# 5.8: The Matrix of a Linear Transformation II

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## Outcomes

1. Find the matrix of a linear transformation with respect to general bases.

We begin this section with an important lemma.

## Lemma $$\PageIndex{1}$$:Mapping of a Basis

Let $$T: \mathbb{R}^n \mapsto \mathbb{R}^n$$ be an isomorphism. Then $$T$$ maps any basis of $$\mathbb{R}^n$$ to another basis for $$\mathbb{R}^n$$.

Conversely, if $$T: \mathbb{R}^n \mapsto \mathbb{R}^n$$ is a linear transformation which maps a basis of $$\mathbb{R}^n$$ to another basis of $$\mathbb{R}^n$$, then it is an isomorphism.

Proof

First, suppose $$T:\mathbb{R}^n \mapsto \mathbb{R}^n$$ is a linear transformation which is one to one and onto. Let $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ be a basis for $$\mathbb{R}^n$$. We wish to show that $$\left\{ T(\vec{v}_{1}),\cdots , T(\vec{v}_{n})\right\}$$ is also a basis for $$\mathbb{R}^n$$.

First consider why it is linearly independent. Suppose $$\sum_{k=1}^{n}a_{k}T(\vec{v}_{k})=\vec{0}$$. Then by linearity we have $$T\left( \sum_{k=1}^{n}a_{k}\vec{v}_{k}\right) =\vec{0}$$ and since $$T$$ is one to one, it follows that $$\sum_{k=1}^{n}a_{k}\vec{v}_{k}=\vec{0}$$. This requires that each $$a_{k}=0$$ because $$\left\{ \vec{v}_{1},\cdots, \vec{v}_{n}\right\}$$ is independent, and it follows that $$\left\{ T(\vec{v}_{1}),\cdots , T(\vec{v}_{n})\right\}$$ is linearly independent.

Next take $$\vec{w}\in \mathbb{R}^n.$$ Since $$T$$ is onto, there exists $$\vec{v}\in \mathbb{R}^n$$ such that $$T(\vec{v})=\vec{w}$$. Since $$\left\{ \vec{v}_{1},\cdots ,\vec{v}_{n}\right\}$$ is a basis, in particular it is a spanning set and there are scalars $$b_{k}$$ such that $$T\left( \sum_{k=1}^{n}b_{k}\vec{v} _{k}\right) =T\left( \vec{v}\right) =\vec{w}$$. Therefore $$\vec{w} =\sum_{k=1}^{n}b_{k}T(\vec{v}_{k})$$ which is in the $$\mathrm{span}\left\{ T(\vec{v}_{1}),\cdots , T(\vec{v}_{n})\right\} .$$ Therefore, $$\left\{ T(\vec{v}_{1}),\cdots ,T(\vec{v}_{n}) \right\}$$ is a basis as claimed.

Suppose now that $$T: \mathbb{R}^n \mapsto \mathbb{R}^n$$ is a linear transformation such that $$T(\vec{v}_{i})=\vec{w}_{i}$$ where $$\left\{\vec{v} _{1},\cdots ,\vec{v}_{n}\right\}$$ and $$\left\{ \vec{w}_{1},\cdots , \vec{w}_{n}\right\}$$ are two bases for $$\mathbb{R}^n$$.

To show that $$T$$ is one to one, let $$T\left( \sum_{k=1}^{n}c_{k}\vec{v}_{k}\right) =\vec{0}$$. Then $$\sum_{k=1}^{n}c_{k}T(\vec{v}_{k})=\sum_{k=1}^{n}c_{k}\vec{w}_{k}=\vec{ 0}$$. It follows that each $$c_{k} = 0$$ because it is given that $$\left\{ \vec{w} _{1},\cdots ,\vec{w}_{n}\right\}$$ is linearly independent. Hence $$T\left( \sum_{k=1}^{n}c_{k}\vec{v}_{k}\right) =\vec{0}$$ implies that $$\sum_{k=1}^{n}c_{k}\vec{v}_{k}=\vec{0}$$ and so $$T$$ is one to one.

