Show that exactly one of the two maps is linear. Find out which by investigating whether they fulfill the two requirements for linearity.
hint
Read definition 12.5 in eNote 12.
hint
You must investigate whether
1. $f((x_1, x_2)+(y_1,y_2))=f((x_1, x_2))+f((y_1,y_2))$
2. $f(k(x_1,x_2))=k\cdot f((x_1,x_2))$
answer
It is $\,f\,$ that is linear. That $\,g\,$ is not linear is most easily shown by a counter example.
B
State the kernel for the linear map found.
hint
To find $\ker(f)$, you must solve the system of equations $(x_1-x_2,-x_1+x_2)=(0,0)\,.$
answer
$(x_1,x_2)=t(1,1)$ where $t \in \reel\,$. In another way: A basis for $\ker (f)$ is given by $\,(\,(1,1)\,)\,.$
C
State the range for the linear map found.
hint
The image space corresponds to the concept of range for an elementary function known from highschool, i.e. all the possible values of the linear map.
hint
Therefore we must find all vectors $\mb=(b_1,b_2)\,$ that has the possibility to be a right-hand side in the equation $(x_1-x_2,-x_1+x_2)=(b_1,b_2)\,.$
answer
$(x_1,x_2)=t(-1,1)$ where $t \in \reel\,.$ In another way: A basis for $\,f(\reel^2)\,$ is given by $\,(\,(-1,1)\,)\,.$
Exercise 2: Investigation of a Linear Map
Let $f:\reel ^4\rightarrow \reel^3$ be given by the expression
Show using main Theorem 12.18 i eNote 12, point 2, that $f$ is linear, and state the mapping matrix $ \matind eFe$ for $f$ with respect to the standard bases in $\reel^4$ and $\reel^3$.
hint
The mapping matrix $ \matind eFe$ must fulfill the equation $ \vekind ey= \matind eFe \cdot \vekind ex$, where $ \vekind ey$ corresponds to $f((x_1,x_2,x_3,x_4))$ and $\vekind ex$ corresponds to $(x_1,x_2,x_3,x_4)$.
hint
State a guess for a mapping matrix using Definition 12.17 in eNote 12. Does the matrix match the expression for for $f\,$?
Find the dimension of the image space and state a basis for the image space.
hint
The number of vectors in the basis equals the dimension of the image space, which in turn are related to the rank of the mapping matrix.
hint
You can find the basis vectors as some of the columns in the mapping matrix, but which and how many?
hint
See Method 12.25 in eNote 12.
answer
The dimension of the image space equals the rank of the mapping matrix, that is 2. Thus the image space is spanned by two vectors. the easy choice is the first two columns in the mapping matrix, that are linearly independent, so $f(\reel^4)=\spanVec {(1,3,2),(1,-1,2)}$. Therefore the basis for the image space is $(\,(1,3,2),(1,-1,2)\,)$.
C
State a basis for the kernel of the map.
hint
The kernel of the map, $\ker(f)$, is a space that consists of all vectors that fulfill the equation $f(\mx)=\mnul$.
hint
Therefore we must solve $f(\mx)=\mnul$, that can also be formulated as $\matind eFe \cdot \vekind ex = \mnul$.
hint
See Method 12.23 in eNote 12: State an augmented matrix, consisting of the matrix $\matind eFe$ complemented with a column of zeros and solve the system of equation using Gauss-Jordan elimination.
answer
Since the rank of $\matind eFe$ is 2 and the number of columns is 4, $\dim (\,\ker (f)\,) =4-2=2$. Therefore the basis must be two linearly independent vectors, spanning the space. Following the Gauss-Jordan elimination it is obvious to choose $(-\frac{5}{4},-\frac{7}{4},1,0)$ and $(-\frac{5}{4},\frac{1}{4},0,1)$.
D
Does $(1,2,3)$ belong to the image space for $f(\reel ^4)\,$?
hint
See Method 12.24 in eNote 12.
answer
No.
E
Solve the vector equation $\,f(\mathbf x)=(2,2,4)\,$.
hint
See Method 12.24 in eNote 12
answer
$\mathbf x = (1,1,0,0)+\ker f\,$.
