A linear transformation $f:\reel^4 \rightarrow\reel^3$ has the following mapping matrix with respect to the standard bases in $\reel^4$ and $\reel^3\,:$
Decide without 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 $\,\mathbf u_1\,$ in the mapping $\,f\,$).
C
Find the dimension of the image space $f(\reel^4)$.
hint
The image space is spanned by the column vectors in $\,\mF\,.$ Therefore the rank of the mapping matrix is decisive.
answer
The rank of the mapping matrix = the dimension of the range = 2.
D
Find without further computations the dimension of ker$(f)\,.$
hint
Use the DimensionsTheorem 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 e.g. at the result in Question a).
answer
Since the dimension of the kernel is 2, we must find two linearly independent vectors in the kernel. Since $\mathbf{u}_2$ and $\mathbf{u}_3$ belong to the kernel and evidently are linearly independent, $\,(\mathbf{u}_2\,,\,\mathbf{u}_3)\,$ is a basis for the kernel.
F
State without further computations a basis for $\,f(\reel^4)\,.$
answer
Since the dimension of the kernel is 2, 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 $\,(\,(1,3,-1)\,,\,(1,0,2)\,)\,.$
G
State without further computations the solution to the vector equation
Read from this a basis for ker$(f)\,$ and state the dimension of the range $f(\reel^3)$.
hint
To find the kernel corresponds to solving a system of linear equations. We already have the solution with $\mathrm{trap}(\mF)\,$ where the right-hand side of only 0’s is left out!
hint
From $\mathrm{trap}(\mF)\,$ 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. Dim$(\,f(\reel ^3)\,)=2\,.$
I
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 Base
If you change the basis, often you can find a mapping matrix that is more simple and therefore easier to work with. We shall see an example of this here where we first start by some exercises in change of base.
In the vector space $\,\reel^2\,$ we consider the standard base $\,e=(\,(1,0),(0,1)\,)\,.$ A new basis $\,a=(\ma_1,\ma_2)\,$ for $\,\reel ^2\,$ are determined by
State the basis shift matrix $\,\matind eMa\,$ that shifts from $a$-coordinates to $e$-coordinates. A vector $\,\mv\,$ has with respect to the basis $a$ the coordinate matrix $\,\vekind av= \begin{matr}{r} -1 \\ 1 \end{matr}\,.$ Determine the coordinate matrix for $\,\mv\,$ with respect to the basis $e\,.$
answer
We read $\matind eMa=\begin{matr}{rr} 1 & 3 \\ 2 & 7 \end{matr}\,.$ And from this we get $\,\vekind ev= \begin{matr}{rr} 1 & 3 \\ 2 & 7 \end{matr}\cdot \begin{matr}{r} -1 \\ 1 \end{matr}=\begin{matr}{r} 2 \\ 5 \end{matr}\,.$
K
State the basis shift matrix $\,\matind aMe\,$ that shifts from $e$-coordinates to $a$-coordinates. A vector $\,\mv\,$ has with respect to basis $e$ the coordinate matrix $\,\vekind ev= \begin{matr}{r} 2 \\ 3 \end{matr}\,.$ Determine the coordinate matrix for $\,\mv\,$ with respect to the basis $a\,.$
answer
$\matind aMe=\matind eMa^{-1}=\begin{matr}{rr} 7 & -3 \\ -2 & 1 \end{matr}\,.$ And from this we get
$\,\vekind ev= \begin{matr}{rr} 7 & -3 \\ -2 & 1 \end{matr}\cdot \begin{matr}{r} 2 \\ 3 \end{matr}=\begin{matr}{r} 5 \\ -1 \end{matr}\,.$
Let $f:\reel ^2\rightarrow\reel ^2$ be a linear transformation that with respect to the standard $e$-basis in $\reel ^2$ has the mapping matrix
Determine the mapping matrix for $f$ with respect to the basis $a\,.$
hint
How can you build $\matind aFa$ from $\matind eFe\,?$
hint
$\matind aFa$ requires changing $\matind eFe\,$ changes both in input and output. Therefore we need to supply $\matind eFe\,$ with basis shift matrices both in front and behind.
The basis shift matrix is with respect to the new basis an upper triangular matrix with all nonsero elements equal to one.
M
A vector $\,\mv\,$ has with respect to the basis $a$ the coordinate vector $\,\vekind av= \begin{matr}{r} m \\ n \end{matr}\,.$ Determine the coordinate vector for $\,f(\mv)\,$ with respect to the basis $a\,.$
answer
$$\,\begin{matr}{r} m+n \\\\ n \end{matr}\,.$$
Exercise 4: Linear Transformation in Abstract Vector Spaces
In this Exercise we work with abstract vector spaces. Thus, we do not know whether it concerns number spaces, matrix spaces polynomial spaces or you name it. This does not prevent us from investigating a linear transfomation that maps vectors in the one vector space onto 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 find the image $\,\mathbf y\,$ of the vector $\,\mathbf x=3\ma_1-\ma_2\,$ using the mapping matrix.
hint
State the definition of the mapping matrix, see 12.17 in eNote 12.
hint
Use the Main Theorem 12.18 in eNote 12 to find the image $\,\mathbf y\,$.
hint
See Example 12.20 in eNote 12.
answer
$$
\matind cFa = \begin{matr}{rr}
1&-2\\\\
-2&4\\\\
1&-2\end{matr}\,.
$$
$\mathbf y = f(\mathbf x)=5\mc_1-10\mc_2+5\mc_3\,.$
O
Which of the vectors $\,\mathbf \ma_1+2\ma_2\,$ and $\,\mathbf 2\ma_1+\ma_2\,$ belong to the kernel for $f\,$? Solve the exercise without determining all of the kernel.
hint
Compute the matrix-vector product of $\matind cFa$ and each of the two given vectors’ coordinate vectors.
answer
$2\ma_1+\ma_2\in \ker(f)\,$.
$\ma_1+2\ma_2 \notin \ker(f)\,$.
P
Determine (readily without making new computations) a basis for the kernel for $f\,$.
answer
The kernel must be 1-dimensional, therefore the vector with the $a$-coordinates $(2,1)$ is a basis for $\ker(f)$.
Q
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 12.22 in eNote 12.
answer
Only $\mc_1-2\mc_2+\mc_3\in f(V),$.
R
State a basis for the range for $f\,$.
answer
The range is 1-dimensional, therefore $\mc_1-2\mc_2+\mc_3$ must be a basis.
Exercise 5: Linear Mapping and Change of Base. 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 for $f$.
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 some given basis for $\reel ^4$. Now let $f:\reel ^2\rightarrow\reel ^4$ be a linear transformation, where