To show that $$T$$ is onto, let $$\vec{w}$$ be an arbitrary vector in $$\mathbb{R}^n$$. This vector can be written as $$\vec{w} = \sum_{k=1}^{n}d_k\vec{w}_k = \sum_{k=1}^{n}d_{k}T(\vec{v}_{k})=T\left( \sum_{k=1}^{n}d_{k} \vec{v}_{k}\right) .$$ Therefore, $$T$$ is also onto.

Consider now an important definition.

## Definition $$\PageIndex{1}$$: Coordinate Vector

Let $$B = \left\{ \vec{v}_1, \vec{v}_2, \cdots, \vec{v}_n \right\}$$ be a basis for $$\mathbb{R}^n$$ and let $$\vec{x}$$ be an arbitrary vector in $$\mathbb{R}^n$$. Then $$\vec{x}$$ is uniquely represented as $$\vec{x} = a_1\vec{v}_1 + a_2\vec{v}_2 + \cdots + a_n\vec{v}_n$$ for scalars $$a_1, \cdots, a_n$$.

The coordinate vector of $$\vec{x}$$ with respect to the basis $$B$$, written $$C_B(\vec{x})$$ or $$[\vec{x}]_B$$, is given by $C_B(\vec{x}) = C_B \left( a_1\vec{v}_1 + a_2\vec{v}_2 + \cdots + a_n\vec{v}_n \right) = \left [ \begin{array}{c} a_1 \\ a_2 \\ \vdots \\ a_n \end{array} \right ]\nonumber$

Below is a video on change of basis between the standard basis and a nonstandard basis.

Below is a video on change of basis between two nonstandard bases.

Consider the following example.

## Example $$\PageIndex{1}$$:Coordinate Vector

Let $$B = \left\{ \left [ \begin{array}{r} 1 \\ 0 \end{array} \right ], \left [ \begin{array}{r} -1 \\ 1 \end{array} \right ] \right\}$$ be a basis of $$\mathbb{R}^2$$ and let $$\vec{x} = \left [ \begin{array}{r} 3 \\ -1 \end{array} \right ]$$ be a vector in $$\mathbb{R}^2$$. Find $$C_B(\vec{x})$$.

Solution

First, note the order of the basis is important so label the vectors in the basis $$B$$ as $B = \left\{ \left [ \begin{array}{r} 1 \\ 0 \end{array} \right ], \left [ \begin{array}{r} -1 \\ 1 \end{array} \right ] \right\} = \left\{ \vec{v}_1, \vec{v}_2 \right\}\nonumber$ Now we need to find $$a_1, a_2$$ such that $$\vec{x} = a_1 \vec{v}_1 + a_2 \vec{v}_2$$, that is: $\left [ \begin{array}{r} 3 \\ -1 \end{array} \right ] = a_1 \left [ \begin{array}{r} 1 \\ 0 \end{array} \right ] + a_2 \left [ \begin{array}{r} -1 \\ 1 \end{array} \right ]\nonumber$ Solving this system gives $$a_1 = 2, a_2 = -1$$. Therefore the coordinate vector of $$\vec{x}$$ with respect to the basis $$B$$ is $C_B(\vec{x}) = \left [ \begin{array}{r} a_1 \\ a_2 \end{array}\right ] = \left [ \begin{array}{r} 2 \\ -1 \end{array} \right ]\nonumber$

Below is a video on finding a vector given basis vectors and its coordinates with respect to this basis.

Below is a video on finding the coordinates of a vector with respect to a given basis knowing its coordinates with respect to the standard basis.

Below is another video on finding a vector given basis vectors and its coordinates with respect to this basis.

Below is a video on finding a vector given basis vectors and its coordinates with respect to this basis.

Below is a video on finding the coordinates of a vector with respect to the standard basis knowing its coordinates with respect to a given basis.

Given any basis $$B$$, one can easily verify that the coordinate function is actually an isomorphism.

## Theorem $$\PageIndex{1}$$:$$C_B$$Transformation is a Linear

For any basis $$B$$ of $$\mathbb{R}^n$$, the coordinate function $C_B: \mathbb{R}^n \rightarrow \mathbb{R}^n\nonumber$ is a linear transformation, and moreover an isomorphism.

We now discuss the main result of this section, that is how to represent a linear transformation with respect to different bases.

