Revisited: $[T]^{\gamma}_{\beta}$ is diagonal? Let $V$ and $W$ be finite-dimensional vector spaces with $\dim(V)=\dim(W)$ and let $T:V \rightarrow W$ be a linear map. How do I prove that there are bases $\beta$ of $V$ and $\gamma$ of $W$ such that the matrix $[T]^{\gamma}_{\beta}$ is diagonal?
\begin{eqnarray}
v_1\mapsto a_{11}w_1\\
v_2\mapsto a_{22}w_2\\
v_3\mapsto a_{33}w_3\\
\vdots\hspace{1cm}
\end{eqnarray}
Right?
 A: Let $V'$ an additional subspace of $\ker T$ in V i.e. such that 
$$V=\ker V\oplus V'$$ then we know that $T_{|V'}:V'\to \mathrm{Im T}$ is an isomorphism and let $(e_1,\ldots,e_r)$ a basis for $V'$ and $(e_{r+1},\ldots,e_n)$ a basis for $\ker V$ then we have $\beta=(e_1,\ldots,e_n)$ is a basis for $V$ and we have $(T(e_1),\ldots,T(e_r))$ is a basis for $\mathrm{Im }T$ which we complete in a basis $\gamma=(T(e_1),\ldots,T(e_r),v_{r+1},\ldots,v_n)$ for $W$ hence
\begin{eqnarray}
[T]_{\beta}^{\gamma} = \begin{pmatrix} I_r & 0 \\ 0 & 0 \end{pmatrix}
\end{eqnarray}
A: Choose any basis $\beta=\{b_1,\ldots,b_n\}$ of $V$. As I discussed in an answer to one of your earlier questions, $[T]_{\beta}^{\gamma}\in\mathrm{M}_{n\times n}(F)$ is the matrix 
$$\begin{bmatrix}
a_{11} & \cdots & a_{1n}\\
\vdots & \ddots & \vdots\\
a_{n1} & \cdots & a_{nn}
\end{bmatrix}$$
where the $a_{ij}$ are the elements of $F$ uniquely determined by
$$T(b_i)=\sum_{j=1}^n a_{ij}\gamma_j.$$
What would the $\gamma_j$'s have to be to make the matrix diagonal? (Note: some of the $\gamma_j$'s might be up to you to choose, i.e. not constrained to be something particular, depending on what $T$ is.)