Exercise 3: Linear Maps in the Plane
We consider in the following an ordinary coordinate system $\,(O, \mathbf i, \mathbf j)\,$ in the plane. All vectors are considered to be drawn from the origin. An arbitrary vector $\,\mathbf x\,$ is drawn in blue, while the image vector $\,\mathbf y\,$ is red. $\,\mathbf F\,$ states the mapping matrix for $f\,$ with respect to the standard base.
by moving the column vectors $\,\mathbf s_1\,$ and $\,\mathbf s_2\,$ using the mouse. Then find the image of $\,(1,2)\,$ by moving $\,\mathbf x\,$ to $\,(1,2)\,$ using the mouse.
Find the image of the basis vector $\,\mathbf i\,$ by pulling $\,\mathbf x\,$ to $(1,0)\,$. Do the same thing with the basis vector $\,\mathbf j\,$. Do the images of the basis vectors fit the numbers in $\,\mathbf F\,$?
What happens to the image vectors when $\,\mathbf x\,$ is moved about?
Compute det$\,(\mathbf F)\,$, and determine the rank of $\,\mathbf F\,$. Determine a basis for the image space.
What should our expectation of the dimension of the kernel be? Determine an equation for the straight line that contains the kernel (Hint: Start by finding a vector that is mapped onto the 0-vector by moving $\,\mx\,$).
The idea is that $\,\mathbf x\,$ is bound to the line segment shown. Move $\,\mathbf x\,$, and follow the image $\,\mathbf y\,$.
Displace the line segment parallel to itself using the mouse and again move $\,\mathbf x\,.$ What happens to the image. Possibly try other settings for $\,\mathbf F\,$. Summarize your observations in a hypothesis.
Introductory exercise: How should $\,\mathbf F\,$ be changed so that the blue house is mapped onto the mirror image in the y-axis? Same question for the x-axis.
It is given that the kernel for $f$ has the dimension 1. Find immediately, just by using your brain, a basis for $\,f(V)\,$.
answer
A possible basis is $\,(\,(1,2,3),(1,0,0)\,)\,$.
C
%\begin{exercise}\label{tn8.opgDimension2}
In space an ordinary coordinate system $\,(O,\mathbf i,\mathbf j,\mathbf k)$ is given. All vectors are imagined to be drawn from the origin. The map $\,p\,$ projects vectors down into the $(X,Y)$-plane in space, see the figure
Show that $\,p\,$ is linear, and state the mapping matrix $\matind ePe$ for $p$ with respect to the standard base $e\,.$ Determine a basis for the kernel and the image space of the projection. Check that the dimension theorem is fulfilled.
A possible basis for $\,\ker(p)\,$ is $\,(\mathbf k)\,$. A possible basis for the image space $\,(\mathbf i,\mathbf j)\,$. The domain has the dimensionen 3, the kernel has the dimension 1 and the image space has the dimension 2. since $3=1+2$ the dimension theorem is fulfilled.
Exercise 6: Mapping Matrices for Reflections
In the plane an ordinary $\,(O,\mathbf i,\mathbf j)$-coordinate system is given, and all vectors are imagined to be drawn from the origin. As mentioned in Exercise 12.3 in eNote 12 reflections in lines through the origin are linear.
Here we consider mirror imaging of vectors in the line $\,y=x\,.$ Let us call this linear map $s\,.$
A
Determine $s(\mathbf i)$ and $s(\mathbf j)$, state the mapping matrix $\matind eSe$ for $s\,$ and determine an expression for the mirror image of an arbitrary plane vector $\,\mathbf u\,$ with the $e$-coordinates $(u_1,u_2)\,$.
answer
$\matind eSe =\begin{matr}{rr}0&1\\1&0\end{matr}\,$. The $e$-coordinate vector for the image is $\,\begin{matr}{r}u_2\\u_1\end{matr}\,$.
We consider a new $\,(O,\mathbf v_1,\mathbf v_2)$-coordinate system in which all vectors are imagined to be drawn from the origin. $\,\mv_1\,$ is a unit vector along the line $\,y=\frac 12\,x\,,$ as shown in the figure, and $\,\mv_2\,$ is the vector perpendicular to $\,\mv_1\,.$
We wish to find the mapping matrix $\matind eRe$ for the linear map $\,r\,$ that reflects vectors in the line $\,y=\frac 12\,x\,.$ We do this in two steps.
B
Determine the mapping matrix $\matind vRv$ for $r\,$ with respect to the base $\,v=(\mv_1,\mv_2)\,.$
hint
What happens to the basis vectors $\,\mv_1\,$ og $\,\mv_2\,$ using this map.
