# 5.6: Isomorphisms

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## Outcomes

1. Determine if a linear transformation is an isomorphism.
2. Determine if two subspaces of $$\mathbb{R}^n$$ are isomorphic.

Recall the definition of a linear transformation. Let $$V$$ and $$W$$ be two subspaces of $$\mathbb{R}^{n}$$ and $$\mathbb{R}^{m}$$ respectively. A mapping $$T:V\rightarrow W$$ is called a linear transformation or linear map if it preserves the algebraic operations of addition and scalar multiplication. Specifically, if $$a,b$$ are scalars and $$\vec{x},\vec{y}$$ are vectors,

$T\left( a\vec{x}+b\vec{y}\right) =aT(\vec{x})+bT(\vec{y})\nonumber$

Consider the following important definition.

## Definition $$\PageIndex{1}$$: Isomorphism

A linear map $$T$$ is called an isomorphism if the following two conditions are satisfied.

• $$T$$ is one to one. That is, if $$T(\vec{x})=T(\vec{y}),$$ then $$\vec{x}=\vec{y}.$$
• $$T$$ is onto. That is, if $$\vec{w}\in W,$$ there exists $$\vec{v}\in V$$ such that $$T(\vec{v})=\vec{w}$$.

Two such subspaces which have an isomorphism as described above are said to be isomorphic.

Consider the following example of an isomorphism.

## Example $$\PageIndex{1}$$:Isomorphism

Let $$T: \mathbb{R}^2 \mapsto \mathbb{R}^2$$ be defined by $T \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} x + y \\ x - y \end{array} \right ] \nonumber$ Show that $$T$$ is an isomorphism.

Solution

To prove that $$T$$ is an isomorphism we must show

1. $$T$$ is a linear transformation;
2. $$T$$ is one to one;
3. $$T$$ is onto.

We proceed as follows.

1. $$T$$ is a linear transformation:

Let $$k, p$$ be scalars. \begin{aligned} T \left( k \left [ \begin{array}{c} x_1 \\ y_1 \end{array} \right ] + p \left [ \begin{array}{c} x_2 \\ y_2 \end{array} \right ] \right) &= T \left( \left [ \begin{array}{c} kx_1 \\ ky_1 \end{array} \right ] + \left [ \begin{array}{c} px_2 \\ py_2 \end{array} \right ] \right) \\ &= T \left( \left [ \begin{array}{c} kx_1 + px_2 \\ ky_1 + py_2 \end{array} \right ] \right) \\ &= \left [ \begin{array}{c} (kx_1 + px_2) + (ky_1 + py_2) \\ (kx_1 + px_2) - (ky_1 + py_2) \end{array} \right ] \\ &= \left [ \begin{array}{c} (kx_1 + ky_1) + (px_2 + py_2) \\ (kx_1 - ky_1) + (px_2 - py_2) \end{array} \right ] \\ &= \left [ \begin{array}{c} kx_1 + ky_1 \\ kx_1 - ky_1 \end{array} \right ] + \left [ \begin{array}{c} px_2 + py_2 \\ px_2 - py_2 \end{array} \right ] \\ &= k \left [ \begin{array}{c} x_1 + y_1 \\ x_1 - y_1 \end{array} \right ] + p \left [ \begin{array}{c} x_2 + y_2 \\ x_2 - y_2 \end{array} \right ] \\ &= k T \left( \left [ \begin{array}{c} x_1 \\ y_1 \end{array} \right ] \right) + p T \left( \left [ \begin{array}{c} x_2 \\ y_2 \end{array} \right ] \right)\end{aligned}

Therefore $$T$$ is linear.

2. $$T$$ is one to one:

We need to show that if $$T (\vec{x}) = \vec{0}$$ for a vector $$\vec{x} \in \mathbb{R}^2$$, then it follows that $$\vec{x} = \vec{0}$$. Let $$\vec{x} = \left [ \begin{array}{c} x \\ y \end{array} \right ]$$.

$T \left( \left [ \begin{array}{c} x \\ y \end{array} \right ] \right) = \left [ \begin{array}{c} x + y\\ x - y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$ This provides a system of equations given by \begin{aligned} x + y &= 0\\ x - y &= 0\end{aligned} You can verify that the solution to this system if $$x = y =0$$. Therefore $\vec{x} = \left [ \begin{array}{c} x \\ y \end{array} \right ] = \left [ \begin{array}{c} 0 \\ 0 \end{array} \right ]\nonumber$ and $$T$$ is one to one.

