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\begin{eqnarray} \text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{eqnarray} Theorem Let $T:V\rightarrow W$ be a linear map with $\scr{B}$ and $\scr{C}$ bases of $V$ and $W$ respectively. Then $\text{rank}(T)=\text{rank}([T]^{\scr{C}}_{\scr{B}})$. \begin{eqnarray} \text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{eqnarray} I've never seen or even heard of a "diagram chase argument," but my professor presented this diagram seen in figure 1 below

\begin{eqnarray} &V&\overset{T}{\longrightarrow}&&W&&\supset R(T)& \\ \cong&\downarrow{\tiny\varphi_{\scr{B}}}&&\cong&\downarrow{\tiny\varphi_{\scr{C}}}&&~~~\downarrow\alpha& \\ &F^n&\underset{L_A}{\longrightarrow}&&F^m&&\supset R(L_A)& , \end{eqnarray}

but how can I use this diagram-chase-argument method to show that $\varphi_{\scr{C}}$ induces an isomorphism, namely via $\alpha$?

\begin{eqnarray} \text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{eqnarray} \begin{eqnarray} \text{IF YOU HAVE A QUESTION, THEN ASK AND I CAN CLARIFY.} \end{eqnarray} \begin{eqnarray} \text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \end{eqnarray}

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  • $\begingroup$ What are $R(T)$ and $R(L_A)$? $\endgroup$ May 1, 2013 at 20:06
  • $\begingroup$ $L_A$ is left multiplication by some matrix $A$? $\endgroup$ May 1, 2013 at 22:07
  • $\begingroup$ $\alpha$ I suspect is transformation between the range of $T$ and the range of $L_A$. Your guess is as good as mine. I am trying to interpret my professor's mode of communication. $\endgroup$
    – Trancot
    May 1, 2013 at 23:50
  • $\begingroup$ @Martin You might want to remove your last comment. Without context they make no sense at all and are genuinely distracting. $\endgroup$
    – Trancot
    May 2, 2013 at 23:37

1 Answer 1

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Note $R(-)$ indicates taking the range of a function. The map $\alpha$ in the diagram is the restriction of the map $\phi_\mathscr{C}$ to $R(T)$. This restriction has image contained in $R(L_A)$.

Here is how a diagram chase shows that $\alpha$ is surjective:

  1. Assume $x \in R(L_A)$
  2. Consider $x \in F^m$
  3. By definition of $R(L_A)$ there is a $y \in F^n$ such that $y \mapsto x$ under $L_A$.
  4. As $\phi_\mathscr{B}$ is an isomorphism choose $z \in V$ such that $z \mapsto y$.
  5. By definition of $R(T)$ the element $T(z) \in W$ is contained in $R(T)$.
  6. As the diagram commutes and $z \mapsto y \mapsto x$ we have $z \mapsto T(z) \mapsto x$.
  7. Thus $\alpha$, being a restriction of $\phi_\mathscr{C}$, maps $T(z) \mapsto x$.
  8. Thus every $x \in R(L_A)$ has a preimage under $\alpha$.

I leave it to you to show that $\alpha$ is injective. Start with $x \in R(T)$ and assume $\alpha$ sends $x \mapsto 0$. Then just start pushing things around the diagram until you get $x = 0$.

That's how diagram chases work, btw. You pretty much aimlessly push things around the diagram until you eventually get what you want. Most of the time there are so few choices for what to do next the proof pretty much writes itself. Hence people never actually write out diagram chases as I've done here, they just say "by a diagram chase such-and-such is true".

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