Vector Space Isomorphisms

# Vector Space Isomorphisms

Definition: Let $V$ and $W$ be vector spaces over the field $\mathbb{F}$. Then $V$ and $W$ are said to be Isomorphic denoted $V \cong W$ if there exists an invertible linear map $T \in \mathcal L (V, W)$. |

We will now look at some important propositions and theorems regarding two vector spaces being isomorphic

Proposition 1: If $V, W$ are vector spaces over the field $\mathbb{F}$ and $V \cong W$, then if either $V$ or $W$ is known to be finite dimensional, then the other vector space is also finite dimensional and $\mathrm{dim} V = \mathrm{dim} W$. |

**Proof:**Let $V \cong W$ and let $T \in \mathcal L (V, W)$ be an invertible linear map from $V$ to $W$ and let $T^{-1} \in \mathcal L (W, V)$ be the inverse of $T$. We have two cases to consider.

**Case 1:**Suppose that $\mathrm{dim} V = n$. Now note that $\mathrm{dim} V = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T))$. Now since $T$ is an invertible linear map, then $T$ is bijective, and more importantly, $T$ is injective, and so $\mathrm{null} (T) = \{ 0 \}$ so $\mathrm{dim} ( \mathrm{null} (T)) = 0$. Thus $\mathrm{dim} V = \mathrm{dim} ( \mathrm{range} (T))$.

- Now since $T$ is surjective, we have that $\mathrm{range} (T) = W$, and so more precisely,

\begin{align} \mathrm{dim} V = n = \mathrm{dim} W \end{align}

**Case 2:**Now instead suppose that $\mathrm{dim} W = n$. Since $T : V \to W$ then $T^{-1} : W \to V$. So we have that $\mathrm{dim} W = \mathrm{dim} (\mathrm{null} (T^{-1})) + \mathrm{dim} ( \mathrm{range} (T^{-1} ))$. Now since $T^{-1}$ is an invertible linear map, then $T^{-1}$ is bijective, and more importantly, $T^{-1}$ is injective, and so $\mathrm{null} (T^{-1}) = \{ 0 \}$ so $\mathrm{dim} ( \mathrm{null} (T)) = 0$. Thus, $\mathrm{dim} W = \mathrm{dim} ( \mathrm{range} (T^{-1}))$.

- Now since $T^{-1}$ is surjective, we have that $\mathrm{range} (T^{-1}) = V$, and so more precisely,

\begin{align} \mathrm{dim} W = n = \mathrm{dim} V \end{align}

- In both cases, we see that if one of $V$ or $W$ is finite dimensional, then the other is also finite dimensional. $\blacksquare$

Theorem 1: Let $V$ and $W$ be finite dimensional vector spaces over the field $\mathbb{F}$. Then $V \cong W$ if and only if $\mathrm{dim} V = \mathrm{dim} W$. |

**Proof:**$\Leftarrow$ Suppose that $V \cong W$. Then there exists an invertible linear map $T \in \mathcal L (V, W)$. Since $T$ is invertible, we have that $T$ is bijective. Therefore $T$ is injective so $\mathrm{null} (T) = \{ 0 \}$ and also $T$ is surjective so $\mathrm{range} (T) = W$ and thus we have that:

\begin{align} \mathrm{dim} V = \mathrm{dim} (\mathrm{null} (T)) + \mathrm{dim} (\mathrm{range} (T)) = 0 + \mathrm{dim} W = \mathrm{dim} W \end{align}

- $\Rightarrow$ Suppose that $\mathrm{dim} V = \mathrm{dim} W$, and let $\{ v_1, v_2, ..., v_n \}$ be a basis of $V$ and let $\{ w_1, w_2, ..., w_n \}$ be a basis of $W$. Then define $T \in \mathcal L (V, W)$ by $T(a_1v_1 + a_2v_2 + ... + a_nv_n) = a_1w_1 + a_2w_2 + ... + a_nw_n$.

- We note that $T$ is surjective since $\{ w_1, w_2, ..., w_n \}$ spans $W$ be the definition of a basis, and also $T$ is injective since $\{ w_1, w_2, ..., w_n \}$ is a linearly independent set, once again by the definition of a basis. Therefore $T$ is bijective and hence invertible, so $V$ is isomorphic to $W$, i.e, $V \cong W$. $\blacksquare$