Find a surjective Linear map $ T: \mathbb{C}^{3} \rightarrow \mathbb{C}^{3} $ such that $ T(1,0,0) = (0,i,0) \space $ and $ T(0,i,0) = (0,0,1) $
We must also verify that $T$ is injective.
I suppose that I must find a transformation matrix such that the image under this transformation is equal to $ \mathbb{C}^{3} $
I can't seem to find this though.
Any ideas?
Let the matrix be $A=\left(\begin{matrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{matrix}\right) $ under $T$ for the canonical basis $e_1=(1,0,1)^T,e_2=(0,1,0)^T,e_3=(0,0,1)^T$, namely for $X\in \mathbb{C}^3$, $$ TX=AX. $$ From $T(1,0,0)^T=(0,i,0)^T$, we obtain $$ (a_{11},a_{21},a_{31})^T=(0,i,0)^T$$ from which we have $a_{11}=a_{31}=0,a_{21}=i$. From $T(0,i,0)^T=(0,0,1)^T$, we obtain $$ (ia_{21},ia_{22},ia_{32})^T=(0,0,1)^T$$ from which we have $a_{21}=a_{22}=0,a_{32}=-i$. Clearly $\det(A)=a_{13}$. In order to make $T$ be surjective, we have to choose $a_{13}$ such that $a_{13}\neq 0$. For simplicity, we choose $a_{13}=1,a_{23}=a_{33}=0$ and hence $T$ is surjective. Now this $T$ is also injective (why?).