5.2: The Matrix of a Linear Transformation I

Outcomes

1. Find the matrix of a linear transformation with respect to the standard basis.
2. Determine the action of a linear transformation on a vector in $\mathbb{R}^n$.

In the above examples, the action of the linear transformations was to multiply by a matrix. It turns out that this is always the case for linear transformations. If $T$ is any linear transformation which maps $\mathbb{R}^{n}$ to $\mathbb{R}^{m},$ there is always an $m\times n$ matrix $A$ with the property that $$T\left(\vec{x}\right) = A\vec{x} \label{matrixoftransf}$$ for all $\vec{x} \in \mathbb{R}^{n}$.

Theorem $\PageIndex{1}$: Matrix of a Linear Transformation

Let $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ be a linear transformation. Then we can find a matrix $A$ such that $T(\vec{x}) = A\vec{x}$. In this case, we say that $T$ is determined or induced by the matrix $A$.

Here is why. Suppose $T:\mathbb{R}^{n}\mapsto \mathbb{R}^{m}$ is a linear transformation and you want to find the matrix defined by this linear transformation as described in $\eqref{matrixoftransf}$. Note that $$\vec{x} =\left[\begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n} \end{array} \right] = x_{1}\left[\begin{array}{c} 1 \\ 0 \\ \vdots \\ 0 \end{array} \right] + x_{2}\left[\begin{array}{c} 0 \\ 1 \\ \vdots \\ 0 \end{array} \right] +\cdots + x_{n}\left[\begin{array}{c} 0 \\ 0 \\ \vdots \\ 1 \end{array} \right] = \sum_{i=1}^{n}x_{i}\vec{e}_{i}\nonumber$$ where $\vec{e}_{i}$ is the $i^{th}$ column of $I_n$, that is the $n \times 1$ vector which has zeros in every slot but the $i^{th}$ and a 1 in this slot.

Then since $T$ is linear, $$\begin{aligned} T\left( \vec{x} \right)&=\sum_{i=1}^{n}x_{i}T\left( \vec{e}_{i}\right) \\ &=\left[\begin{array}{ccc} | & & | \\ T\left( \vec{e}_{1}\right) & \cdots & T\left( \vec{e}_{n}\right) \\ | & & | \end{array} \right] \left[\begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right] \\ &= A\left[\begin{array}{c} x_{1} \\ \vdots \\ x_{n} \end{array} \right]\end{aligned}$$ The desired matrix is obtained from constructing the $i^{th}$ column as $T\left( \vec{e}_{i}\right) .$ Recall that the set $\left\{ \vec{e}_1, \vec{e}_2, \cdots, \vec{e}_n \right\}$ is called the standard basis of $\mathbb{R}^n$. Therefore the matrix of $T$ is found by applying $T$ to the standard basis. We state this formally as the following theorem.

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

Let $T: \mathbb{R}^{n} \mapsto \mathbb{R}^{m}$ be a linear transformation. Then the matrix $A$ satisfying $T\left(\vec{x}\right)=A\vec{x}$ is given by $$A= \left[\begin{array}{ccc} | & & | \\ T\left( \vec{e}_{1}\right) & \cdots & T\left( \vec{e}_{n}\right) \\ | & & | \end{array} \right]\nonumber$$ where $\vec{e}_{i}$ is the $i^{th}$ column of $I_n$, and then $T\left( \vec{e}_{i} \right)$ is the $i^{th}$ column of $A$.

The following Corollary is an essential result.

Corollary $\PageIndex{1}$: Matrix and Linear Transformation

A transformation $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is a linear transformation if and only if it is a matrix transformation.

Consider the following example.

Example $\PageIndex{1}$: The Matrix of a Linear Transformation

Suppose $T$ is a linear transformation, $T:\mathbb{R}^{3}\rightarrow \mathbb{ R}^{2}$ where $$T\left[\begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right] =\left[\begin{array}{r} 1 \\ 2 \end{array} \right] ,\ T\left[\begin{array}{r} 0 \\ 1 \\ 0 \end{array} \right] =\left[\begin{array}{r} 9 \\ -3 \end{array} \right] ,\ T\left[\begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right] =\left[\begin{array}{r} 1 \\ 1 \end{array} \right]\nonumber$$ Find the matrix $A$ of $T$ such that $T \left( \vec{x} \right)=A\vec{x}$ for all $\vec{x}$.

Solution

By Theorem $\PageIndex{2}$ we construct $A$ as follows: $$A = \left[\begin{array}{ccc} | & & | \\ T\left( \vec{e}_{1}\right) & \cdots & T\left( \vec{e}_{n}\right) \\ | & & | \end{array} \right]\nonumber$$

In this case, $A$ will be a $2 \times 3$ matrix, so we need to find $T \left(\vec{e}_1 \right), T \left(\vec{e}_2 \right),$ and $T \left(\vec{e}_3 \right)$. Luckily, we have been given these values so we can fill in $A$ as needed, using these vectors as the columns of $A$. Hence, $$A=\left[\begin{array}{rrr} 1 & 9 & 1 \\ 2 & -3 & 1 \end{array} \right]\nonumber$$

In this example, we were given the resulting vectors of $T \left(\vec{e}_1 \right), T \left(\vec{e}_2 \right),$ and $T \left(\vec{e}_3 \right)$. Constructing the matrix $A$ was simple, as we could simply use these vectors as the columns of $A$. The next example shows how to find $A$ when we are not given the $T \left(\vec{e}_i \right)$ so clearly.

