6.2: The Matrix of a Linear Transformation I
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Outcomes
- Find the matrix of a linear transformation with respect to the standard basis.
- 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 \(\vec{x} \in \mathbb{R}^{n}\).
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.
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 \(\vec{v} \in \mathbb{R}^3\).
- Does this transformation come from multiplication by a matrix?
- If so, what is the matrix?
Solution
- 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 \(T(\vec{x}) = A\vec{x}\).
- 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 \[ \frac{1}{14}\left[\begin{array}{rrr} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array} \right]\nonumber\]