What is this dimension?

Question

What is the real dimension of the cone of $2$ by $2$ Hermitain matrices with at lease one eigenvalue that is $0$?

Answer 1

Let $A=\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ be a Hermitian matrix; we see that $a$ and $d$ are real and $b=\overline{c}$. Therefore, the space of $2\times 2$ Hermitian matrices depends on 3 real parameters. Now $A$ has eigenvalue 0 if and only if $\det A=0$, which means that $ad-|b|^2=0$, or in other words $ad=|b|^2$. Take the open set $\{a\neq0\}$; the dimension of this set intersected with the set you are looking for has the same dimension. In this set we have that $d=\frac{|b|^2}{a}$, and so $$A=\left(\begin{array}{cc}a&b\\\overline{b}&\frac{|b|^2}{a}\end{array}\right).$$ This means that the space depends on $2$ parameters. This is just a sketch of a proof, you'll have to write out all the details.