Example $\PageIndex{2}$: The Matrix of Linear Transformation: Inconveniently Defined

Suppose $T$ is a linear transformation, $T:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ and $$T\left[\begin{array}{r} 1 \\ 1 \end{array} \right] =\left[\begin{array}{r} 1 \\ 2 \end{array} \right] ,\ T\left[\begin{array}{r} 0 \\ -1 \end{array} \right] =\left[\begin{array}{r} 3 \\ 2 \end{array} \right]\nonumber$$ Find the matrix $A$ of $T$ such that $T \left( \vec{x} \right)=A\vec{x}$ for all $\vec{x}$.

Solution

By Theorem $\PageIndex{2}$ to find this matrix, we need to determine the action of $T$ on $\vec{e}_{1}$ and $\vec{e}_{2}$. In Example 9.9.2, we were given these resulting vectors. However, in this example, we have been given $T$ of two different vectors. How can we find out the action of $T$ on $\vec{e}_{1}$ and $\vec{e}_{2}$? In particular for $\vec{e}_{1}$, suppose there exist $x$ and $y$ such that $$\left[\begin{array}{r} 1 \\ 0 \end{array} \right] = x\left[\begin{array}{r} 1\\ 1 \end{array} \right] +y\left[\begin{array}{r} 0 \\ -1 \end{array} \right] \label{matrixvalues}$$

Then, since $T$ is linear, $$T\left[\begin{array}{r} 1 \\ 0 \end{array} \right] = x T\left[\begin{array}{r} 1 \\ 1 \end{array} \right] +y T\left[\begin{array}{r} 0 \\ -1 \end{array} \right]\nonumber$$

Substituting in values, this sum becomes $$T\left[\begin{array}{r} 1 \\ 0 \end{array} \right] = x\left[\begin{array}{r} 1 \\ 2 \end{array} \right] +y\left[\begin{array}{r} 3 \\ 2 \end{array} \right] \label{matrixvalues2}$$

Therefore, if we know the values of $x$ and $y$ which satisfy $\eqref{matrixvalues}$, we can substitute these into equation $\eqref{matrixvalues2}$. By doing so, we find $T\left(\vec{e}_1\right)$ which is the first column of the matrix $A$.

We proceed to find $x$ and $y$. We do so by solving $\eqref{matrixvalues}$, which can be done by solving the system $$\begin{array}{c} x = 1 \\ x - y = 0 \end{array}\nonumber$$

We see that $x=1$ and $y=1$ is the solution to this system. Substituting these values into equation $\eqref{matrixvalues2}$, we have $$T\left[\begin{array}{r} 1 \\ 0 \end{array} \right] = 1 \left[\begin{array}{r} 1 \\ 2 \end{array} \right] + 1 \left[\begin{array}{r} 3 \\ 2 \end{array} \right] = \left[\begin{array}{r} 1 \\ 2 \end{array} \right] + \left[\begin{array}{r} 3 \\ 2 \end{array} \right] = \left[\begin{array}{r} 4 \\ 4 \end{array} \right]\nonumber$$

Therefore $\left[\begin{array}{r} 4 \\ 4 \end{array} \right]$ is the first column of $A$.

Computing the second column is done in the same way, and is left as an exercise.

The resulting matrix $A$ is given by $$A = \left[\begin{array}{rr} 4 & -3 \\ 4 & -2 \end{array} \right]\nonumber$$

This example illustrates a very long procedure for finding the matrix of $A$. While this method is reliable and will always result in the correct matrix $A$, the following procedure provides an alternative method.

Procedure $\PageIndex{1}$: Finding the Matrix of Inconveniently Defined Linear Transformation

Suppose $T:\mathbb{R}^{n}\rightarrow \mathbb{R}^{m}$ is a linear transformation. Suppose there exist vectors $\left\{ \vec{a}_{1},\cdots ,\vec{a}_{n}\right\}$ in $\mathbb {R}^{n}$ such that $\left[\begin{array}{ccc} \vec{a}_{1} & \cdots & \vec{a}_{n} \end{array} \right] ^{-1}$ exists, and $$T \left(\vec{a}_{i}\right)=\vec{b}_{i}\nonumber$$ Then the matrix of $T$ must be of the form $$\left[\begin{array}{ccc} \vec{b}_{1} & \cdots & \vec{b}_{n} \end{array} \right] \left[\begin{array}{ccc} \vec{a}_{1} & \cdots & \vec{a}_{n} \end{array} \right] ^{-1}\nonumber$$

We will illustrate this procedure in the following example. You may also find it useful to work through Example $\PageIndex{2}$ using this procedure.

Example $\PageIndex{3}$: Matrix of a Linear Transformation Given Inconveniently

Suppose $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ is a linear transformation and $$T\left[\begin{array}{r} 1 \\ 3 \\ 1 \end{array} \right] =\left[\begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right] ,T\left[\begin{array}{r} 0 \\ 1 \\ 1 \end{array} \right] =\left[\begin{array}{r} 2 \\ 1 \\ 3 \end{array} \right] ,T\left[\begin{array}{r} 1 \\ 1 \\ 0 \end{array} \right] =\left[\begin{array}{r} 0 \\ 0 \\ 1 \end{array} \right]\nonumber$$ Find the matrix of this linear transformation.

Solution

By Procedure $\PageIndex{1}$, $A= \left[\begin{array}{rrr} 1 & 0 & 1 \\ 3 & 1 & 1 \\ 1 & 1 & 0 \end{array} \right] ^{-1}$ and $B=\left[\begin{array}{rrr} 0 & 2 & 0 \\ 1 & 1 & 0 \\ 1 & 3 & 1 \end{array} \right]$

Then, Procedure $\PageIndex{1}$ claims that the matrix of $T$ is $$C= BA^{-1} =\left[\begin{array}{rrr} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \right]\nonumber$$

Indeed you can first verify that $T(\vec{x})=C\vec{x}$ for the 3 vectors above:

$$\left[\begin{array}{ccc} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \right] \left[\begin{array}{c} 1 \\ 3 \\ 1 \end{array} \right] =\left[\begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right] ,\ \left[\begin{array}{ccc} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \right] \left[\begin{array}{c} 0 \\ 1 \\ 1 \end{array} \right] =\left[\begin{array}{c} 2 \\ 1 \\ 3 \end{array} \right]\nonumber$$ $$\left[\begin{array}{ccc} 2 & -2 & 4 \\ 0 & 0 & 1 \\ 4 & -3 & 6 \end{array} \right] \left[\begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right] =\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]\nonumber$$

But more generally $T(\vec{x})= C\vec{x}$ for any $\vec{x}$. To see this, let $\vec{y}=A^{-1}\vec{x}$ and then using linearity of $T$: $$T(\vec{x})= T(A\vec{y}) = T \left( \sum_i \vec{y}_i\vec{a}_i \right) = \sum \vec{y}_i T(\vec{a}_i) \sum \vec{y}_i \vec{b}_i = B\vec{y} = BA^{-1}\vec{x} = C\vec{x}\nonumber$$

Recall the dot product discussed earlier. Consider the map $\vec{v}$$\mapsto$ $\mathrm{proj}_{\vec{u}}\left( \vec{v}\right)$ which takes a vector a transforms it to its projection onto a given vector $\vec{u}$. It turns out that this map is linear, a result which follows from the properties of the dot product. This is shown as follows. $$\begin{aligned} \mathrm{proj}_{\vec{u}}\left( k \vec{v}+ p \vec{w}\right) &=\left( \frac{(k \vec{v}+ p \vec{w})\bullet \vec{u}}{ \vec{u}\bullet \vec{u}}\right) \vec{u} \\ &= k \left( \frac{ \vec{v}\bullet \vec{u}}{\vec{u}\bullet \vec{u}}\right) \vec{u}+p \left( { 0.05in}\frac{\vec{w}\bullet \vec{u}}{\vec{u}\bullet \vec{u}}\right) \vec{u} \\ &= k \; \mathrm{proj}_{\vec{u}}\left( \vec{v}\right) +p \; \mathrm{proj} _{\vec{u}}\left( \vec{w}\right) \end{aligned}$$

Consider the following example.

Example $\PageIndex{4}$: Matrix of a Projection Map

Let $\vec{u} = \left[\begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]$ and let $T$ be the projection map $T: \mathbb{R}^3 \mapsto \mathbb{R}^3$ defined by $$T(\vec{v}) = \mathrm{proj}_{\vec{u}}\left( \vec{v}\right)\nonumber$$ for any $\vec{v} \in \mathbb{R}^3$.

1. Does this transformation come from multiplication by a matrix?
2. If so, what is the matrix?

Solution

1. First, we have just seen that $T (\vec{v}) = \mathrm{proj}_{\vec{u}}\left( \vec{v}\right)$ is linear. Therefore by Theorem $\PageIndex{1}$, we can find a matrix $A$ such that $T(\vec{x}) = A\vec{x}$.
2. The columns of the matrix for $T$ are defined above as $T(\vec{e}_{i})$. It follows that $T(\vec{e}_{i}) = \mathrm{proj} _{\vec{u}}\left( \vec{e}_{i}\right)$ gives the $i^{th}$ column of the desired matrix. Therefore, we need to find $$\mathrm{proj}_{\vec{u}}\left( \vec{e}_{i}\right) = \left( \frac{\vec{e}_{i}\bullet \vec{u}}{\vec{u}\bullet \vec{u}}\right) \vec{u}\nonumber$$ For the given vector $\vec{u}$, this implies the columns of the desired matrix are $$\frac{1}{14}\left[\begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] , \frac{2}{14}\left[\begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] , \frac{3}{14}\left[\begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]\nonumber$$ which you can verify. Hence the matrix of $T$ is $$\frac{1}{14}\left[\begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array} \right]\nonumber$$