Elementary properties of matrix represention of a linear map Definition: Let $V,W$ be finite dimensional vector space over field $F$. Let $B=\{\alpha_1,…\alpha_n\}$ and $B’=\{\beta_1,…,\beta_m\}$ be ordered basis of $V$ and $W$, respectively. Let $T:V\to W$ be a linear map. Matrix of $T$ relative to ordered bases $B$, $B’$ is defined as $[T]_{B}^{B’}$ $=([T(\alpha_i)]_{B’}\dotsb [T(\alpha_n)]_{B’})\in M_{m\times n}(F)$, where $[T(\alpha_i)]_{B’}$ is coordinate matrix of $T(\alpha_i)\in W$ relative to ordered basis $B’$.


Let $V,W$ be finite dimensional vector space over field $F$. Let $B=\{\alpha_1,…\alpha_n\}$ and $B’=\{\beta_1,…,\beta_m\}$ be ordered basis of $V$ and $W$, respectively. If $T,U:V\to W$ is linear map, then:
$(1)$ $T=U$$\iff$$[U]_{B}^{B’}=[T]_{B}^{B’}$.
$(2)$ $[T+U]_{B}^{B’}=[T]_{B}^{B’}+[U]_{B}^{B’}$.
$(3)$ $[c\cdot T]_{B}^{B’}=c\cdot [T]_{B}^{B’}$, for all $c\in F$.

My attempt: $(1)$ Suppose $U=T$. Since $U=T$, we have $T(\alpha_i)=U(\alpha_i)\in W$, $\forall i\in J_n$. So $[T(\alpha_i)]_{B’}=[U(\alpha_i)]_{B’}$, $\forall i\in J_n$. Thus $[T]_{B}^{B’}$ $=([T(\alpha_i)]_{B’}\dotsb [T(\alpha_n)]_{B’})$ $= ([U(\alpha_i)]_{B’}\dotsb [U(\alpha_n)]_{B’})$ $=[U]_B^{B’}$. Hence $[U]_{B}^{B’}=[T]_{B}^{B’}$. Conversely, let $[U]_{B}^{B’}=[T]_{B}^{B’}$. Then $[T(\alpha_i)]_{B’}=[U(\alpha_i)]_{B’}$, $\forall i\in J_n$. Since $f:W\to F^{m\times 1}$ defined by $f(w)=[w]_{B’}$ is bijective (here is proof), we have $T(\alpha_i)=U(\alpha_i)$, $\forall i\in J_n$. Let $\alpha \in V$. Since $\mathrm{span}(B)=V$, we have $\alpha=\sum_{i\in J_n}a_i\cdot_V \alpha_i$. Since $T,U$ is linear map, $T(\alpha)$ $=T(\sum_{i\in J_n}a_i\cdot_V \alpha_i)$ $= \sum_{i\in J_n}a_i\cdot_W T(\alpha_i)$ $= \sum_{i\in J_n}a_i\cdot_W U(\alpha_i)$ $= U(\sum_{i\in J_n}a_i\cdot_V \alpha_i)$ $=U(\alpha)$. Thus $T(\alpha)=U(\alpha)$, $\forall \alpha \in V$. Hence $T=U$.
$(2)$ $[T+U]_{B}^{B’}$ $=([T+U(\alpha_1)]_{B’}\dotsb [T+U(\alpha_n)]_{B’})$ $= ([T(\alpha_1)+U(\alpha_1)]_{B’}\dotsb [T(\alpha_n)+U(\alpha_n)]_{B’})$. Since $f:W\to F^{m\times 1}$ defined by $f(w)=[w]_{B’}$ is linear, (here is proof), we have $[T(\alpha_i)+U(\alpha_i)]_{B’}$ $= [T(\alpha_i)]_{B’}+ [U(\alpha_i)]_{B’}$, $\forall i\in J_n$. So $([T(\alpha_1)+U(\alpha_1)]_{B’}\dotsb [T(\alpha_n)+U(\alpha_n)]_{B’})$ $= ([T(\alpha_1)]_{B’}+ [U(\alpha_1)]_{B’}\dotsb [T(\alpha_n)]_{B’}+ [U(\alpha_n)]_{B’})$ $=[T]_B^{B’}+[U]_B^{B’}$. Hence $[T+U]_{B}^{B’}=[T]_{B}^{B’}+[U]_{B}^{B’}$.
$(3)$ Let $c\in F$. Then $[c\cdot T]_{B}^{B’}$ $=([c\cdot T(\alpha_1)]_{B’}\dotsb [c \cdot T(\alpha_n)]_{B’})$ $= ([c\cdot_W T(\alpha_1)]_{B’}\dotsb [c \cdot_W T(\alpha_n)]_{B’})$. Since $f:W\to F^{m\times 1}$ defined by $f(w)=[w]_{B’}$ is linear, (here is proof), we have $[c\cdot_W T(\alpha_i)]_{B’}$ $=c \cdot [T(\alpha_i)]_{B’}$, $\forall i\in J_n$. So $([c\cdot_W T(\alpha_1)]_{B’}\dotsb [c \cdot_W T(\alpha_n)]_{B’})$ $= (c\cdot [T(\alpha_1)]_{B’}\dotsb c \cdot [T(\alpha_n)]_{B’})$ $=c\cdot ([T(\alpha_1)]_{B’}\dotsb [T(\alpha_n)]_{B’})$ $=c\cdot [T]_{B}^{B’}$. Hence $[c\cdot T]_{B}^{B’}=c\cdot [T]_{B}^{B’}$, for all $c\in F$. Is my proof correct?
 A: This is very thorough, well thought out, and logically rigorous. You leave me no doubt whatsoever that you completely understand every step of the proof, which is the most important thing when writing proofs in elementary courses. Little touches like distinguishing the three scalar multiplications in the proof (applying to $V$, to $W$, and to $F^n$), citing external results (which I saw that you wrote, I didn't check, but I know to be true), and helpfully providing relevant definitions before the proof, all show me that you are completely on top of this. If I were assessing this proof, I wouldn't be able to take a half-mark off, even if I wanted to!
So, since you put so much effort into this, let me scrape together some pointers to improve your proof:

*

*In the proof of (1), the step "Since $\mathrm{span}(B)=V$, we have $\alpha=\sum_{i\in J_n}a_i\cdot_V \alpha_i$." could use a little touch-up. Specifically, you introduce new pronumerals $a_1, \ldots, a_n$ without a quantifier. I would write, instead, "Since $\mathrm{span}(B)=V$, there exist $a_1, \ldots, a_n \in F$ such that $\alpha=\sum_{i\in J_n}a_i\cdot_V \alpha_i$." As it is written, a reader could conceivably wonder where these new variables $a_i$ came from.

*The notation $J_n$ is not standard. While I'm sure it's a standard notation in whatever textbook you're using, this may not be clear to a general reader. I understood it to be $\{1, 2, \ldots, n\}$, and this made sense in the proof. But, I've also seen $J_n$ refer to the $n \times n$ matrix whose entries are all $1$s in a linear algebra context. I wouldn't necessarily worry if this is standard in your course, but since you presented this proof for general consumption on this site, I think it would be worth adding $J_n$ to the definitions at the top of your proof.

*Despite said helpful definitions, I did briefly struggle to understand what $\cdot_V$ and $\cdot_W$ referred to. Writing them like this makes it obvious, but in situ, writing something like
$$\alpha=\sum_{i\in J_n}a_i\cdot_V \alpha_i$$
made it not immediately clear. I wasn't sure that the $_V$ was attached to the $\cdot$. I was wondering if the $\cdot_V$ was attached to the subscript $i$ of the $a$ somehow? As you can see, I did work it out (and without too much consternation), but it might be helpful to do one of two things: either point out previously in he proof what $\cdot_V$, $\cdot_W$, and $\cdot$ mean (e.g. rather than introduce $V$, you could introduce $(V, +_V, \cdot_V)$), or (as you will, and should, do eventually) denote them all as $\cdot$, and trust the reader to understand which scalar multiplication is meant, from context.

*This brings me to a final point: readability. You present your proofs, on this site at least, in a very dense fashion. Give the words and formulas some breathing room! It will make it much easier for the reader to follow your arguments. For example, here is how I would present your proof for (1), without changing a single word or formula:


Suppose $U=T$. Since $U=T$, we have
$$T(\alpha_i)=U(\alpha_i)\in W, \forall i\in J_n.$$
So
$$[T(\alpha_i)]_{B’}=[U(\alpha_i)]_{B’}, \forall i\in J_n.$$
Thus
$$[T]_{B}^{B’}=([T(\alpha_i)]_{B’}\dotsb [T(\alpha_n)]_{B’})= ([U(\alpha_i)]_{B’}\dotsb [U(\alpha_n)]_{B’})=[U]_B^{B’}.$$
Hence $[U]_{B}^{B’}=[T]_{B}^{B’}$.
Conversely, let $[U]_{B}^{B’}=[T]_{B}^{B’}$. Then
$$[T(\alpha_i)]_{B’}=[U(\alpha_i)]_{B’}, \forall i\in J_n.$$
Since $f:W\to F^{m\times 1}$ defined by $f(w)=[w]_{B’}$ is bijective (here is proof), we have $$T(\alpha_i)=U(\alpha_i), \forall i\in J_n.$$
Let $\alpha \in V$. Since $\mathrm{span}(B)=V$, we have
$$\alpha=\sum_{i\in J_n}a_i\cdot_V \alpha_i.$$
Since $T,U$ is linear map,
\begin{align}
T(\alpha)&=T\left(\sum_{i\in J_n}a_i\cdot_V \alpha_i\right)\\
&= \sum_{i\in J_n}a_i\cdot_W T(\alpha_i)\\
&= \sum_{i\in J_n}a_i\cdot_W U(\alpha_i)\\
&= U\left(\sum_{i\in J_n}a_i\cdot_V \alpha_i\right)\\
&=U(\alpha).
\end{align}
Thus $T(\alpha)=U(\alpha)$, $\forall \alpha \in V$. Hence $T=U$.

Yes, it takes much more space, but it's easier to read.
