A linear transformation $f:\reel^4 \rightarrow\reel^3$ has the following mapping matrix with respect to the standard bases for $\reel^4$ and $\reel^3$:
State without further computations whether the vector $\,\mb=(2,9,-5)\,$ belongs to the range $f(\reel^4)$.
answer
We saw above that $\,f(\mathbf u_1)=\mb\,,$ which shows that $\,\mb\,$ is an image vector (the image of $\,\mathbf u_1\,$ when undergoing the mapping $\,f\,$).
C
Compute the dimension of the image space $f(\reel^4)$.
hint
The image space is spanned by the column vectors of $\,\mF\,.$ Therefore the rank of the mapping matrix is decisive.
answer
$$2$$
D
State without further computations the dimension of $\mathrm{ker}(f)\,.$
hint
Use the Dimension Theorem 12.26 in eNote 12.
answer
$$\dim (\ker f)=\dim(\reel^4)-\rho (\mF)=4-2=2$$
E
State without further computations a basis for $\ker(f)$.
hint
Look at the result from a).
answer
The dimension of the kernel is 2 so we need just two linearly independent vectors in the kernel. Since $\mathbf{u}_2$ and $\mathbf{u}_3$ belong to the kernel and evidently are linearly independent, then $\,(\mathbf{u}_2\,,\,\mathbf{u}_3)\,$ is a basis for the kernel.
F
State without further computations a basis for $\,f(\reel^4)\,.$
answer
The dimension of the range (the image space) is 2, so we must find two linearly independent vectors in the range. Since the range is spanned by the columns of the mapping matrix we can use two columns that are linearly independent, e.g. the first two. A basis for the range is thus $\,(\,(1,3,-1)\,,\,(1,0,2)\,)\,.$
G
State without further computations the solution to the vector equation
Read from this a basis for $\mathrm{ker}(f)\,$ and state the dimension of the range $f(\reel^3)$.
hint
Finding the kernel corresponds to solving a system of linear equations. We already have the solution from $\mathrm{rref}(\mF)\,$ where the right-hand side of only 0’s is left out.
hint
From $\mathrm{rref}(\mF)\,$ we read $\,v_1=-3v_3\,$ and $\,v_2=-v_3\,\,.$
hint
Written in standard parametric form the kernel is given by:
The vector $\,(-3,-1,1)\,$ is a basis for the kernel.
$$\mathrm{dim}(\,f(\reel ^3)\,)=2$$
B
Is it also possible to determine a basis for the range?
answer
Not in this case since we do not know the mapping matrix but only its reduced row-echelon form.
Exercise 3: New Mapping Matrix from Change of Basis
If you change the basis, often you can find a mapping matrix that is more simple and therefore easier to work with.
In the vector space $\,\reel^2\,$ we consider the standard basis $\,e=(\,(1,0),(0,1)\,)\,.$ A new basis $\,a=(\ma_1,\ma_2)\,$ for $\,\reel ^2\,$ is given by
State the change-of-basis matrix $\,\matind eMa\,$ that shifts from $a$-coordinates to $e$-coordinates. A vector $\,\mv\,$ has, with respect to basis $a$, the coordinates $\,\vekind av= \begin{matr}{r} -1 \newline 1 \end{matr}\,.$ Determine the coordinates of $\,\mv\,$ with respect to basis $e\,.$
State the change-of-basis matrix $\,\matind aMe\,$ that shifts from $e$-coordinates to $a$-coordinates. A vector $\,\mv\,$ has, with respect to basis $e$, the coordinates $\,\vekind ev= \begin{matr}{r} 2 \newline 3 \end{matr}\,.$ Determine the coordinates of $\,\mv\,$ with respect to basis $a\,.$
Determine the mapping matrix of $f$ with respect to basis $a\,.$
hint
How can you build $\matind aFa$ from $\matind eFe\,?$
hint
$\matind aFa$ requires changing $\matind eFe\,$ for both input and output. For that we need to supplement $\matind eFe\,$ with change-of-basis matrices from either side.
A vector $\,\mv\,$ has, with respect to basis $a$, the coordinate vector $\,\vekind av= \begin{matr}{r} m \newline n \end{matr}\,.$ Determine the coordinate vector of $\,f(\mv)\,$ with respect to basis $a\,.$
answer
$$\begin{matr} m+n \newline n \end{matr}$$
Exercise 4: Linear Transformations of Abstract Vector Spaces
In this exercise we work with abstract vector spaces. We do not know whether we are dealing with number spaces, matrix spaces, polynomial spaces or some entirely different vector space. But this won’t prevent us from investigating a linear transfomation that maps vectors from one vector space to vectors in the other vector space.
A 2-dimensional vector space $V$ has a basis $a=(\ma_1,\ma_2)\,$, and a 3-dimensional vector space $W$ has a basis $c=(\mc_1,\mc_2,\mc_3)\,$. A linear transformation $\,f:V\rightarrow W\,$ is given by
State the mapping matrix $\,\matind cFa\,$, and compute the image $\,\mathbf y\,$ of the vector $\,\mathbf x=3\ma_1-\ma_2\,$ using this mapping matrix.
hint
Look up the definition of the mapping matrix, see 12.17 in eNote 12.
hint
Use Theorem 12.18 in eNote 12 to find the image $\,\mathbf y\,$.
hint
See Example 12.20 in eNote 12.
answer
$$
\matind cFa = \begin{matr}{rr}
1&-2\newline
-2&4\newline
1&-2\end{matr}
$$
$$\mathbf y = f(\mathbf x)=5\mc_1-10\mc_2+5\mc_3$$
B
Which of the vectors $\,\mathbf \ma_1+2\ma_2\,$ and $\,\mathbf 2\ma_1+\ma_2\,$ belong to the kernel of $f\,$? Solve the exercise without determining all of the kernel.
hint
Compute the matrix-vector product of $\matind cFa$ with each of the two given vectors.
answer
$$2\ma_1+\ma_2\in \ker(f)$$
$$\ma_1+2\ma_2 \notin \ker(f)$$
C
Determine without further computations a basis for the kernel of $f\,$.
answer
The kernel must be 1-dimensional, therefore the vector with $a$-coordinates $(2,1)$ is a basis for $\ker(f)$.
D
Which of the vectors $\,\mc_1-2\mc_2+\mc_3\,$ and $\,2\mc_1-\mc_2+2\mc_3\,$ belong to $f(V)\,$?
hint
See Method 12.24 in eNote 12.
answer
Only $\mc_1-2\mc_2+\mc_3\in f(V)$.
E
State a basis for the range of $f\,$.
answer
The range is 1-dimensional, therefore $\mc_1-2\mc_2+\mc_3$ is a basis.
Exercise 5: Linear Mapping and Change of Basis. Maple
Show that the set $\,v=(\mv_1,\mv_2,\mv_3)\,$ constitutes a basis for $\,\reel^3\,$ and that the set $\,w=(\mw_1,\mw_2,\mw_3,\mw_4)\,$ constitutes a basis for $\,\reel^4\,.$
Since $(\mv_1,\mv_2,\mv_3)$ and $(\mw_1,\mw_2,\mw_3,\mw_4)$ constitute bases for $\reel^3$ and $\reel^4$, respectively, we can state $\matind wFv$ using the coefficients from the expression of $f$.
Exercise 6: Extra Training Exercise in Linear Mappings
Let $(\mathbf{e}_1,\mathbf{e}_2)$ denote the standard basis for $\reel ^2$ and let $c=(\mathbf{c}_1,\mathbf{c}_2,\mathbf{c}_3,\mathbf{c}_4)$ denote a basis for $\reel ^4$. Now let $f:\reel ^2\rightarrow\reel ^4$ be a linear transformation, where
Solve the linear equation $\,f(\mx) =5\mc_1+3\mc_2-3\mc_3+17\mc_4\,.$
hint
See Method 12.24.
hint
What is the coordinate vector $\,\vekind cb\,$ of $\,\mb=5\mc_1+3\mc_2-3\mc_3+17\mc_4\,$?
hint
Solve the matrix equation $\matind cFe \cdot \vekind ex = \vekind cb\,$ by computing the reduced row-echelon form of the corresponding augmented matrix.
answer
$\vekind ex =\begin{matr}{rr} 3 \newline 2 \end{matr}\,$ or $\,\mathbf x= 3\mathbf e_1+2\mathbf e_2\,.$