3. $$T$$ is onto:

Let $$a,b$$ be scalars. We want to check if there is always a solution to $T \left( \left [ \begin{array}{c} x \\ y \end{array} \right ] \right) = \left [ \begin{array}{c} x + y\\ x - y \end{array} \right ] = \left [ \begin{array}{c} a \\ b \end{array} \right ]\nonumber$

This can be represented as the system of equations \begin{aligned} x + y &= a\\ x - y &= b\end{aligned}

Setting up the augmented matrix and row reducing gives $\left [ \begin{array}{cc|c} 1 & 1 & a \\ 1 & -1 & b \end{array} \right ] \rightarrow \cdots \rightarrow \left [ \begin{array}{cc|c} 1 & 0 & \frac{a+b}{2} \\ 0 & 1 & \frac{a-b}{2} \end{array} \right ]\nonumber$ This has a solution for all $$a,b$$ and therefore $$T$$ is onto.

Therefore $$T$$ is an isomorphism.

Below is a video on one to one and onto functions (isomorphisms).

An important property of isomorphisms is that its inverse is also an isomorphism.

## Proposition $$\PageIndex{1}$$:Inverse of an Isomorphism

Let $$T:V\rightarrow W$$ be an isomorphism and $$V,W$$ be subspaces of $$\mathbb{R}^n$$. Then $$T^{-1}:W\rightarrow V$$ is also an isomorphism.

Proof

Let $$T$$ be an isomorphism. Since $$T$$ is onto, a typical vector in $$W$$ is of the form $$T(\vec{v})$$ where $$\vec{v} \in V$$. Consider then for $$a,b$$ scalars, $T^{-1}\left( aT(\vec{v}_{1})+bT(\vec{v}_{2})\right)\nonumber$ where $$\vec{v}_{1}, \vec{v}_2 \in V$$. Is this equal to $aT^{-1}\left( T (\vec{v}_{1})\right) +bT^{-1}\left( T(\vec{v}_{2})\right) =a\vec{v}_{1}+b\vec{v}_{2}?\nonumber$ Since $$T$$ is one to one, this will be so if $T\left( a\vec{v}_{1}+b\vec{v}_{2}\right) =T\left( T^{-1}\left( aT(\vec{v}_{1})+bT(\vec{v}_{2})\right) \right) =aT(\vec{v}_{1})+bT(\vec{v}_{2}).\nonumber$ However, the above statement is just the condition that $$T$$ is a linear map. Thus $$T^{-1}$$ is indeed a linear map. If $$\vec{v} \in V$$ is given, then $$\vec{v}=T^{-1}\left( T(\vec{v})\right)$$ and so $$T^{-1}$$ is onto. If $$T^{-1} (\vec{v})=0,$$ then $\vec{v}=T\left( T^{-1}(\vec{v})\right) =T(\vec{0})=\vec{0}\nonumber$ and so $$T^{-1}$$ is one to one.

Another important result is that the composition of multiple isomorphisms is also an isomorphism.

## Proposition $$\PageIndex{2}$$:Composition of Isomorphisms

Let $$T:V\rightarrow W$$ and $$S:W\rightarrow Z$$ be isomorphisms where $$V,W,Z$$ are subspaces of $$\mathbb{R}^n$$. Then $$S\circ T$$ defined by $$\left( S\circ T\right) \left( \vec{v} \right) = S\left( T\left( \vec{v} \right) \right)$$ is also an isomorphism.

Proof

Suppose $$T:V\rightarrow W$$ and $$S:W\rightarrow Z$$ are isomorphisms. Why is $$S\circ T$$ a linear map? For $$a,b$$ scalars,

\begin{aligned} S\circ T\left( a\vec{v}_{1}+b(\vec{v}_{2})\right) &= S\left( T\left(a\vec{v}_{1}+b\vec{v}_{2}\right) \right) =S\left( aT\vec{v}_{1}+bT\vec{v}_{2}\right) \\ &=aS\left( T\vec{v}_{1}\right) +bS\left( T\vec{v}_{2}\right) = a\left( S\circ T\right) \left( \vec{v}_{1}\right) +b\left( S\circ T\right) \left( \vec{v}_{2}\right)\end{aligned}\nonumber

