Let denote by $M_{3,2}(\mathbb C) $ the space of all $(3\times2)$-matrix of complex-dimension equal $6$ with basis $(E_{1},E_{2},E_{3},E_{4},E_{5},E_{6})$. Let $f$ a $\mathbb R$-bilinear skew-symmetric map given by $$\begin{array}{crcl} f : &M_{3,2}(\mathbb C) \times M_{3,2}(\mathbb C) & \longrightarrow& M_{2,2}(\mathbb C)\\ &(A,B) & \longmapsto &f(A,B) = i(A^{*} B - B^{*} A), \end{array}$$ where $M^{*}$ denote the conjugate transpose of a matrix $M$ (i.e., $M^{*}= \, ^{t}\bar{M} $).
I like to know is what $f$ is degenerate or non-degenerate ?
Recall that: a $\mathbb R$-bilinear map $f :V \times V \longrightarrow W$ is $$\mbox{non-degenerate} \, \Leftrightarrow \, \forall x\in V, \mbox{if} \, f(x,y)=0, \, \forall \, y\in V,\, \mbox{then} \, x=0\, \Leftrightarrow\, \ker{f}=\{0\}, $$ where $\ker{f}=\{x\in V / \, f(x,y) = 0 \, \mbox{for all}\, y\in V \}$ the kernel subspace of $f$.
Noted that in our case: $\ker{f}=\{A\in M_{3,2}(\mathbb C)/ \, A^{*} B =B^{*} A \, \mbox{for all}\, B\in M_{3,2}(\mathbb C)\}$
Thank you in advance
Let $V, W$ are vectors spaces over the same field $K$ and $B$ a base of $V$, a bilinear map $f:V\times V\rightarrow W$ is nondegenerate iff there is no $v\not=0$ in $V$ s.t $f(v,e)=0 $ for all $e\in B$.
So $f$ is well a $\Bbb{R}$-bilinear map, normaly if we want to check that $f$ is nondegenrate we must resolve $f(A,e)=0$ for all $e $ in the canonical real bases of $M_{32}(\Bbb{C})$ as $\Bbb{R}$-vector space. But in this particular case it is enough to solve it for $e\in S=\{E_{ij}, i=1,2,3 \;j=1,2\}$ the canonical $\Bbb{C}$-base of $M_{32}(\Bbb{C})$, because $f(A,E_{ij})=0$ for all $E_{ij}\in S $ get then $A=0$. So by resolving this 6 matricial equation we get that $A$ must be $0$,and so $f$ nondegenerate bilinear map .
Example of resolving: fixe a variable matrix as $A=\left( \begin{array}{cc} x & t \\ y & u \\ z & v \end{array} \right) $ if we resolve $f(A,E_{11})=f(\left( \begin{array}{cc} x & t \\ y & u \\ z & v \end{array} \right),\left( \begin{array}{cc} 1 & 0 \\ 0 & 0 \\ 0 & 0 \end{array} \right))=0$ we arrive that $t=0$
and also if we resolve $f(A,E_{12})=0$ we arrive that $u=0$. So on we come to the conclusion that $A=0$ and $f$ nondegenerate.