## Theorem $$\PageIndex{2}$$:The Matrix of a Linear

Let $$T: \mathbb{R}^n \mapsto \mathbb{R}^m$$ be a linear transformation, and let $$B_1$$ and $$B_2$$ be bases of $$\mathbb{R}^{n}$$ and $$\mathbb{R}^{m}$$ respectively.

Then the following holds $C_{B_2} T = M_{B_{2} B_{1}} C_{B_1} \label{matrixequation}$ where $$M_{B_{2} B_{1}}$$ is a unique $$m \times n$$ matrix.

If the basis $$B_1$$ is given by $$B_1=\{ \vec{v}_1, \cdots, \vec{v}_n \}$$ in this order, then $M_{B_{2} B_{1}} = \left [ C_{B_2}(T(\vec{v}_1)) \; C_{B_2}(T(\vec{v}_2)) \; \cdots C_{B_2}(T(\vec{v}_n)) \right ]\nonumber$

Proof

The above equation $$\eqref{matrixequation}$$ can be represented by the following diagram. $\begin{array}{rrcll} & & T & & \\ & \mathbb{R}^n & \rightarrow & \mathbb{R}^m & \\ & C_{B_{1} }\downarrow & \circ & \downarrow C_{B_{2} } & \\ & \mathbb{R}^{n} & \rightarrow & \mathbb{R}^{m} & \\ & & M_{B_{2} B_{1} } & & \end{array}\nonumber$

Since $$C_{B_1}$$ is an isomorphism, then the matrix we are looking for is the matrix of the linear transformation $C_{B_2} T C^{-1}_{B_1} : \mathbb{R}^n \mapsto \mathbb{R}^m.\nonumber$ By Theorem 5.2.2, the columns are given by the image of the standard basis $$\left\{ \vec{e}_1, \vec{e}_2, \cdots, \vec{e}_n \right\}$$. But since $$C^{-1}_{B_1}( \vec{e}_i) = \vec{v}_i$$, we readily obtain that $\begin{array}{ll} M_{B_{2} B_{1}} & = \left [ C_{B_2}T C^{-1}_{B_1} (\vec{e}_1) \;\; C_{B_2}T C^{-1}_{B_1} (\vec{2}_2) \;\; \cdots \;\; C_{B_2}T C^{-1}_{B_1} (\vec{e}_n) \right ] \\ & = \left [ C_{B_2}(T(\vec{v}_1)) \;\; C_{B_2}(T(\vec{v}_2)) \;\; \cdots \;\; C_{B_2}(T(\vec{v}_n)) \right ] \end{array}\nonumber$ and this completes the proof.

Consider the following example.

## Example $$\PageIndex{2}$$:Matrix of a Linear

Let $$T: \mathbb{R}^2 \mapsto \mathbb{R}^2$$ be a linear transformation defined by $$T \left( \left [ \begin{array}{r} a \\ b \end{array} \right ] \right) = \left [ \begin{array}{r} b \\ a \end{array} \right ]$$.

Consider the two bases $B_1 = \left\{ \vec{v}_{1}, \vec{v}_{2} \right\} = \left\{ \left [ \begin{array}{r} 1 \\ 0 \end{array}\right ], \left [ \begin{array}{r} -1 \\ 1 \end{array} \right ] \right\}\nonumber$ and $B_2 = \left\{ \left [ \begin{array}{r} 1 \\ 1 \end{array} \right ], \left [ \begin{array}{r} 1 \\ -1 \end{array} \right ] \right\}\nonumber$

Find the matrix $$M_{B_2,B_1}$$ of $$T$$ with respect to the bases $$B_1$$ and $$B_2$$.

Solution

By Theorem $$\PageIndex{2}$$, the columns of $$M_{B_{2} B_{1}}$$ are the coordinate vectors of $$T(\vec{v}_{1}), T(\vec{v}_{2})$$ with respect to $$B_2$$.

Since $T \left( \left [ \begin{array}{r} 1 \\ 0 \end{array}\right ] \right) = \left [ \begin{array}{r} 0 \\ 1 \end{array}\right ] ,\nonumber$ a standard calculation yields $\left [ \begin{array}{r} 0 \\ 1 \end{array}\right ] = \left(\frac{1}{2} \right)\left [ \begin{array}{r} 1 \\ 1 \end{array} \right ] + \left(-\frac{1}{2} \right) \left [ \begin{array}{r} 1 \\ -1 \end{array} \right ],\nonumber$ the first column of $$M_{B_{2} B_{1}}$$ is $$\left [ \begin{array}{r} \frac{1}{2}\\ -\frac{1}{2} \end{array}\right ]$$.