Hence $$S\circ T$$ is a linear map. If $$\left( S\circ T\right) \left( \vec{v} \right) =0,$$ then $$S\left( T\left( \vec{v} \right) \right) =0$$ and it follows that $$T(\vec{v})=\vec{0}$$ and hence by this lemma again, $$\vec{v}=\vec{0}$$. Thus $$S\circ T$$ is one to one. It remains to verify that it is onto. Let $$\vec{z} \in Z$$. Then since $$S$$ is onto, there exists $$\vec{w} \in W$$ such that $$S(\vec{w})=\vec{z}.$$ Also, since $$T$$ is onto, there exists $$\vec{v}\in V$$ such that $$T(\vec{v})=\vec{w}.$$ It follows that $$S\left( T\left( \vec{v}\right) \right) =\vec{z}$$ and so $$S\circ T$$ is also onto.

Consider two subspaces $$V$$ and $$W$$, and suppose there exists an isomorphism mapping one to the other. In this way the two subspaces are related, which we can write as $$V \sim W$$. Then the previous two propositions together claim that $$\sim$$ is an equivalence relation. That is: $$\sim$$ satisfies the following conditions:

• $$V\sim V$$
• If $$V\sim W,$$ it follows that $$W\sim V$$
• If $$V\sim W$$ and $$W\sim Z,$$ then $$V\sim Z$$

We leave the verification of these conditions as an exercise.

Consider the following example.

## Example $$\PageIndex{2}$$:Matrix Isomorphism

Let $$T:\mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$$ be defined by $$T(\vec{x}) = A(\vec{x})$$ where $$A$$ is an invertible $$n\times n$$ matrix. Then $$T$$ is an isomorphism.

Solution

The reason for this is that, since $$A$$ is invertible, the only vector it sends to $$\vec{0}$$ is the zero vector. Hence if $$A(\vec{x})=A(\vec{y}),$$ then $$A\left( \vec{x}-\vec{y}\right) =\vec{0}$$ and so $$\vec{x}=\vec{y}$$. It is onto because if

$\vec{y}\in \mathbb{R}^{n},A\left( A^{-1} (\vec{y})\right) =\left( AA^{-1}\right) (\vec{y}) =\vec{y}. \nonumber$

In fact, all isomorphisms from $$\mathbb{R}^{n}$$ to $$\mathbb{R}^{n}$$ can be expressed as $$T(\vec{x}) = A(\vec{x})$$ where $$A$$ is an invertible $$n \times n$$ matrix. One simply considers the matrix whose $$i^{th}$$ column is $$T\vec{e}_{i}$$.

Recall that a basis of a subspace $$V$$ is a set of linearly independent vectors which span $$V$$. The following fundamental lemma describes the relation between bases and isomorphisms.

## Lemma $$\PageIndex{1}$$:Mapping Bases

Let $$T:V\rightarrow W$$ be a linear transformation where $$V,W$$ are subspaces of $$\mathbb{R}^n$$. If $$T$$ is one to one, then it has the property that if $$\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}$$ is linearly independent, so is $$\left\{ T(\vec{u}_{1}),\cdots ,T(\vec{u}_{k})\right\}$$.

More generally, $$T$$ is an isomorphism if and only if whenever $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ is a basis for $$V,$$ it follows that $$\left\{ T (\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\}$$ is a basis for $$W$$.

Proof

First suppose that $$T$$ is a linear transformation and is one to one and $$\left\{ \vec{u}_{1},\cdots ,\vec{u}_{k}\right\}$$ is linearly independent. It is required to show that $$\left\{ T(\vec{u}_{1}),\cdots ,T(\vec{ u}_{k})\right\}$$ is also linearly independent. Suppose then that $\sum_{i=1}^{k}c_{i}T(\vec{u}_{i})=\vec{0}\nonumber$ Then, since $$T$$ is linear, $T\left( \sum_{i=1}^{n}c_{i}\vec{u}_{i}\right) =\vec{0}\nonumber$ Since $$T$$ is one to one, it follows that $\sum_{i=1}^{n}c_{i}\vec{u}_{i}=0\nonumber$ Now the fact that $$\left\{ \vec{u}_{1},\cdots ,\vec{u}_{n}\right\}$$ is linearly independent implies that each $$c_{i}=0$$. Hence $$\left\{ T(\vec{u} _{1}),\cdots ,T(\vec{u}_{n})\right\}$$ is linearly independent.

