Diagonalization problem in linear algebra

Question

Why we talk about diagonalization problem in case of linear operator only not in case T: V->W where dimV =dimW

Answer 1

A linear map $T:\>V\to W$ between different spaces can always be diagonalized, because we can choose two bases in $V$ and $W$ separately.

Begin by choosing a basis $(f_1,\ldots, f_r)$ of the subspace $U:=T(V)\subset W$, and supplement these $r$ vectors to a basis $(f_1,\ldots, f_r,f_{r+1},\ldots, f_n)$ of all of $W$. Choose $r$ vectors $e_i\in V$ $(1\leq i\leq r)$ with $Te_i=f_i$. Note that the rank of $T$ is $r$. It follows that ${\rm dim}({\rm ker}\,T)=n-r$. We now end up with choosing a basis $(e_{r+1},\ldots, e_n)$ of ${\rm ker}\,T\subset V$. I claim that $(e_1,\ldots,e_r,e_{r+1},\ldots, e_n)$ is a basis of $V$.

Proof. If $s:=\sum_{i=1}^n \lambda_i\,e_i=0$ then $Ts=\sum_{i=1}^r \lambda_i f_i=0$, hence $\lambda_i=0$ $(1\leq i\leq r)$, since the$f_i$ are linearly independent. From $\sum_{i=r+1}^n\lambda_i\,e_i=0$ it then follows that the $\lambda_i$ with $i>r$ are $=0$ as well.

The matrix of $T$ with respect to these bases has all zeros, except at the places $(i,i)$ with $1\leq i\leq r$, where there is a $1$.