Isomorphism

Definition

Definition 1(Definition).

An invertible linear transformation $A:V\to W$ is called an isomorphism.
Two vector spaces $V$ and $W$ are called isomorphic (denoted $V\cong W$) if there is an isomorphism $A:V\to W$.

Isomorphic spaces can be considered as different representations of the same space, meaning that all properties and constructions involving vector space operations are preserved under isomorphism.

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Properties

An isomorphism maps a basis to a basis

Theorem 2(Theorem).

Let $A:V\to W$ be an isomorphism, and let $S=\{{\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_n\}$ be a basis in $V$. Then $S^{\prime }=\{A{\mathbf{v}}_1,A{\mathbf{v}}_2,\dots A{\mathbf{v}}_n\}$ is a basis in $W$.

LADW, 6.6

Proof

$S^{\prime }$ is a spanning set in W

Since $S$ is a basis in $V$, $S$ must be a spanning set in $V$, and thus must permit a representation of every vector in $V$ as a linear combination of its elements.

$$\mathbf{x}=\sum _{k=1}^n{\alpha }_k{\mathbf{v}}_k\forall \mathbf{x}\in V$$ $$⟹A\mathbf{x}=\sum _{k=1}^n{\alpha }_{k}A{\mathbf{v}}_k\forall \mathbf{x}\in V$$

Also, since $A$ is invertible, $A$ must be surjective. Thus, for every $\mathbf{y}\in W$, $\exists \mathbf{x}\in V$ such that $\mathbf{y}=A\mathbf{x}$. Thus, $S^{\prime }$ is a spanning set in $W$.

$S^{\prime }$ is linearly independent

Say $S^{\prime }$ is not linearly independent. Then, some vector $\mathbf{w}\in W$ must permit representation as two distinct linear combinations of the vectors in $S^{\prime }$.

$$\mathbf{w}=\sum _{i=1}^n{\beta }_i(A{\mathbf{v}}_i)=\sum _{i=1}^n{\gamma }_i(A{\mathbf{v}}_i),$$

where ${\beta }_i\ne {\gamma }_i$ for some $i$. Now,

$$\begin{array}{rlr}& \sum _{i=1}^n{\beta }_i(A{\mathbf{v}}_i)=\sum _{i=1}^n{\gamma }_i(A{\mathbf{v}}_i)& \\ ⟹& A\left(\sum _{i=1}^n{\beta }_i{\mathbf{v}}_i\right)=A\left(\sum _{i=1}^n{\gamma }_i{\mathbf{v}}_i\right)& \\ ⟹& \sum _{i=1}^n{\beta }_i{\mathbf{v}}_i=\sum _{i=1}^n{\gamma }_i{\mathbf{v}}_i,& \text{A is injective}\\ ⟹& {\beta }_i={\gamma }_i\forall 1\le i\le n,& \text{S is a basis in V}\end{array}$$

We have reached a contradiction. Thus, $S^{\prime }$ must be linearly independent. ❏

In the above theorem, one can replace “basis” by “linearly independent”, “spanning”, or “linearly dependent” - all these properties are preserved under isomorphisms.

Also, If $A$ is an isomorphism, so is $A^{-1}$. Therefore, in the above theorem we can say that $S$ is a basis if and only if $S^{\prime }$ is a basis.

Basis to basis maps are isomorphisms

Theorem 3(Theorem).

Let $A:V\to W$ be a linear map, and let $S=\{{\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_n\}$ be a basis in $V$. If $A(S)$ is a basis in $W$, $A$ is an isomorphism.

LADW, 6.7

Proof
One way to prove this is to show $A$ is injective and surjective, which is easily done.

An alternative is to show $A$ is invertible. As we know, a linear map is defined by its values on a basis. Let $S^{\prime }=A(S)$. Define $A^{-1}$ by $A^{-1}(S^{\prime })=S$. It is easy to verify that $A^{-1}$ works as both a left inverse and right inverse of $A$. Thus, $A$ must be an isomorphism. ❏

Example 4(Example).

• ${\mathbb{P}}_n^{\mathbb{F}}\cong {\mathbb{F}}^{n+1}$: (${\mathbb{P}}_n^{\mathbb{F}}$ is the set of polynomials with degree at most $n$ and with coefficients in $\mathbb{F}$.) The standard basis in ${\mathbb{P}}_n^{\mathbb{F}}$ is $\{1,x,x^2,\dots ,x^n\}$. Let $A:{\mathbb{F}}^{n+1}\to {\mathbb{P}}_n^{\mathbb{F}}$ be defined by $A{\mathbf{e}}_1=1,A{\mathbf{e}}_2=x,\dots ,A{\mathbf{e}}_{n+1}=x^n$, where ${\mathbf{e}}_i$ are the standard basis vectors in ${\mathbb{F}}^{n+1}$. By the previous theorem, $A$ is an isomorphism, and the assertion follows.
• Any vector space over $\mathbb{F}$ is isomorphic to ${\mathbb{F}}^n$ for some $n$.
• $M_{m\times n}^{\mathbb{F}}\cong {\mathbb{F}}^{mn}$

Thus, to show two vector spaces are isomorphic, it is sufficient to construct a LT which maps the basis of one vector space onto another.

Putting it all together

Let $A:V\to W$ be a linear map. Consider the statements:

1. $S$ is a basis in $V$
2. $A(S)$ is a basis in $W$.
3. $A$ is an isomorphism (remember, this is just fancy talk for saying $A$ is invertible).

Theorem 2 states $1\&3⟹2$
Theorem 3 states $1\&2⟹3$.

Thus, the two theorems effectively say given $1$, we have $2⟺3$.

Here is a concise thought trail (Given $1$):

$$\begin{array}{rl}A\text{ is invertible}\underset{\text{defn}}{⟺}& (\text{Left inverse exists})\text{ and }(\text{Right inverse exists})\\ \underset{\ast }{⟺}& (A\text{ is injective})\text{ and }(A\text{ is surjective})\\ \underset{\ast \ast }{⟺}& (A(S)\text{ is linearly independent})\text{ and }\\ & (A(S)\text{ is a spanning set})\\ \underset{\text{defn}}{⟺}& A(S)\text{ is a basis in }W\end{array}$$

(*) is easy to show. Refer this.
(**) is important. Proof goes exactly like the proof of Theorem 2.

Speaking in terms of matrices, we get the following corollary to the two theorems:

Corollary

An $m\times n$ matrix is invertible if and only if its columns form a basis in ${\mathbb{F}}^m$.