Now suppose that $$T$$ is an isomorphism and $$\left\{ \vec{v}_{1},\cdots ,\vec{ v}_{n}\right\}$$ is a basis for $$V$$. It was just shown that $$\left\{ T(\vec{v} _{1}),\cdots ,T(\vec{v}_{n})\right\}$$ is linearly independent. It remains to verify that span$$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\}=W$$. If $$\vec{w}\in W,$$ then since $$T$$ is onto there exists $$\vec{v}\in V$$ such that $$T(\vec{v})=\vec{w}$$. Since $$\left\{ \vec{v} _{1},\cdots ,\vec{v}_{n}\right\}$$ is a basis, it follows that there exists scalars $$\left\{ c_{i}\right\} _{i=1}^{n}$$ such that $\sum_{i=1}^{n}c_{i}\vec{v}_{i}=\vec{v}.\nonumber$ Hence, $\vec{w}=T(\vec{v})=T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right) =\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})\nonumber$ It follows that span$$\left\{ T(\vec{v}_{1}),\cdots , T(\vec{v}_{n})\right\} =W$$ showing that this set of vectors is a basis for $$W$$.

Next suppose that $$T$$ is a linear transformation which takes a basis to a basis. This means that if $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ is a basis for $$V,$$ it follows $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\}$$ is a basis for $$W.$$ Then if $$w\in W,$$ there exist scalars $$c_{i}$$ such that $$w=\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})=T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right)$$ showing that $$T$$ is onto. If $$T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right) =\vec{0}$$ then $$\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})=\vec{0}$$ and since the vectors $$\left\{ T(\vec{v} _{1}),\cdots ,T(\vec{v}_{n})\right\}$$ are linearly independent, it follows that each $$c_{i}=0.$$ Since $$\sum_{i=1}^{n}c_{i}\vec{v}_{i}$$ is a typical vector in $$V$$, this has shown that if $$T(\vec{v})=\vec{0}$$ then $$\vec{v}=\vec{0}$$ and so $$T$$ is also one to one. Thus $$T$$ is an isomorphism.

The following theorem illustrates a very useful idea for defining an isomorphism. Basically, if you know what it does to a basis, then you can construct the isomorphism.

## Theorem $$\PageIndex{1}$$:Isomorphic Subspaces

Suppose $$V$$ and $$W$$ are two subspaces of $$\mathbb{R}^n$$. Then the two subspaces are isomorphic if and only if they have the same dimension. In the case that the two subspaces have the same dimension, then for a linear map $$T:V\rightarrow W$$, the following are equivalent.

1. $$T$$ is one to one.
2. $$T$$ is onto.
3. $$T$$ is an isomorphism.
Proof

Suppose first that these two subspaces have the same dimension. Let a basis for $$V$$ be $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ and let a basis for $$W$$ be $$\left\{ \vec{w}_{1},\cdots ,\vec{w}_{n}\right\}$$. Now define $$T$$ as follows. $T(\vec{v}_{i})=\vec{w}_{i}\nonumber$ for $$\sum_{i=1}^{n}c_{i}\vec{v}_{i}$$ an arbitrary vector of $$V,$$ $T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right) = \sum_{i=1}^{n}c_{i}T \vec{v}_{i}=\sum_{i=1}^{n}c_{i}\vec{w}_{i}.\nonumber$ It is necessary to verify that this is well defined. Suppose then that $\sum_{i=1}^{n}c_{i}\vec{v}_{i}=\sum_{i=1}^{n}\hat{c}_{i}\vec{v}_{i}\nonumber$ Then $\sum_{i=1}^{n}\left( c_{i}-\hat{c}_{i}\right) \vec{v}_{i}=\vec{0}\nonumber$ and since $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ is a basis, $$c_{i}=\hat{c}_{i}$$ for each $$i$$. Hence $\sum_{i=1}^{n}c_{i}\vec{w}_{i}=\sum_{i=1}^{n}\hat{c}_{i}\vec{w}_{i}\nonumber$ and so the mapping is well defined. Also if $$a,b$$ are scalars, \begin{aligned} T\left( a\sum_{i=1}^{n}c_{i}\vec{v}_{i}+b\sum_{i=1}^{n}\hat{c}_{i}\vec{v}_{i}\right) &=T\left( \sum_{i=1}^{n}\left( ac_{i}+b\hat{c}_{i}\right) \vec{v}_{i}\right) =\sum_{i=1}^{n}\left( ac_{i}+b\hat{c}_{i}\right) \vec{w}_{i} \\ &=a\sum_{i=1}^{n}c_{i}\vec{w}_{i}+b\sum_{i=1}^{n}\hat{c}_{i}\vec{w}_{i} \\ &=aT\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right) +bT\left( \sum_{i=1}^{n} \hat{c}_{i}\vec{v}_{i}\right)\end{aligned} Thus $$T$$ is a linear transformation.

Now if $T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right) =\sum_{i=1}^{n}c_{i}\vec{w}_{i}=\vec{0},\nonumber$ then since the $$\left\{ \vec{w}_{1},\cdots ,\vec{w}_{n}\right\}$$ are independent, each $$c_{i}=0$$ and so $$\sum_{i=1}^{n}c_{i}\vec{v}_{i}=\vec{0}$$ also. Hence $$T$$ is one to one. If $$\sum_{i=1}^{n}c_{i}\vec{w}_{i}$$ is a vector in $$W,$$ then it equals $\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})=T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right)\nonumber$ showing that $$T$$ is also onto. Hence $$T$$ is an isomorphism and so $$V$$ and $$W$$ are isomorphic.

Next suppose $$T:V \mapsto W$$ is an isomorphism, so these two subspaces are isomorphic. Then for $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ a basis for $$V$$, it follows that a basis for $$W$$ is $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\}$$ showing that the two subspaces have the same dimension.

Now suppose the two subspaces have the same dimension. Consider the three claimed equivalences.

First consider the claim that $$1.)\Rightarrow 2.).$$ If $$T$$ is one to one and if $$\left\{ \vec{v}_{1},\cdots ,\vec{v} _{n}\right\}$$ is a basis for $$V,$$ then $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v }_{n})\right\}$$ is linearly independent. If it is not a basis, then it must fail to span $$W$$. But then there would exist $$\vec{w}\notin \mathrm{span} \left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\}$$ and it follows that $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n}),\vec{w} \right\}$$ would be linearly independent which is impossible because there exists a basis for $$W$$ of $$n$$ vectors.

Hence $$\mathrm{span}\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\} =W$$ and so $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n})\right\}$$ is a basis. If $$\vec{w}\in W,$$ there exist scalars $$c_{i}$$ such that $\vec{w}=\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})=T\left( \sum_{i=1}^{n}c_{i}\vec{v} _{i}\right)\nonumber$ showing that $$T$$ is onto. This shows that $$1.)\Rightarrow 2.).$$

Next consider the claim that $$2.)\Rightarrow 3.).$$ Since $$2.)$$ holds, it follows that $$T$$ is onto. It remains to verify that $$T$$ is one to one. Since $$T$$ is onto, there exists a basis of the form $$\left\{ T(\vec{v}_{i}),\cdots ,T(\vec{v}_{n})\right\} .$$ Then it follows that $$\left\{ \vec{v}_{1},\cdots , \vec{v}_{n}\right\}$$ is linearly independent. Suppose $\sum_{i=1}^{n}c_{i}\vec{v}_{i}=\vec{0}\nonumber$ Then $\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})=\vec{0}\nonumber$ Hence each $$c_{i}=0$$ and so, $$\left\{ \vec{v}_{1},\cdots ,\vec{v} _{n}\right\}$$ is a basis for $$V$$. Now it follows that a typical vector in $$V$$ is of the form $$\sum_{i=1}^{n}c_{i}\vec{v}_{i}$$. If $$T\left( \sum_{i=1}^{n}c_{i}\vec{v}_{i}\right) =\vec{0},$$ it follows that $\sum_{i=1}^{n}c_{i}T(\vec{v}_{i})=\vec{0}\nonumber$ and so, since $$\left\{ T(\vec{v}_{i}),\cdots ,T(\vec{v}_{n})\right\}$$ is independent, it follows each $$c_{i}=0$$ and hence $$\sum_{i=1}^{n}c_{i}\vec{v} _{i}=\vec{0}$$. Thus $$T$$ is one to one as well as onto and so it is an isomorphism.

If $$T$$ is an isomorphism, it is both one to one and onto by definition so $$3.)$$ implies both $$1.)$$ and $$2.)$$.

Note the interesting way of defining a linear transformation in the first part of the argument by describing what it does to a basis and then “extending it linearly” to the entire subspace.

