Isomorphisms

Definition. A linear transformation $T: V \to W$ is an isomorphism if there is a linear transformation, usually denoted by $T^{-1}: W \to V$, satisfying $T^{-1}(T(v)) = v ~\text{for all}~ v \in V$ and $T(T^{-1}(w)) = w ~\text{for all}~ w \in W$ In other words, $T^{-1} \circ T = \operatorname{Id}_V$ and $T \circ T^{-1} = \operatorname{Id}_W$

Proposition. A linear transformation $T: V \to W$ is an isomorphism if and only if $T$ is a bijection.

Proof. (Expand to view)

If $T$ is an isomorphism, then $T^{-1}$ is the inverse function to $T$. So $T$ is a bijection.

Assume that $T$ is a linear transformation and also a bijection. Since it is a bijection, there is an inverse function $S : W \to V$ of sets. We want to see that $S$ is also a natural transformation.

Pick $w \in W$. Since $T$ is surjective, we can write it as $w = T(v)$. Then,
$S(cw) = S(cT(v)) = S(T(cv)) = cv = cS(w)$ Similarly, if we have $w,w^\prime \in W$, we can write $w = T(v)$ and $w^\prime = T(v^\prime)$. Then, $S(w + w^\prime) = S(T(v) + T(v^\prime)) = ST(v+v^\prime) = v + v^\prime = S(w) + S(w^\prime)$ Thus, $S$ is a linear transformation and $T$ is an isomorphism. ■

Two vector spaces that are related by an isomorphism are called isomorphic and we write $V \cong W.$ Isomorphic vector spaces are not the same, ie equal. But, since they differ by a linear relabeling, they are the same by any measure of linear algebra.

Example. For a field $k$, the vector spaces of $m\times n$ matrices, $\operatorname{Mat}_{m,n}(k)$, and $k^{mn}$ are isomorphic.

We consider the function $\begin{aligned} T : \operatorname{Mat}_{m,n}(k) & \to k^{mn} \\ (A_{ij}) & \mapsto \begin{pmatrix} A_{11} & \cdots & A_{mn} \end{pmatrix}^T \end{aligned}$ which just lists the entries of the matrix reading left to right and then moving down to the next row.

This is a bijection since we can take the length $mn$ list and break it up into $m$ intervals of length $n$ to get the rows of the matrix.

Since $c(A_{ij})= (cA_{ij})$ and $(A_{ij}) + (B_{ij}) = (A_{ij}+B_{ij})$ are just componentwise scaling and addition, we see the result is the same whether we scale or add before or after applying $T$. Thus, we have an isomorphism.

Example. Let $A$ be an $n \times n$ matrix and take $\begin{aligned} T : k^n & \to k^n \\ \mathbf{v} & \mapsto A \mathbf{v} \end{aligned}$ Then, $T$ is an isomorphism if and only if $A$ is an invertible matrix.

Indeed, if we have an inverse matrix $A^{-1}$, multiplication by $A^{-1}$ gives the inverse function $T^{-1}(\mathbf{v}) := A^{-1}\mathbf{v}$. $A(A^{-1}\mathbf{v}) = I \mathbf{v} = \mathbf{v} \\ A^{-1}(A\mathbf{v}) = I \mathbf{v} = \mathbf{v} \\$

In the other direction, assume we have an inverse linear transformation $T^{-1} : k^n \to k^n$. Using the standard basis for both copies of $k^n$ we get a matrix representation $B$ of $T^{-1}$. This satisfies $\mathbf{v} = T(T^{-1}(\mathbf{v})) = AB\mathbf{v}$ for all $\mathbf{v}$ and $\mathbf{v} = T^{-1}(T(\mathbf{v}) = BA\mathbf{v}$ Plugging in $\mathbf{e}_i$, $AB\mathbf{e}_i$ is the i-th column of $AB$. Since $AB\mathbf{e}_i = \mathbf{e}_i$ for all $i$, we see that $AB = I$. Similarly, $BA = I$.

Proposition. Isomorphic vector spaces have equal dimensions.

Proof. (Expand to view)

Let $T: V \to W$ be an isomorphism of vector spaces. And let $v_i, i \in B$ be a basis for $V$. We claim that $T(v_i), i \in B$ is a basis for $W$. Assuming this claim for a moment, since $T : \lbrace v_i \mid i \in B \rbrace \to \lbrace T(v_i) \mid i \in B \rbrace$ is a bijection, then number of $v_i$ equals the number of $T(v_i)$. Thus, the dimensions of $V$ and $W$ coincide.

Now let’s check that $T(v_i), i \in B$ forms a basis for $W$. Assume we have a linear relation $0 = \sum_{i \in B} c_i T(v_i)$ Then, since $0=\sum_{i \in B} c_i T(v_i) = T\left( \sum_{i \in B} c_iv_i \right)$ $T(0) = 0$ and $T$ is injective, we have $0 = \sum_{i \in B} c_iv_i$ This is a linear relation amongst elements of a basis. Thus $c_i = 0$ for all $i$. So $T(v_i), i \in $ is a linearly independent set.

Now pick $w \in W$. Since $T$ is surjective, we know there is some $v \in V$ with $T(v) = w$. Since $v_i$ span, we can write $v = \sum_{i \in B} c_i v_i$ Applying $T$ we have $w = T(v) = T\left( \sum_{i \in B} c_i v_i \right) = \sum_{i \in B} c_i T(v_i)$ So $w$ is in the span of the $T(v_i)$. ■

The converse statement also holds.

Proposition. If two $k$-vector spaces have the same dimension, then they are isomorphic.

Proof. (Expand to view)

Let $V$ and $W$ be $k$-vector spaces with $\dim V = \dim W$. Pick bases $v_1,\ldots,v_d$ and $w_1,\ldots,w_d$ for $V$ and for $W$. Then we have seen that there exists unique linear transformation $T: V \to W$ with $T(v_i) = w_i$ Similarly, there exists a unique line $S:W \to V$ with $S(w_i) = v_i$ Note that for each $i$ $ST(v_i) = v_i$ Thus, since they both act the same on a basis, we have $ST = \operatorname{Id}_V$. Similarly we have $$TS = \operatorname{Id}_W$. So we see that $T$ is an isomorphism. ■

Combining the two statements, we have the following.

Theorem. Two vector spaces are isomorphic if and only if they have the same dimension.

This theorem says that the dimension of $V$ captures all the information about it.