I think).
$df=[a_{11}-a_{22}, \bar a_{12},a_{12},-(a_{11}-a_{22})]=0$ iff $a_{11}=a_{22}$ and $\bar a_{12}=0=a_{12}$.
Thus if we let $C$ be the subset of $M$ such that $a_{12}=0$, then on $M-C$ $0$ is a reg value of $f$ and so $W-C=f^{-1}(0)$ is a submanifold. 
Am I making any mistakes?
Update: 
Taking "squirrel's" comments into account:
Equal eigenvalues requires: $(a_{11}-a_{22})^2-4(a_{11}a_{22}-|a_{12}|)^2=0$. If we seperate this into real and complex components, letting $a_{11}=a+bi, a_{12}=c+di, a_{22}=e+fi$, we get two equations:
$(a+e)^2 -(b+f)^2 -c^2-d^2=0$ and $(a+e)(b+f)=0$
Set $F,G:\mathbb{R}^6 \rightarrow \mathbb{R}$ as 
$F(a,b,c,d,e,f)=(a+e)^2 -(b+f)^2 -c^2-d^2$ and $G(a,b,c,d,e,f)=(a+e)(b+f)$. Then 
$dF=[2(a+e) \hspace{2mm} -2(b+f) \hspace{2mm}  -2c \hspace{2mm} -2d \hspace{2mm} 2(a+e) \hspace{2mm} -2(b+f)]$
and $dG=[(b+f) \hspace{2mm} (a+e) \hspace{2mm}  0 \hspace{2mm} 0 \hspace{2mm} (b+f) \hspace{2mm} (a+e) ]$. 
Now I'm not sure if $W=F^{-1}(0) \cap G^{-1}(0)$ is a submanifold or not, but if I can restrict $F$ and $G$ from some subset $C\subset W$ (to make $0$ a regular value) and then show that $F^{-1}(0)$ and $G^{-1}(0)$ (now restricted) are transverse then I would be done...I think...
 A: The set $M\subset\mathbb R^8\equiv\mathbb C^4$ is a real $6$-plane, not a complex $3$ plane: conjugation makes complex discussion impossible, as already remarked. Now $W$ is well described by the two equations $F,G$ (except that my computations give $\ -4c^2-4d^2$ as written below), and it is a singular real algebraic set $Z$. It is irreducible and diffeomorphic to $C\times\mathbb R^2$, where $C$ is a cone in $\mathbb R^3$. The singular locus of such model is a plane $\Pi=\{0\}\times\mathbb R^2$ and the regular locus $C\setminus\Pi$ is a $4$-manifold, union of two connected components. 
To see this quickly, split the system defining $Z$ into the following two:
$$
Z_1:\begin{cases}F(a,b,c,d,e,f)=0\\a+e=0,\end{cases}\equiv
    \begin{cases}-(b+f)^2-4c^2-4d^2=0\\a+e=0,\end{cases}\equiv \{b+f=c=d=a+e=0\}, 
$$
and
$$
Z_2:\begin{cases}F(a,b,c,d,e,f)=0\\b+f=0,\end{cases}\equiv
    \begin{cases}(a+e)^2-4c^2-4d^2=0\\b+f=0,\end{cases}.
$$
Thus $Z_1$ is a plane contained in $Z_2$, hence $Z=Z_2$. We can eliminate $f$ in the system for $Z_2$ to get the equation $(a+e)^2-4c^2-4d^2=0$ in the five variables $(a,b,c,d,e)$. Make the linear change $a+e=u$ and you have the cone
$$
C=\{u^2-4c^2-4d^2=0\}\subset\mathbb R^3,
$$
such that $Z=C\times\mathbb R^2$ (the $\mathbb R^2$ corresponding to the variables $(a,e)$, say). Note that the plane $Z_1$ is exactly the singular locus of $Z=Z_1$. In transversality terms, the equations defining $Z=Z_2$ are transversal to $(0,0)$ exactly at the points in $Z_2\setminus Z_1$.
