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Let $T : V \to W$ be a linear transformation from an $n$-dimensional vector space $V$ to an $m$-dimensional vector space $W$. Let $\beta$ and $\gamma$ be ordered bases for $V$ and $W$, respectively. Prove that $\operatorname{rank}(T) = \operatorname{rank}(L_A)$ and $\operatorname{nul} (T) = \operatorname{nul}(L_A)$, where $A = [T]_{\beta}^{\gamma}$.

Proof: Note that $L_A : F^n \to F^m$ is defined by $L_A(x) = Ax$, and recall that $f_{\beta} : W \to F^m$ defined by $f_{\gamma}(w) = [w]_\gamma$ is an isomorphism and therefore preserves dimension under images ($f_{\beta}$ is defined similarly). Because $R(T)$ is a subspace of $W$, $f_{\gamma}(R(T))$ will be a subspace of $F^m$ and have the same dimension as $R(T)$. But note that

\begin{align*} f_{\gamma}(R(T)) &= \{f_{\gamma}(T(v)) \mid v \in V\} \\ &= \{ [T(v)]_{\gamma} \mid v \in V\} \\ &= \{[T]_{\beta}^{\gamma} [v]_{\beta} \mid v \in V\} \\ &= \{A[v]_{\beta} \mid v \in V\}. \\ \end{align*}

Since $f_{\beta} : V \to F^n$ is an isomorphism, the $v$'s in $V$ can be put in a one-to-one correspondence with the $x$'s in $F^n$. Therefore,

$$f_{\gamma}(R(T)) = \{Ax \mid x \in F^n\} = R(L_A),$$

and finally $\dim(R(T)) = \dim(R(L_A))$ or $\operatorname{rank}(T) = \operatorname{rank}(L_A)$. By the rank-nullity theorem we have

$$\operatorname{rank}(T) + \operatorname{nul}(T) = \dim(V) = n$$

and

$$\operatorname{rank}(L_A) + \operatorname{nul}(L_A) = \dim(F^n) = n$$

and therefore $\operatorname{nul}(T) = \operatorname{nul}(L_A)$. $\Box$

How does this sound?

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  • $\begingroup$ I guess that $R(T)$ is the range of $T$, right? $\endgroup$
    – Masacroso
    Jul 15, 2017 at 23:13
  • $\begingroup$ @Masacroso Yes. That is right. $\endgroup$
    – user193319
    Jul 16, 2017 at 11:37

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Your proof is correct. But it is a bit misleading to say that “[s]ince $f_{\beta} : V \to F^n$ is an isomorphism, the $v$'s in $V$ can be put in a one-to-one correspondence with the $x$'s in $F^n$” (emphasis by me) since you don’t just want some one-to-one correspondence, but a very specific one. It is therefore probably better to say something like “since $f_{\beta} = [\,\cdot\,]_\beta$ is a one-to-one correspondence, it follows that […]”.

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  • $\begingroup$ I realize that the question may not be of interest anymore to the OP, but I think it still deserves an answer. $\endgroup$ Sep 7, 2018 at 15:27

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