Let $\,f:\reel^3\rightarrow\reel^3\,$ be the mapping that with respect to the standard basis for $\,\reel^3\,$ has the mapping matrix
\begin{equation}
\mA=\begin{matr}{rrr} 3 & 4 & 4 \\ 6 & 6 & 6 \\ -6 & -7 & -7 \end{matr}.
\end{equation}
A
What is the easiest way to check whether the vectors $\,\mv_1=(1,0,-1)\,$, $\,\mv_2=(0,1,-1)\,$ and $\,\mv_3=(1,2,-2)\,$ are eigenvectors for $\,f\,?$ Do this, and state the corresponding eigenvectors!
hint
Since we already have some suggested eigenvectors, we do not need to compute the characteristic polynomial, etc.
hint
Compute $\,\mA\cdot\mv_1\,.$ Perhaps the result looks a bit like $\,\mv_1\,$….
answer
$\mA\cdot\mv_1=\begin{matr}{rrr} -1 \\ 0 \\ 1 \end{matr}=-1\cdot\mv_1\,$, so $\,\lambda_1=-1\,$.
$\mA\cdot\mv_2=\begin{matr}{rrr} 0 \\ 0 \\ 0 \end{matr}=0\cdot\mv_2\,$, so $\,\lambda_2=0\,$.
$\mA\cdot\mv_3=\begin{matr}{rrr} 3 \\ 6 \\ -6 \end{matr}=3\cdot\mv_3\,$, so $\,\lambda_3=3\,$.
B
How can we most easily argue that $\mv_1,\,\mv_2$ and $\mv_3$ are linearly independent.
hint
Apply with advantage Theorem 13.11, point 1, in eNote 13.
C
How can we most easily show, that $\,f(\reel^3)=\spanVec{\mv_1,\mv_3}\,?$
hint
Use that $(\mv_1,\mv_2,\mv_3)$ is a basis for $\reel^3$ consisting of eigenvectors for $f\,,$ an eigenbasis. The image space $f(\reel^3)$ is (as always) spanned by the images of the basis vectors.
hint
$f(\mv_1)$, $f(\mv_2)$ and $f(\mv_3)$ we have just computed.
A standard $\,(O,\mathbf i,\mathbf j,\mathbf k)$-coordinate system is given. All vectors are thought to be drawn from the origin. The map $\,p\,$ projects vectors down into the $(X,Y)$-plane in space, see the Figure. It is given that $\,p\,$ is linear (need not be proved).
Now we shall consider the eigenvalue problem for the projection down into the $\,(x,y)\,$-plane.
D
Determine all eigenvalues for $\,p\,$ and the eigen-spaces that belong to the eigenvalues, solely by brain power (pondering).
hint
There are two eigenvalues that are now given to you. The first one is the number 1. So think about: Which vectors are fix-vectors for $\,p\,,$ that is, they are mapped onto themselves? The other one is the number 0. So think about: Which vectors are mapped onto a single point the ($\,\mnul$-vector).
answer
$\,E_1\,$ is all vectors in the $\,(x,y)$-plane in space. They do not move at all by the projection down onto the $\,(x,y)$-plane. $\,E_0\,$ are all vectors on the $\,z$-axis, they are mapped onto the origin.
E
Choose two different eigenbases (these are bases consisting of eigenvectors for $\,p\,$) and determine in each of the two cases the diagonal matrix that becomes the mapping matrix for $\,p\,$ with respect to the chosen basis.
hint
First use the standard base $\,(\mathbf i,\mathbf j,\mathbf k)\,$. How?
answer
Using the standard base we find the mapping matrix:
Using e.g. the eigenbasis $\,\big(\,(1,1,0),(1,-1,0),(0,0,-1)\,\big)\,$ we get the same mapping matrix. Find yourself an eigenbasis that do not have the same mapping matrix.
State eigenvalues and all corresponding eigenvectors for the linear map $\,f:\reel^3\rightarrow\reel^3\,$ that with respect to the standard basis $\,e\,$ in $\,\reel^3\,$ has the mapping matrix $\,\mA\,.$
answer
It appears from the Mapleoutput that $\,\mA\,$ has the eigenvalues 4, -2 and 3, all with algebraic and geometric multiplicity 1.
The corresponding eigenvector spaces is also apparent from the output, since $\,E_4=\spanVec{(-2,-2,1)}\,,$$\,E_3=\spanVec{(1,1,0)}\,$ and $\,E_{-2}=\spanVec{(-1,-2,4)}\,.$
C
Find a basis $\,v=(\mv_1,\mv_2,\mv_3)\, $ for $\,\reel^3\,$ consisting of eigenvectors for $\,f\,.$
hint
Have another look at the important Theorem 13.11 in eNote 13.
answer
$\mv_1=(-2,-2,1)\in E_4\,$, $\,\mv_2=(1,1,0)\in E_{3}\,$ and $\,\mv_3=(-1,-2,4)\in E_{-2}\,$ can according to Theorem 13.11, point 1, in eNote 13 be a basis for $\,\reel^3\,.$
D
Find the mapping matrix for $\,f\,$ with respect to the basis $\,v\,$ found in the preceding chapter.
Diagonalization is possible, because two linearly independent eigenvectors exist for $\,\mA\,.$
More correct answers are possible, here is one of them:
Note that 0 is one of the eigenvalues.
Diagonalization is possible, since three linearly independent eigenvectors exist for $\,\mB\,.$
More correct answers are possible, here is one of them:
Show that $\mA$ and $\mB$ are similar with the same diagonal matrix
answer
If we set $\,\mV=\begin{matr}{cc} -i&i\\1&1\end{matr}\,,$$\,\mU=\begin{matr}{cc} i&-i\\1&1\end{matr}\,$ and $\,\mathbf{\Lambda}=\begin{matr}{cc} i&0\\0&-i\end{matr}\,$ then it applies that:
Advanced:
Now we consider $\,\mA\,$ to be a mapping matrix for a linear map $\,f:\reel^2\rightarrow\reel^2\,$ with respect to the standard base in $\,\reel^2\,.\,$ Determine a new basis $\,m\,$ for $\,\reel^2\,$ with respect to which $\,f\,$ is represented by the mapping matrix $\,\mB\,.$
Find eigenvalues and the corresponding complex eigenvector spaces for $\,\mA\,.$
answer
The eigenvalues are $1+i$, $1-i$ og 3. All with algebraic multiplicity 1.
The eigenvectors corresponding to $\lambda=1+i$ are $\mx=t_1\cdot\begin{matr}{rrr} 0 \\ i \\ 1 \end{matr}$, where $t_1\in\mathbb{C}$,
the eigenvectors corresponding to $\lambda=1-i$ are $\mx=t_2\cdot\begin{matr}{rrr} 0 \\ -i \\ 1 \end{matr}$, where $t_2\in\mathbb{C}$ and
the eigenvectors corresponding to $\lambda=3$ are $\mx=t_3\cdot\begin{matr}{rrr} 1 \\ -1 \\ 2 \end{matr}$, where $t_3\in\mathbb{C}$.
E
Diagonalize $\,\mA\,$ by a similarity transformation.
answer
If we set $\,\mV=\begin{matr}{ccc} 0&0&1\\i&-i&-1\\1&1&2\end{matr}\,$ and $\,\mathbf{\Lambda}=\begin{matr}{ccc} 1+i&0&0\\0&1-i&0\\0&0&3\end{matr}\,$