Determine the mapping matrix $\matind eRe$ for $r\,$ with respect to the standard base. Let $\,\mathbf u\,$ be an arbitrary vector in the plane with the $e$-coordinates $(u_1,u_2)\,.$ Determine an expression for the mirror image of $\,\mathbf u\,$ in the line $\,y=\frac 12\,x\,.$
hint
You will need the basis shift matrix $\,\matind eMv\,.$ And possibly even its inverse.
answer
$$\matind eRe =\begin{matr}{rr}\frac35&\frac45\\\\ \frac45&-\frac35\end{matr}\,.$$
Exercise 7: Play with Mapping Matrices by Change of Base
Given the vectors $\,\ma_1=(1,2)\,$ and $\,\ma_2=(3,7)\,$ in $\,\reel^2\,$ and $\,\mc_1=(1,2,2)\,,$$\,\mc_2=(2,3,1)\,$ and $\,\mc_3=(1,2,1)\,$ in $\,\reel^3\,$. Let the linear map $\,f:\reel^2\rightarrow\reel^3\,$ be given by
Show that $\,\ma_1\,$ and $\,\ma_2\,$ constitute a basis for $\,\reel^2\,$ and that $\,\mc_1\,$, $\,\mc_2\,$ and $\,\mc_3\,$ constitute a basis for $\,\reel^3\,.$
hint
Basis for a vector space is a set of vectors, that spans the space and that are mutually, linearly independent.
hint
Enter the two (three) vectors as columns in a matrix. If this matrix is regular, the vectors are linearly independent.
answer
Since $\ma_1$ and $\ma_2$ are linearly independent and span $\reel^2$, they constitute a basis for $\reel^2$, and since $\mc_1$, $\mc_2$ and $\mc_3$ are linearly independent and span $\reel^3$, they constitute a basis for $\reel^3$.
B
State the mapping matrix for $\,f\,$ with respect to the basis $\,(\ma_1,\ma_2)\,$ in $\,\reel^2\,$ and the base $\,(\mc_1,\mc_2,\mc_3)\,$ in $\,\reel^3\,$.
hint
You shall state a matrix $\matind cFa$, such that $\vekind cy=\matind cFa\cdot\vekind ax$.
Remember that in relation to the base $(\ma_1,\ma_2)$, we have $\ma_1=\begin{matr}{rr} 1 \\ 0 \end{matr}$ and $\ma_2=\begin{matr}{rr} 0 \\ 1 \end{matr}$.
hint
The elements in $_\mathrm cf(\ma_1))$ are the coefficients from $f(\ma_1)$.
The set of second degree polynomials $\,P_2(\reel)\,$ can be viewed as a 3-dimensional vector space. The real numbers $\,\reel\,$ is a 1-dimensional vector space. We investigate maps from the first vector space to the second.
%Hvis man vælger standard monomie-basen $(1,x,x^2)\,$, kan ethvert andengradspolynomium beskrives ved en koordinatvektor, hvor polynomiets koefficienter udgør vektorens koordinater. For eksempel har $P(x)=3x^2+4x-1$ koordinatvektoren $(-1,4,3)$ i forhold til monomie-basen.
A map $\,f:P_2(\reel)\rightarrow \reel\,$ is given by
$$\,f(P(x))=P\,'(1)\,.$$
We illustrate with a couple of examples:
F
Determine $\,f(x^2)\,$ and $\,f(-x^2+2x-2)\,,$ see the figure.
G
Show that $f$ is linear.
hint
You must show
1. $f(P(x)+Q(x))=f(P(x))+f(Q(x))$
2. $f(k(P(x))=k\cdot f(P(x))$
answer
See example 12.8 in eNote 8.
H
One of the two polynomials in the figure belongs to the kernel for $\,f\,,$ which? Determine a basis for $\,\ker (f)\,.$
hint
What must the coefficients in a polynomial that belongs to ker$(f)\,$ fulfill?
answer
A basis for the kernel is given by $\,(1\,,\,x^2-2x)\,.$
I
Show that the image space $\,f(P_2(\reel))\,$ for $\,f\,$ is equal to the codomain for $\,f\,.$
A map $\,g:P_2(\reel)\rightarrow \reel\,$ is given by