## Example $$\PageIndex{4}$$: Isomorphic Subspaces

Let $$V=\mathbb{R}^{3}$$ and let $$W$$ denote $\mathrm{span}\left\{ \left [ \begin{array}{r} 1 \\ 2 \\ 1 \\ 1 \end{array} \right ] ,\left [ \begin{array}{r} 0 \\ 1 \\ 0 \\ 1 \end{array} \right ] ,\left [ \begin{array}{r} 1 \\ 1 \\ 2 \\ 0 \end{array} \right ] \right\}\nonumber$ Show that $$V$$ and $$W$$ are isomorphic.

Solution

First observe that these subspaces are both of dimension 3 and so they are isomorphic by Theorem $$\PageIndex{1}$$. The three vectors which span $$W$$ are easily seen to be linearly independent by making them the columns of a matrix and row reducing to the reduced row-echelon form.

You can exhibit an isomorphism of these two spaces as follows. $T(\vec{e}_{1})=\left [ \begin{array}{c} 1 \\ 2 \\ 1 \\ 1 \end{array} \right ], T(\vec{e}_{2})=\left [ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right ], T(\vec{e}_{3})=\left [ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array} \right ]\nonumber$ and extend linearly. Recall that the matrix of this linear transformation is just the matrix having these vectors as columns. Thus the matrix of this isomorphism is $\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{array} \right ]\nonumber$ You should check that multiplication on the left by this matrix does reproduce the claimed effect resulting from an application by $$T$$.

Consider the following example.

## Example $$\PageIndex{5}$$:Finding the Matrix of an Isomorphism

Let $$V=\mathbb{R}^{3}$$ and let $$W$$ denote

$\mathrm{span}\left\{ \left [ \begin{array}{c} 1 \\ 2 \\ 1 \\ 1 \end{array} \right ] ,\left [ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right ] ,\left [ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array} \right ] \right\}\nonumber$

Let $$T: V \mapsto W$$ be defined as follows. $T\left [ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 2 \\ 1 \\ 1 \end{array} \right ] ,T\left [ \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right ] =\left [ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right ] ,T\left [ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 1 \\ 2 \\ 0 \end{array} \right ]\nonumber$ Find the matrix of this isomorphism $$T$$.

Solution

First note that the vectors $\left [ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right ] ,\left [ \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right ] ,\left [ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ]\nonumber$ are indeed a basis for $$\mathbb{R}^{3}$$ as can be seen by making them the columns of a matrix and using the reduced row-echelon form.

Now recall the matrix of $$T$$ is a $$4\times 3$$ matrix $$A$$ which gives the same effect as $$T.$$ Thus, from the way we multiply matrices, $A\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ] =\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{array} \right ]\nonumber$ Hence, $A=\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 2 & 1 & 1 \\ 1 & 0 & 2 \\ 1 & 1 & 0 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ] ^{-1}=\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 2 & -1 \\ 2 & -1 & 1 \\ -1 & 2 & -1 \end{array} \right ]\nonumber$

Note how the span of the columns of this new matrix must be the same as the span of the vectors defining $$W$$.

This idea of defining a linear transformation by what it does on a basis works for linear maps which are not necessarily isomorphisms.

## Example $$\PageIndex{6}$$: Finding the Matrix of an Isomorphism

Let $$V=\mathbb{R}^{3}$$ and let $$W$$ denote $\mathrm{span}\left\{ \left [ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right ] ,\left [ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right ] ,\left [ \begin{array}{c} 1 \\ 1 \\ 1 \\ 2 \end{array} \right ] \right\}\nonumber$ Let $$T: V \mapsto W$$ be defined as follows. $T\left [ \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right ] = \left [ \begin{array}{c} 1 \\ 0 \\ 1 \\ 1 \end{array} \right ] ,T\left [ \begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right ] =\left [ \begin{array}{c} 0 \\ 1 \\ 0 \\ 1 \end{array} \right ] ,T\left [ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ] =\left [ \begin{array}{c} 1 \\ 1 \\ 1 \\ 2 \end{array} \right ]\nonumber$ Find the matrix of this linear transformation.

Solution

Note that in this case, the three vectors which span $$W$$ are not linearly independent. Nevertheless the above procedure will still work. The reasoning is the same as before. If $$A$$ is this matrix, then $A\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ] =\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 2 \end{array} \right ]\nonumber$ and so $A=\left [ \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 2 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 0 & 1 & 1 \end{array} \right ] ^{-1}=\left [ \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ 1 & 0 & 1 \end{array} \right ]\nonumber$

The columns of this last matrix are obviously not linearly independent.

Below is a video on determining a function output given an isomorphic transformation.

This page titled 5.6: Isomorphisms is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .