Let $\,f:\reel^3\rightarrow\reel^3\,$ be the linear map that with respect to the standard basis for $\,\reel^3\,$ has the mapping matrix
\begin{equation}
\mA=\begin{matr}{rrr} 3 & 4 & 4 \newline 6 & 6 & 6 \newline -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 of $\,f\,?$ Do the check, and state the corresponding eigenvalues.
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
$\mA\cdot\mv_1=\begin{matr}{rrr} -1 \newline 0 \newline 1 \end{matr}=-1\cdot\mv_1\,$, so $\,\lambda_1=-1\,$.
$\mA\cdot\mv_2=\begin{matr}{rrr} 0 \newline 0 \newline 0 \end{matr}=0\cdot\mv_2\,$, so $\,\lambda_2=0\,$.
$\mA\cdot\mv_3=\begin{matr}{rrr} 3 \newline 6 \newline -6 \end{matr}=3\cdot\mv_3\,$, so $\,\lambda_3=3\,$.
B
How can we most easily argue that $\mv_1,\,\mv_2$ and $\mv_3$ are linearly independent?
hint
See Theorem 13.11, bullet 1 in eNote 13.
C
How can we most easily show that $\,f(\reel^3)=\spanVec \lbrace \mv_1,\mv_3 \rbrace \,?$
hint
Use the fact that $(\mv_1,\mv_2,\mv_3)$ is a basis for $\reel^3$ consisting of eigenvectors of $f\,,$ so an eigenbasis. The image space $f(\reel^3)$ is (as always) spanned by the images of the basis vectors.
hint
We have just determined $f(\mv_1)$, $f(\mv_2)$ and $f(\mv_3)$.
A standard $\,(O,\mathbf i,\mathbf j,\mathbf k)$ coordinate system is given. All vectors are drawn from the origin. The map $\,p\,$ projects vectors down into the $(X,Y)$ plane in 3D space as shown on the figure. We are furthermore informed that $\,p\,$ is linear (this was shown in an exercise from a previous week).
A
Determine all eigenvalues of $\,p\,$ and the eigenspaces that correspond to these eigenvalues. Solve this solely by mental work.
hint
First look for the eigenvectors. Find these by looking at the figure and thinking about which vectors that either do not change when projected down or that are scaled up or down (elongated/compressed) when projected down.
hint
One eigenvalue is $1$. This corresponds to so-called fix vectors that do not change when mapped, meaning they are mapped onto themselves. Which vectors are fix-vectors of $\,p\,$?
The other eigenvalue is $0$. Which vectors are mapped onto a single point being the $\,\mnul$-vector?
answer
Eigenspace $\,E_1\,$ contains all vectors in the $\,(X,Y)$ plane. They are not moved nor distorted at all when projected down onto the $\,(X,Y)$ plane, because they are already there.
Eigenspace $\,E_0\,$ contains all vectors on the $\,z$-axis that all are mapped onto the origin.
B
Choose two different eigenbases (bases consisting of eigenvectors of $\,p\,$) and determine in each of the two cases the diagonal matrix that becomes the mapping matrix of $\,p\,$ with respect to the chosen basis.
hint
The first and most obvious choice of eigenbasis would be the standard base $\,(\mathbf i,\mathbf j,\mathbf k)\,$. How come? Is this an eigenvector basis?
answer
Choosing the standard basis as an eigenbasis we set up the mapping matrix:
State the eigenvalues and their corresponding eigenvectors of the linear map $\,f:\reel^3\rightarrow\reel^3\,$ that with respect to the standard basis $\,e\,$ in $\,\reel^3\,$ has the mapping matrix $\,\mA\,.$
hint
All information can be extracted from the Maple output without further calculations.
answer
The eigenvalues are $4$, $-2$ and $3$, all with algebraic and geometric multiplicities of $1$.
The corresponding eigenspaces are:
$$E_4=\spanVec \lbrace (-2,-2,1)\rbrace$$
$$E_3=\spanVec \lbrace (1,1,0) \rbrace$$
$$E_{-2}=\spanVec \lbrace (-1,-2,4) \rbrace$$
C
Provide a basis $\,v=(\mv_1,\mv_2,\mv_3)\, $ for $\,\reel^3\,$ consisting of eigenvectors of $\,f\,.$
hint
Have another look at the important Theorem 13.11 in eNote 13.
answer
Since $\mv_1=(-2,-2,1)\in E_4\,$, $\,\mv_2=(1,1,0)\in E_{3}\,$ and $\,\mv_3=(-1,-2,4)\in E_{-2}\,,$ then the three vectors can according to Theorem 13.11, bullet 1 in eNote 13 constitute a basis for $\,\reel^3\,$ because different eigenspaces always are linearly independent.
D
Determine the mapping matrix of $\,f\,$ with respect to the basis $\,v\,$ that was defined in the previous question.
Investigate whether $\,\mA\,$ can be diagonalized and, if so, construct a regular matrix $\,\mV\,$ and a diagonal matrix $\,\mathbf{\Lambda}\,$ such that
Investigate whether $\,\mB\,$ can be diagonalized and construct in that case a regular matrix $\,\mV\,$ and a diagonal matrix $\,\mathbf{\Lambda}\,$ such that
Show that $\mA$ and $\mB$ are similar to the same diagonal matrix.
answer
If we set $\,\mV=\begin{matr}{cc} -i&i\newline 1&1\end{matr}\,,$$\,\mU=\begin{matr}{cc} i&-i\newline 1&1\end{matr}\,$ and $\,\mathbf{\Lambda}=\begin{matr}{cc} i&0\newline 0&-i\end{matr}\,$ then:
Now we consider $\,\mA\,$ to be tthe mapping matrix of a linear map $\,f:\reel^2\rightarrow\reel^2\,$ with respect to the standard basis in $\,\reel^2\,.\,$ Determine a new basis $\,m\,$ for $\,\reel^2\,$ with respect to which $\,f\,$ is represented by the mapping matrix $\,\mB\,.$
hint
See today’s Maple demo.
answer
$$m=(\,(-1,0),(0,1))$$
Exercise 6: Complex Diagonalization
We are given the matrix
\begin{equation}
\mA=\begin{matr}{rrr} 3 & 0 & 0 \newline 0 & 1 & -1 \newline 5 & 1 & 1 \end{matr}\,.
\end{equation}
A
Compute the eigenvalues of $\,\mA\,$ and their corresponding complex eigenspaces.
hint
Remember that imaginary eigenvalues and eigenvectors come in complex conjugated pairs.
answer
The eigenvalues are $1+i$, $1-i$ and $3$, all with algebraic multiplicity 1.
The eigenvectors corresponding to $\lambda=1+i$ are $\mx=t_1\cdot\begin{matr}{rrr} 0 \newline i \newline 1 \end{matr}$, where $t_1\in\mathbb{C}$.
The eigenvectors corresponding to $\lambda=1-i$ are $\mx=t_2\cdot\begin{matr}{rrr} 0 \newline -i \newline 1 \end{matr}$, where $t_2\in\mathbb{C}$.
the eigenvectors corresponding to $\lambda=3$ are $\mx=t_3\cdot\begin{matr}{rrr} 1 \newline -1 \newline 2 \end{matr}$, where $t_3\in\mathbb{C}$.
B
Diagonalize $\,\mA\,$ by a similarity transformation.
answer
If we set $\,\mV=\begin{matr}{ccc} 0&0&1\newline i&-i&-1\newline 1&1&2\end{matr}\,$ and $\,\mathbf{\Lambda}=\begin{matr}{ccc} 1+i&0&0\newline 0&1-i&0\newline 0&0&3\end{matr}\,,$ then we have achieved the relationship that defines a similarity transformation:
