Houjun Liu


An isomorphism is an invertable Linear Map. Two vector spaces are called isomorphic if there is an isomorphism from one to another.

“A linear map that maintain the correct structure of the structure.”

This makes the vector spaces that are isomorphic “equivalent”, because the isomorphism is the equivalence relationship. Of course, they are still not equal.

Generally, isomorphisms can only be built between vector spaces over the same field.

additional information


We know we can represent Linear Maps as matricies.

So, given some \(A\), we have an inverse \(A^{-1}\).


\begin{equation} A A^{-1} = I = A^{-1} A \end{equation}

In this case, the \(I\) is the identity map: \(Iv = v\).

two vector spaces are isomorphic IFF they have the same dimension

note: this relationship works over the SAME field \(\mathbb{F}\), otherwise lin comb can’t work

Given vector spaces \(I,W\) isomorphic, we desire \(dim V = dim W\)

Suppose \(V\) and \(W\) are finite-dimensional vector spaces that are isomorphic. There means that there is an isomorphism, an invertable Linear Map between them which we will name \(T \in \mathcal{L}(V,W)\).

Because \(T\) is invertable, and injectivity and surjectivity implies invertability, so \(null\ T = \{0\}\) and \(range\ T = W\).

Lastly, we have that:

\begin{align} \dim V &= \dim null\ T + \dim range\ T \\ &= 0 + dim\ W \\ &= dim\ W \end{align}

as desired.

Given \(dim V = dim W\), show the vector spaces are isomorphic

Take \(v_1, \dots v_{n}\) a basis of \(V\), and \(w_1 \dots w_{n}\) a basis of \(W\).

Define a map by basis of domain mapping \(Tv_{j} = w_{j}\), that is, \(T(c_1v_1 + \dots + c_{n}v_{n}) = c_1 w_1 + \dots + c_{n} w_{n}\).

Because \(w_1 \dots w_{n}\) spans \(W\) (it is a basis after all), \(T\) is surjective.

An input with some set of \(c_{j}\) is in the null space of \(T\) if \(c_1 w_1 + \dots + c_{n}w_{n}\) adds up to \(0\) (by definition, as that’s the output of \(T\)).

Because \(w_1 \dots w_{n}\) is a basis, the only linear combination thereof which makes \(0\) is by taking all \(c_1 = \dots c_{n} = 0\). This make it so that the only valid input to \(T\) that will map to \(0\) requires \(c_1=\dots c_{n} = 0\), making \(null\ T = \{0\}\), showing that \(T\) is injective.

Having shown \(T\) is injective and surjective, it is an isomorphism, as desired. \(\blacksquare\)

matricies and Linear Maps from the right dimensions are isomorphic

Formally: suppose \(v_1 \dots v_{n}\) is a basis of \(V\), and \(w_1 \dots w_{m}\) is a basis of \(W\), then, \(\mathcal{M}\) the matrixify operation that takes Linear Maps and turn them into matricies is an isomorphism between \(\mathcal{L}(V,W)\) and \(\mathbb{F}^{m,n}\).

The matrixify operation \(\mathcal{M}\) is linear, because matricies are linear. The only thing that \(\mathcal{M}\) will turn into the zero matrix is the zero Linear Map (i.e. \(\mathcal{M}(t)=0 \implies T v_{k} = 0\ \forall k 1 \dots n\) by construction of matricies, and because the \(v_{k}\) are a basis, \(T v_{k} =0 \implies T=0\)), so the null space of \(\mathcal{M}\) is \(\{0\}\), making \(\mathcal{M}\) injective.

Now, because of the fact one can construct a matrix based on the scalars applied to map the input basis to the output basis; i.e. that, for any map \(T \in \mathcal{L}(V,W)\):

\begin{equation} Tv_{k} = \sum_{j=i}^{m}A_{j,k} w_{j} \end{equation}

for some matrix \(\mathcal{M}(T) = A \in \mathbb{F}^{m,n}\), we have that \(\mathcal{M}\) can be used to produce any map between \(V\) and \(W\). This makes \(\mathcal{M}\) surjective.

\(\dim \mathcal{L}(V,W) = (\dim V)(\dim W)\)

\(\mathcal{L}(V,W)\) is isomorphic to the set of matricies \(\mathbb{F}^{m,n}\) where \(w_1 \dots w_{m}\) is a basis for \(W\) and \(v_1 \dots v_{n}\) is a basis for \(V\). two vector spaces are isomorphic IFF they have the same dimension, so \(\dim \mathcal{L}(V,W) = \dim \mathbb{F}^{m,n} = m\cdot n\) (see \(\mathbb{F}^{m,n}\)).

Having claimed that \(w_1 \dots w_{m}\) is a basis of \(W\) and \(v_1 \dots v_{n}\) is a basis of \(V\), \(W\) and \(V\) have dimensions \(m\) and \(n\) respectively. So \((\dim V)(\dim W) = n \cdot m = m\cdot n = \dim \mathbb{F}^{m,n} = \dim \mathcal{L}(V,W)\), as desired.