The second column is found in a similar way. We have $T \left( \left [ \begin{array}{r} -1 \\ 1 \end{array} \right ] \right) = \left [ \begin{array}{r} 1 \\ -1 \end{array}\right ] ,\nonumber$ and with respect to $$B_2$$ calculate: $\left [ \begin{array}{r} 1 \\ -1 \end{array}\right ] = 0 \left [ \begin{array}{r} 1 \\ 1 \end{array} \right ] + 1 \left [ \begin{array}{r} 1 \\ -1 \end{array} \right ]\nonumber$ Hence the second column of $$M_{B_{2} B_{1}}$$ is given by $$\left [ \begin{array}{r} 0 \\ 1 \end{array} \right ]$$. We thus obtain $M_{B_{2} B_{1}} = \left [ \begin{array}{rr} \frac{1}{2} & 0 \\ -\frac{1}{2} & 1 \end{array} \right ]\nonumber$

We can verify that this is the correct matrix $$M_{B_{2} B_{1}}$$ on the specific example $\vec{v} = \left [ \begin{array}{r} 3 \\ -1 \end{array} \right ]\nonumber$ First applying $$T$$ gives $T( \vec{v} ) = T \left( \left [ \begin{array}{r} 3 \\ -1 \end{array} \right ] \right) = \left [ \begin{array}{r} -1\\ 3 \end{array} \right ]\nonumber$ and one can compute that $C_{B_2} \left( \left [ \begin{array}{r} -1 \\ 3 \end{array} \right ] \right) = \left [ \begin{array}{r} 1\\ -2 \end{array} \right ] .\nonumber$

On the other hand, one compute $$C_{B_1}( \vec{v})$$ as $C_{B_1} \left( \left [ \begin{array}{r} 3 \\ -1 \end{array} \right ] \right) = \left [ \begin{array}{r} 2\\ -1 \end{array} \right ] ,\nonumber$ and finally applying $$M_{B_1 B_2}$$ gives $\left [ \begin{array}{rr} \frac{1}{2} & 0 \\ -\frac{1}{2} & 1 \end{array} \right ] \left [ \begin{array}{r} 2 \\ -1 \end{array}\right ] = \left [ \begin{array}{r} 1 \\ -2 \end{array} \right ]\nonumber$ as above.

We see that the same vector results from either method, as suggested by Theorem $$\PageIndex{2}$$.

If the bases $$B_1$$ and $$B_2$$ are equal, say $$B$$, then we write $$M_{B}$$ instead of $$M_{B B}$$. The following example illustrates how to compute such a matrix. Note that this is what we did earlier when we considered only $$B_1=B_2$$ to be the standard basis.

## Example $$\PageIndex{3}$$:Matrix of a Linear Transformation with respect to an Arbitrary

Consider the basis $$B$$ of $$\mathbb{R}^3$$ given by $B = \{\vec{v}_1 , \vec{v}_2, \vec{v}_3 \} = \left\{ \left [ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right ] ,\left [ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right ] ,\left [ \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \right ] \right\}\nonumber$ And let $$T :\mathbb{R}^{3}\mapsto \mathbb{R}^{3}$$ be the linear transformation defined on $$B$$ as: $T\left [ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right ] =\left [ \begin{array}{r} 1 \\ -1 \\ 1 \end{array} \right ] ,T \left [ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ] =\left [ \begin{array}{r} 1 \\ 2 \\ -1 \end{array} \right ] ,T\left [ \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \right ] =\left [ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right ]\nonumber$

1. Find the matrix $$M_{B}$$ of $$T$$ relative to the basis $$B$$.
2. Then find the usual matrix of $$T$$ with respect to the standard basis of $$\mathbb{R}^{3}$$.

Solution

Equation $$\eqref{matrixequation}$$ gives $$C_BT=M_{B}C_B$$, and thus $$M_{B} = C_BTC^{-1}_B$$.

