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Exercise: Exercise: Suppose $V$ and $W$ are finite-dimensional and $T \in L(V, W)$. Prove that there exist a basis of $V$ and a basis of $W$ such that with respect to these bases, all entries of $M(T)$ are $0$ except that the entries in row $j$ , column $j$ , equal $1$ for $1 \le j \le \dim \text{range }T$.

Proof: Let dim range $T=m$. By the fundamental theorem of linear maps we have that dim null $T=n-m$ where dim $V=n$. Thus there exist $m$ basis vectors of $V$ that do not get mapped to $0$. Let $v_1,\dots,v_m, \dots,v_n$ be that basis in which those $m$ vector exist and where the remaining $n-m$ vectors get mapped to zero. Using the theorem that if $v_1,\dots,v_n$ spans $V$ then $Tv_1,\dots,Tv_n$ spans range $T$, we see that $Tv_1,\dots,Tv_n$ is a spanning list in range $T$. Using the theorem that a spanning list can be reduced to a basis, we can reduce the list $Tv_1,\dots,Tv_n$ to a basis of range $T$ by removing those vectors that are in the span of the previous ones. In doing so, all the $n-m$ vectors that map to $0$ get removed.

The remaining vectors $Tv_{1},\dots,Tv_m$ span range $T$. To show that this is a basis of range $T$, it suffices to show that $Tv_j$ is not in the span of the previous vectors.

Suppose there exist scalars $a_{1},\dots,a_{j-1}$ such that $a_{1}Tv_{1}+\dots+a_{j-1}Tv_{j-1}=Tv_j$. Subtracting $Tv_j$ from both sides and using the linearity of $T$ we get $T(a_{1}v_{1}+\dots+a_{j-1}v_{j-1}-v_j)=0$. Because none of the vectors in the list $Tv_{1},\dots,Tv_m$ get mapped to zero, the independence of $v_{1},\dots,v_m$ and the above representation imply that the scalars $a_{1}=\dots=a_m=0$. Thus, the list $Tv_{1},\dots,Tv_m$ is a basis of range $T$.

Extending this basis to a basis of $W$ we see that $Tv_1,\dots,Tv_m$ is part of the basis of $W$. For the basis $v_1,\dots,v_m,\dots,v_n$ of $V$ and the basis $Tv_1,\dots,Tv_m,\dots,w_p$ of $W$ we have that $M(T)$ has a $1$ on row $j$ and column $j$ for $1\le j\le \dim \text{range }T$ as $Tv_j=1Tv_j$. All the other entries equal zero as the rest of the vectors in the basis get mapped to zero.

Is this proof correct?

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    $\begingroup$ Your proof is confused, as when you say "There exist $m$ basis vectors of $V$" it is not clear which basis you are talking about: no basis has been chosen yet. The whole point of this exercise is to choose appropriate bases, and you must be very clear about the process of choosing, notably about the order of choices. Since the situation is rather abstract you will need to allow the choices to be made by general principle like the fact that an independent family of vectors can be completed to a basis, but you need to be clear about the order in which things are done. $\endgroup$ Aug 11, 2022 at 5:53
  • $\begingroup$ @MarcvanLeeuwen Yeah I didn’t notice that. Thanks for pointing it out. What I was think when I wrote was that there are vectors that there exist vectors that don’t get mapped to 0. These have to be the vectors in some basis of $V$ as if that were not true, then every other vector would get mapped to zero. Which is false. After that I agree I should have said something as follows. Given that these vectors are linearly independent, we can extend it to a basis of $V$. Then proceed with the remaining argument. $\endgroup$
    – Seeker
    Aug 11, 2022 at 6:03
  • $\begingroup$ @MarcvanLeeuwen Considering the case $m > 0$ As it holds in the case $m=0$. $\endgroup$
    – Seeker
    Aug 11, 2022 at 6:09

2 Answers 2

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You've got the right general idea but your proof is overly complicated. Just observe that $Tv_1,\ldots,Tv_m$ are linearly independent since if $$\alpha_1Tv_1+\cdots+\alpha_m Tv_m=0$$ then $$T(\alpha_1v_1+\cdots+\alpha_mv_m)=0$$ by linearity of $T$, which implies that $\alpha_1v_1+\cdots+\alpha_mv_m$ is in the kernel of $T$. But this means $$\alpha_1v_1+\cdots+\alpha_mv_m=\alpha_{m+1}v_{m+1}+\cdots+\alpha_nv_n$$ for some scalars $\alpha_{m+1},\cdots,\alpha_n$. Subtracting and using linear independence of $v_1,\ldots,v_n$ it follows that $\alpha_1=\cdots\alpha_m=0$.

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Here is rapidly how one can get those bases. Let $n=\dim V$ and $\def\rk{\operatorname{rk}}\def\im{\operatorname{im}}r=\rk T=\dim\im(T)$ (that's your $m$), and $p=\dim W$. Choose a basis of $\ker(T)$ which by rank-nullity has $n-r$ elements so we can label them $v_{r+1},\ldots,v_n$. Complete this (with $r$ initial vectors) to a basis $v_1,\ldots,v_n$ of $V$. The images $T(v_i)$ span $\im(T)$, but the last $n-r$ of them are zero, so $T(v_1),\ldots,T(v_r)$ already span $\im(T)$, and since their number equals the dimension of $\im(T)$, they form a basis of that subspace*. Put $w_i=T(v_i)$ for $i=1,\ldots,r$, and complete to a basis $w_1,\ldots,w_p$ of $W$. Since $T(v_i)=w_i$ for $i\leq r$ and $T(v_i)=0$ otherwise, we get the desired matrix.

*One can alternatively show directly that $T(v_1),\ldots,T(v_r)$ are linearly independent: if some linear combination of these vectors is zero, then by linearity of $T$ the corresponding linear combination of $v_1,\ldots,v_r$ lies in $\ker(T)$, but then it is also a linear combination of the remaining basis vectors $v_{r+1},\ldots,v_n$, which is only possible for the trivial linear combination. Using this argument one can avoid the initial use of rank-nullity (just put $\dim \ker(T)=n-k$ for some $k$, use $k$ instead of $r$, and after the given argument conclude that $k$ was in fact the rank of $T$), which then in fact provides a proof of rank-nullity in passing.

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  • $\begingroup$ That is a really nice approach. Thanks! $\endgroup$
    – Seeker
    Aug 11, 2022 at 8:03

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