# Question on proof for why $\operatorname{rank}(T) = \operatorname{rank}(LA)$

I'm curious about why an isomorphism between vector spaces would preserve the rank (dimension of range). I'm aware and understand why vector spaces can only be isomorphic if $\dim(V) = \dim(W)$. I certainly believe that the isomorphism would preserve rank and have included my reasoning below. I would appreciate comments and review, thanks!

First:

Show that an isomorphism from $F:V \to W$ maps a basis vector to a basis vector.
i.e: $F(v) = w$ where $v$ is a basis vector in $V$ and $w$ is a basis vector in $W$.

Now given the above is true, if we have $T: V \to W$, then if $\operatorname{rank}(T) = \dim(W)$ then we are done since we know there exists an isomorphism $G: W \to F^M$ where $M$ is the dimension of $W$ and $F^M$ is the standard representation of $W$ with respect to a basis $B$ in $W$. Hence $\dim(W) = \dim(F^M)$. So rank is preserved.

Now if $\operatorname{rank}(T) < \dim(W)$ then we know the isomorphism $G:W \to F^M$ will reflect the decreased rank since $G$ will only map over basis elements to basis elements. So looking at this range of $I$, we will preserve the dimension of the $\operatorname{range}(T)$ since fewer basis elements are mapped to representation in $F^M$.

Note: Is an easier way to show isomorphisms preserves rank by looking at subspaces?

Thanks!

Your basic idea is correct, but it is worthwhile to be more precise about what structure an isomorphism preserves. First, note that there isn't a notion of a "basis vector". A basis for $V$ is always collection of linearly independent vectors that spans $V$. The most important thing to remember is that a vector space $V$ has many different bases so it makes no sense to talk about a "basis vector" without specifying some specific basis for $V$.

Now, let $T \colon V \rightarrow W$ be an isomorphism of vector spaces. Then we have the following useful properties:

1. A collection $(v_1, \dots, v_k)$ of vectors in $V$ is linearly independent (in $V$) if and only if $(Tv_1, \dots, Tv_k)$ is linearly independent (in $W$).
2. A collection $(v_1, \dots, v_k)$ of vectors in $V$ spans $V$ if and only if $(Tv_1, \dots, Tv_k)$ spans $W$.

The properties above immediately imply the following:

1. Note that $(1) + (2)$ imply that if is an isomorphism and $(v_1, \dots, v_n)$ is some basis for $V$ then $(Tv_1, \dots, Tv_n)$ is a basis of $W$. In particular, we must have $\dim V = \dim W$.
2. Let $U \subseteq V$ be a subspace. We can restrict $T$ to $U$ and get a map $T|_{U} \colon U \rightarrow T(U)$. Since $T$ was an isomorphism (one-to-one and onto), so is $T|_{U}$ is also an isomorphism and we have $\dim U = \dim T|_{U}(U) = \dim T(U)$. That is, isomorphisms preserve the dimension of subspaces.

Now, let $T \colon V \rightarrow W$ be a linear map (not neccesary and isomorphism) and choose some bases $\gamma = (v_1,\dots,v_n)$ for $V$ and $\beta = (w_1,\dots,w_m)$ for $W$. Using $\gamma$, we can define a linear map $\gamma \colon \mathbb{F}^n \rightarrow V$ (also denoted by $\gamma$ in order to not introduce new notation) by requiring that $\gamma(e_i) = v_i$ where $(e_1,\dots,e_n)$ is the standard basis of $\mathbb{F}^n$. Since $(v_1,\dots,v_n)$ is a basis, the map $\gamma$ is an isomorphism. Similarly, $\beta$ defines a map $\beta \colon \mathbb{F}^m \rightarrow W$. Using the notation above, the map $L_{[T]^{\gamma}_{\beta}} \colon \mathbb{F}^n \rightarrow \mathbb{F}^m$ is precisely the map $\beta^{-1} \circ T \circ \gamma$.

Let us prove that $\operatorname{rank}(T) = \operatorname{rank}(\beta^{-1} \circ T \circ \gamma)$. We have

$$\operatorname{rank}(T) = \dim T(V) = \dim T(\gamma(\mathbb{F}^n)) = \dim \beta^{-1}(T(\gamma(\mathbb{F}^n))) \\ = \dim (\beta^{-1} \circ T \circ \gamma)(\mathbb{F}^n) = \operatorname{rank}(\beta^{-1} \circ T \circ \gamma).$$

Make sure you can justify each equality using the definitions and the properties mentioned above. Then try to show to run the same argument for $\ker$.