Standard Matrix of a Linear Transformation from to
In Matrix Transformations and Introduction to Linear Transformations we learned several important properties of matrix transformations of and subspaces of . Let’s summarize the main points.
For a matrix transformation , induced by an matrix we
have the following results:
- is linear. (Theorem th:matrixtran) This means that for vectors and in and scalars and in .
- Columns of are the images of the standard unit vectors of under . (Observation obs:imagesOfijk) \begin{equation*} \label{eq:matlintrans} A=\begin{bmatrix} a_{11} & a_{12}&\dots &a_{1n}\\ a_{21}&a_{22} &\dots &a_{2n}\\ \vdots & \vdots &\ddots &\vdots \\ a_{m1}&\dots &\dots &a_{mn} \end{bmatrix} = \begin{bmatrix} | & |& &|\\ T(\vec{e}_1) & T(\vec{e}_2)&\dots &T(\vec{e}_n)\\ |&| & &| \end{bmatrix} \end{equation*}
- The action of on all of the elements of is completely determined by where maps the standard unit vectors. (See Examples ex:imageOfBasisVectors and ex:imageofatransformation)
The last point in the summary is so important that it is worth illustrating again.
Let be a linear transformation. Suppose that the only information we have about
this transformation is that , and . Is this information sufficient to determine the
image of ?
In Example ex:imageofatransformation, there was nothing special about the vector . Any vector of can be written as a unique linear combination of the standard unit vectors . Therefore, the image of any vector under a linear transformation is uniquely determined by the images of . Knowing allows us to construct a matrix , with as columns, that induces transformation . We formalize this idea in a theorem.
Let be a linear transformation. Then is a matrix transformation with
\begin{equation*} \label{matlintrans} A=\begin{bmatrix} | & |& &|\\ T(\vec{e}_1) & T(\vec{e}_2)&\dots &T(\vec{e}_n)\\ |&| & &| \end{bmatrix} \end{equation*}
as a matrix that induces .
- Proof
- Observe that \begin{align*} \vec{x}=\begin{bmatrix}x_1\\x_2\\\vdots \\x_n\end{bmatrix}=x_1\begin{bmatrix}1\\0\\\vdots \\0\end{bmatrix}+x_2\begin{bmatrix}0\\1\\\vdots \\0\end{bmatrix}+\dots +x_n\begin{bmatrix}0\\0\\\vdots \\1\end{bmatrix}=x_1\vec{e}_1+x_2\vec{e}_2+\dots +x_n\vec{e}_n \end{align*}
Because is linear, we have \begin{align*} T(\vec{x})&=T(x_1\vec{e}_1+x_2\vec{e}_2+\dots +x_n\vec{e}_n)=x_1T(\vec{e}_1)+x_2T(\vec{e}_2)+\dots +x_nT(\vec{e}_n)\\ &=\begin{bmatrix} | & |& &|\\ T(\vec{e}_1) & T(\vec{e}_2)&\dots &T(\vec{e}_n)\\ |&| & &| \end{bmatrix}\begin{bmatrix}x_1\\x_2\\\vdots \\x_n\end{bmatrix}=A\vec{x} \end{align*}
Thus, for every in , we have .
Theorem th:matrixtran shows that every matrix transformation is linear. Theorem th:matlin states that every linear transformation from into is a matrix transformation. We combine these results in a corollary.
The results of this section rely on the fact that every vector of can be written as a unique linear combination of the standard unit vectors . These vectors form the standard basis for . We will see in Matrix of a Linear Transformation with Respect to Arbitrary Bases that the matrix used to represent a linear transformation depends on a choice of basis. Because we are using the standard basis, it is natural to name the matrix in Theorem th:matlin accordingly.
Find the standard matrix of a linear transformation if and .
In this example we are
not given the images of the standard basis vectors and . However, we can find the
images of and by expressing and as linear combinations of and , then apply the
fact that is linear.
Let’s start with the easy one. Therefore, by linearity of , we have: This gives us the first column of the standard matrix for .
You can solve the vector equation to express as a linear combination of and as follows: By linearity of , \begin{align*} T(\vec{j})&=T\left (\begin{bmatrix}3\\1\end{bmatrix}+\frac{3}{2}\begin{bmatrix}-2\\0\end{bmatrix}\right )=T\left (\begin{bmatrix}3\\1\end{bmatrix}\right )+\frac{3}{2}T\left (\begin{bmatrix}-2\\0\end{bmatrix}\right )\\ &=\begin{bmatrix}6\\1\\13\\-1\end{bmatrix}+\frac{3}{2}\begin{bmatrix}-2\\0\\-8\\2\end{bmatrix}=\begin{bmatrix}3\\1\\1\\2\end{bmatrix} \end{align*}
This gives us the second column of the standard matrix. Putting all of the information together, we get the following standard matrix for :
Practice Problems
Problems prob:standardmatrix1-prob:standardmatrix4.
Find the standard matrix of each linear transformation described below.
2024-09-07 16:10:34