\[T\Big (\begin{bmatrix}4\\-1\end{bmatrix}\Big )=\begin{bmatrix}\answer {-16}\\\answer {9}\\\answer {-1}\end{bmatrix}\]
Standard Matrix of a Linear Transformation from \(\RR ^n\) to \(\RR ^m\)
In Matrix Transformations and Introduction to Linear Transformations we learned several important properties of matrix transformations of \(\RR ^n\) and subspaces of \(\RR ^n\). Let’s summarize the main points.
For a matrix transformation \(T:\RR ^n\rightarrow \RR ^m\), induced by an \(m\times n\) matrix \(A\) we
have the following results:
-
\(T\) is linear. (Theorem th:matrixtran) This means that
\[T(k_1\vec {u}+k_2\vec {v})= k_1T(\vec {u})+k_2T(\vec {v})\]for vectors \(\vec {u}\) and \(\vec {v}\) in \(\RR ^n\) and scalars \(k_1\) and \(k_2\) in \(\RR \). -
Columns of \(A\) are the images of the standard unit vectors of \(\RR ^n\) under \(T\). (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 \(T\) on all of the elements of \(\RR ^n\) is completely determined by where \(T\) 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 \(T:\RR ^3\rightarrow \RR ^2\) be a linear transformation. Suppose that the only information we have about
this transformation is that \(T(\vec {i})=\begin{bmatrix}3\\-1\end{bmatrix}\), \(T(\vec {j})=\begin{bmatrix}0\\4\end{bmatrix}\) and \(T(\vec {k})=\begin{bmatrix}-2\\1\end{bmatrix}\). Is this information sufficient to determine the
image of \(\vec {w}=\begin{bmatrix}1\\-3\\6\end{bmatrix}\)?
Observe that
\[\vec {w}=\vec {i}-3\vec {j}+6\vec {k}\]
We find \(T(\vec {w})\) by using the fact that \(T\) is linear.
\[T(\vec {w})=T(\vec {i}-3\vec {j}+6\vec {k})=T(\vec {i})-3T(\vec {j})+6T(\vec {k})=\begin{bmatrix}3\\-1\end{bmatrix}-3\begin{bmatrix}0\\4\end{bmatrix}+6\begin{bmatrix}-2\\1\end{bmatrix}=\begin{bmatrix}-9\\-7\end{bmatrix}\]
Because of properties of linear transformations, the information about the images of
the standard unit vectors proved to be sufficient for us to determine the image of \(\vec {w}\). In Example ex:imageofatransformation, there was nothing special about the vector \(\vec {w}\). Any vector \(\vec {x}\) of \(\RR ^n\) can be written as a unique linear combination of the standard unit vectors \(\vec {e}_1,\ldots , \vec {e}_n\). Therefore, the image of any vector \(\vec {x}\) under a linear transformation \(T:\RR ^n\rightarrow \RR ^m\) is uniquely determined by the images of \(\vec {e}_1, \ldots , \vec {e}_n\). Knowing \(T(\vec {e}_1),\ldots , T(\vec {e}_n)\) allows us to construct a matrix \(A\), with \(T(\vec {e}_1),\ldots , T(\vec {e}_n)\) as columns, that induces transformation \(T\). We formalize this idea in a theorem.
Let \(T:\RR ^n\rightarrow \RR ^m\) be a linear transformation. Then \(T\) 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 \(T\). \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 \(T\) 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 \(\vec {x}\) in \(\RR ^n\), we have \(T(\vec {x})=A\vec {x}\).
Theorem th:matrixtran shows that every matrix transformation is linear. Theorem th:matlin states that every linear transformation from \(\RR ^n\) into \(\RR ^m\) is a matrix transformation. We combine these results in a corollary.
A transformation \(T:\RR ^n\rightarrow \RR ^m\) is a linear transformation if and only if it is a matrix
transformation.
The results of this section rely on the fact that every vector of \(\RR ^n\) can be written as a unique linear combination of the standard unit vectors \(\vec {e}_1,\vec {e}_2,\dots ,\vec {e}_n\). These vectors form the standard basis for \(\RR ^n\). 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.
The standard matrix of a linear transformation \(T:\RR ^3\rightarrow \RR ^2\) such that \(T(\vec {i})=\begin{bmatrix}2\\-1\end{bmatrix}\), \(T(\vec {j})=\begin{bmatrix}-1\\3\end{bmatrix}\) and \(T(\vec {k})=\begin{bmatrix}0\\4\end{bmatrix}\) is
\[A=\begin{bmatrix}2&-1&0\\-1&3&4\end{bmatrix}\]
Find the standard matrix of a linear transformation \(T:\RR ^2\rightarrow \RR ^2\) such that \(T(\vec {i})=2\vec {i}\) and \(T(\vec {j})=2\vec {j}\).