Now $$C_B(\vec{v}_i) = \vec{e}_i$$, so the matrix of $$C_B^{-1}$$ (with respect to the standard basis) is given by $\left [ C_B^{-1}(\vec{e}_1) \;\; C_B^{-1}(\vec{e}_2) \;\; C_B^{-1}(\vec{e}_2) \right ] = \left [ \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right ]\nonumber$ Moreover the matrix of $$T C_B^{-1}$$ is given by $\left [ TC_B^{-1}(\vec{e}_1) \;\; TC_B^{-1}(\vec{e}_2) \;\; TC_B^{-1}(\vec{e}_2) \right ] = \left [ \begin{array}{rrr} 1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1 \end{array} \right ]\nonumber$ Thus $\begin{array}{ll} M_{B} & = C_BTC^{-1}_B = [C^{-1}_B]^{-1} [TC^{-1}_B] \\ & = \left [ \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right ] ^{-1}\left [ \begin{array}{rrr} 1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1 \end{array} \right ] \\ &=\left [ \begin{array}{rrr} 2 & -5 & 1 \\ -1 & 4 & 0 \\ 0 & -2 & 1 \end{array} \right ] \end{array}\nonumber$

Consider how this works. Let $$\vec{b} = \left [ \begin{array}{r} b_1 \\ b_2 \\ b_3 \end{array} \right ]$$ be an arbitrary vector in $$\mathbb{R}^3$$.

Apply $$C^{-1}_{B}$$ to $$\vec{b}$$ to get $b_1\left [ \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right ] + b_2\left [ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right ] + b_3\left [ \begin{array}{r} -1 \\ 1 \\ 0 \end{array} \right ]\nonumber$ Apply $$T$$ to this linear combination to obtain $b_1\left [ \begin{array}{r} 1 \\ -1 \\ 1 \end{array} \right ] + b_2\left [ \begin{array}{r} 1 \\ 2 \\ -1 \end{array} \right ] + b_3\left [ \begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right ] =\left [ \begin{array}{c} b_1+b_2 \\ -b_1 + 2b_2+ b_3 \\ b_1-b_2+b_3 \end{array} \right ]\nonumber$ Now take the matrix $$M_{B}$$ of the transformation (as found above) and multiply it by $$\vec{b}$$. $\left [ \begin{array}{rrr} 2 & -5 & 1 \\ -1 & 4 & 0 \\ 0 & -2 & 1 \end{array} \right ] \left [ \begin{array}{c} b_1 \\ b_2 \\ b_3 \end{array} \right ] =\left [ \begin{array}{c} 2b_1-5b_2+b_3 \\ -b_1 + 4b_2 \\ -2b_2 + b_3 \end{array} \right ]\nonumber$ Is this the coordinate vector of the above relative to the given basis? We check as follows. $\left( 2b_1-5b_2+b_3\right) \left [ \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right ] +\left( -b_1 + 4b_2\right) \left [ \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ] +\left( -2b_2+b_3\right) \left [ \begin{array}{c} -1 \\ 1 \\ 0 \end{array} \right ]\nonumber$ $= \left [ \begin{array}{c} b_1+b_2 \\ -b_1 + 2b_2+b_3 \\ b_1-b_2+b_3 \end{array} \right ]\nonumber$ You see it is the same thing.

Now lets find the matrix of $$T$$ with respect to the standard basis. Let $$A$$ be this matrix. That is, multiplication by $$A$$ is the same as doing $$T$$. Thus $A\left [ \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right ] =\left [ \begin{array}{rrr} 1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1 \end{array} \right ]\nonumber$ Hence $A=\left [ \begin{array}{rrr} 1 & 1 & 0 \\ -1 & 2 & 1 \\ 1 & -1 & 1 \end{array} \right ] \left [ \begin{array}{rrr} 1 & 1 & -1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right ] ^{-1}=\left [ \begin{array}{rrr} 0 & 0 & 1 \\ 2 & 3 & -3 \\ -3 & -2 & 4 \end{array} \right ]\nonumber$ Of course this is a very different matrix than the matrix of the linear transformation with respect to the non standard basis.

This page titled 5.8: The Matrix of a Linear Transformation II is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .