# 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!

## 1 Answer

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 $\ker$.