Find the standard matrix of a linear transformation \(T:\RR ^2\rightarrow \RR ^4\) if \(T\left (\begin{bmatrix}3\\1\end{bmatrix}\right )=\begin{bmatrix}6\\1\\13\\-1\end{bmatrix}\) and \(T\left (\begin{bmatrix}-2\\0\end{bmatrix}\right )=\begin{bmatrix}-2\\0\\-8\\2\end{bmatrix}\).
In this example we are
not given the images of the standard basis vectors \(\vec {i}\) and \(\vec {j}\). However, we can find the
images of \(\vec {i}\) and \(\vec {j}\) by expressing \(\vec {i}\) and \(\vec {j}\) as linear combinations of \(\begin{bmatrix}3\\1\end{bmatrix}\) and \(\begin{bmatrix}-2\\0\end{bmatrix}\), then apply the
fact that \(T\) is linear.
Let’s start with the easy one.
\[\vec {i}=-\frac {1}{2}\begin{bmatrix}-2\\0\end{bmatrix}\]
Therefore, by linearity of \(T\), we have:
\[T(\vec {i})=T\left (-\frac {1}{2}\begin{bmatrix}-2\\0\end{bmatrix}\right )=-\frac {1}{2}T\left (\begin{bmatrix}-2\\0\end{bmatrix}\right )=-\frac {1}{2}\begin{bmatrix}-2\\0\\-8\\2\end{bmatrix}=\begin{bmatrix}1\\0\\4\\-1\end{bmatrix}\]
This gives us the first column of the standard matrix for \(T\).
You can solve the vector equation
\[a\begin{bmatrix}3\\1\end{bmatrix}+b\begin{bmatrix}-2\\0\end{bmatrix}=\vec {j}\]
to express \(\vec {j}\) as a linear combination of \(\begin{bmatrix}3\\1\end{bmatrix}\) and \(\begin{bmatrix}-2\\0\end{bmatrix}\) as follows:
\[\vec {j}=\begin{bmatrix}3\\1\end{bmatrix}+\frac {3}{2}\begin{bmatrix}-2\\0\end{bmatrix}\]
By linearity of \(T\), \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 \(T\):
\[A=\begin{bmatrix}1&3\\0&1\\4&1\\-1&2\end{bmatrix}\]
Practice Problems
Suppose that a linear transformation \(T:\RR ^2\rightarrow \RR ^3\) is such that \(T(\vec {i})=\begin{bmatrix}-4\\2\\1\end{bmatrix}\) and \(T(\vec {j})=\begin{bmatrix}0\\-1\\5\end{bmatrix}\). Find \(T\Big (\begin{bmatrix}4\\-1\end{bmatrix}\Big )\).
Suppose that a linear transformation \(T:\RR ^2\rightarrow \RR ^3\) is such that \(T\Big (\begin{bmatrix}1\\-1\end{bmatrix}\Big )=\begin{bmatrix}1\\4\\-1\end{bmatrix}\) and \(T\Big (\begin{bmatrix}2\\0\end{bmatrix}\Big )=\begin{bmatrix}0\\6\\4\end{bmatrix}\). Find the standard matrix \(A\)
of \(T\).
\[A=\begin{bmatrix}\answer {0}&\answer {-1}\\\answer {3}&\answer {-1}\\\answer {2}&\answer {3}\end{bmatrix}\]
Find the standard matrix \(A\) of the linear transformation \(T:\RR ^2\rightarrow \RR ^2\) if \(T\) doubles the \(x\) component
of every vector and triples the \(y\) component.
\[A=\begin{bmatrix}\answer {2}&\answer {0}\\\answer {0}&\answer {3}\end{bmatrix}\]
Find the standard matrix \(A\) of the linear transformation \(T:\RR ^2\rightarrow \RR ^2\) if \(T\) reverses the direction of
each vector.
\[A=\begin{bmatrix}\answer {-1}&\answer {0}\\\answer {0}&\answer {-1}\end{bmatrix}\]
Find the standard matrix \(A\) of the linear transformation \(T:\RR ^2\rightarrow \RR ^2\) if \(T\) doubles the length of
each vector.
\[A=\begin{bmatrix}\answer {2}&\answer {0}\\\answer {0}&\answer {2}\end{bmatrix}\]
Find the standard matrix \(A\) of the linear transformation \(T:\RR ^2\rightarrow \RR ^2\) if \(T\) projects each vector
onto the \(x\)-axis. (e.g. \(T\left (\begin{bmatrix}4\\5\end{bmatrix}\right )=\begin{bmatrix}4\\0\end{bmatrix}\))
\[A=\begin{bmatrix}\answer {1}&\answer {0}\\\answer {0}&\answer {0}\end{bmatrix}\]
Find the standard matrix \(A\) of the linear transformation \(T:\RR ^2\rightarrow \RR ^2\) if \(T\) projects each vector
onto the \(y\)-axis. (e.g. \(T\left (\begin{bmatrix}4\\5\end{bmatrix}\right )=\begin{bmatrix}0\\5\end{bmatrix}\))
\[A=\begin{bmatrix}\answer {0}&\answer {0}\\\answer {0}&\answer {1}\end{bmatrix}\]