We consider the set of plane vectors in a standard $\,(O, \mathbf i, \mathbf j)\,$-coordinate system. All vectors are considered to be drawn from the origin. $\,\mathbf F\,$ constitutes the mapping matrix of a linear map $f\,$ with respect to the standard basis. An arbitrary vector $\,\mathbf x\,$ is drawn in blue, while the image vector $\,\mathbf y=f(\mathbf x)\,$ is red.
Right click (on Mac, Ctrl+click) on $\,\mathbf x\,$ and choose Animation on. How many times is $\,\mathbf y=f(\mathbf x)\,$ parallel to $\,\mathbf x\,$ during a passage of the circle?
Stop the animation with the undo-button in the tool bar. Move (using the cursor) $\,\mathbf x\,$ to the first position where the two vectors are parallel, and determine the ratio between the length of $\,\mathbf y\,$ and the length of $\,\mathbf x\,$. Use the same procedure on the other positions where the two vectors are parallel.
Explain that we (in general) can determine all eigenvalues of $f$ by letting $\,\mathbf x\,$ pass a semicircle of (e.g.) radius $1\,$.
Rotate $\,\mathbf x\,$ in a semicircle and find all eigenvalues. Furthermore state for each eigenvalue a corresponding (integer) eigenvector.
Check that the eigenvalues found are roots of the characteristic polynomial (by hand).
Check using paper and pencil that the eigenvectors found are the right ones.
You can change $\,\mF\,$ by moving the column vectors $\,\mathbf s1\,$ and $\,\mathbf s2\,$. Repeat the experiment for the points $1,\,2$ and $3$ above using the following settings of $\,\mF\,$: $\begin{matr}{rr}1&0\newline 2&-3\end{matr}\,$, $\,\begin{matr}{rr}3&-1\newline 1&1\end{matr}\,$ and $\,\begin{matr}{rr}-2&4\newline 1&-2\end{matr}\,.$ What are the characteristic differences in each of the three scenarios?
Set $\,\mF\,$ to $\,\begin{matr}{rr}2&2\newline -1&4\end{matr}\,.\,$ Rotate $\,\mathbf x\,$ through the semicircle and read all real eigenvalues.
The chacteristic polynomial is the determinant of the characteristic matrix.
C
Set up the characteristic equation of $\mA$ and compute from this the eigenvalues of $\mA\,.$
hint
The chacteristic equation is simply the characteristic polynomial set equal to zero.
answer
$$\lambda_1=\,3+i \,,\, \lambda_2=3-i$$
D
Now set up the characteristic matrix of $\mA$ corresponding to one of the eigenvalues, and find using this the eigenspace corresponding to the eigenvalue.
hint
Give the characteristic matrix a zero-right-hand side after having substituted in the eigenvalue, and solve it. For this, you may use Maple.
answer
The eigenvalue $\lambda_1=3+i$ corresponds to the eigenspace:
Note that you can choose many equivalent basis vectors (e.g. the chosen basis vector can be multiplied by any (complex) number which will give another possible basis vector.)
E
State without further calculations the eigenspace that corresponds to the other eigenvalue.
hint
Eigenvalues and eigenspaces come in complex conjugated pairs.
answer
Depending on your answer to the previous question, the answer could be:
Check the results using Maple’s Eigenvectors command. Use the argument output=list within the command and explain the meaning of each result in the output.
hint
Sometimes it can be a challenge to decide whether two given eigenspaces are identical. Is $\,\mathrm{span}\lbrace{(2,1+i)\rbrace}=\mathrm{span}\lbrace{(1-i,1)\rbrace}\,?$
hint
Note how Maple’s Eigenvectors command can provide all information about the eigenproblem in one single output: both the eigenvalues, the corresponding eigenvectors as well as the algebraic and geometric multiplicities.
Exercise 3: Eigenvalues and Eigenvectors. By Hand
A linear map $\,f: \reel^3\rightarrow\reel^3\,$ is with respect to the standard basis in $\,\reel^3\,$ given by the mapping matrix
\begin{equation}
\mA=\begin{matr}{rrr} 1 & -1 & 1 \newline 2 & 4 & -1 \newline 0 & 0 & 3 \end{matr}\,.
\end{equation}
A
Determine the characteristic polynomial and find the eigenvalues of $\,f\,$. State the algebraic multiplicity of the eigenvalues. Determine the real eigenspaces that correspond to each of the eigenvalues, and state the geometric multiplicity of the eigenvalues.
hint
For the characteristic matrix and polynomial and the characteristic equation, see Theorem 13.27 in eNote 13. For multiplicity, see Definition 13.33 in eNote 13.
hint
The algebraic multiplicity is the number of times the eigenvalue $\,\lambda\,$ is a root of the characteristic polynomial, while the geometric multiplicity is the dimensionen of the subspace spanned by $\,\lambda$’s eigenvectors.
hint
See Example 13.36 in eNote 13.
answer
The eigenvalues are $2$ and $3$ with $\mathrm{gm}(2) = \mathrm{am}(2) = 1$ and $\mathrm{gm}(3) = \mathrm{am}(3) = 2$.
A basis for $\,E_2\,$ is $\,\big(\,(-1,1,0)\,\big)\,$.
A basis for $\,E_3\,$ is $\,\big(\,(1/2,0,1),(-1/2,1,0)\,\big)\,.$
B
If possible choose a basis for $\reel^3$ with respect to which the mapping matrix of $f$ becomes a diagonal matrix. State this diagonal matrix.
hint
It must be an eigenbasis, i.e. a basis consisting of eigenvectors of $\,\mA\,.$
hint
A possible eigenbasis is $\,v=\big(\mv_1,\mv_2,\mv_3\big)=\big(\,(-1,1,0),(1/2,0,1),(-1/2,1,0)\,\big)\,.$
Note that the order of the eigenvalues matches the order of the eigenvectors in the chosen eigenbasis.
Now we consider the matrix
\begin{equation}
\mB=\begin{matr}{rrr} 1 & 1 & 0 \newline 2 & -1 & -1 \newline 0 & 2 & 1 \end{matr}.
\end{equation}
C
Compute the eigenvalues of $\mB$ and state their algebraic multiplicities. Determine the real eigenspaces corresponding to each of the eigenvalues, and state the geometric multiplicities of the eigenvalues.
answer
The eigenvalues are $1$ and $-1$ with $\mathrm{gm}(1) = 1 < 2 = \mathrm{am}(1)$ and $\mathrm{gm}(-1) = \mathrm{am}(-1) = 1$.
A basis for $\,E_{-1}\,$ is $\,\big(\,(1/2,-1,1)\,\big)\,$.
A basis for $\,E_1\,$ is $\,\big(\,(1/2,0,1)\,\big)\, .$
D
If possible create a regular matrix $\,\mV\,$ and a diagonal matrix $\,\mathbf{\Lambda}\,$ that satisfy
If a basis for $\reel^3$ can consist of eigenvectors of $\,\mB\,,$ then $\,\mV\,$ is the matrix whose columns are these eigenbasis vectors.
answer
Since the sum of the geometric multiplicities of the eigenvalues of $\,\mB\,$ is $2$, i.e. less than $\mathrm{dim}(\reel^3)=3\,,$ then it is not possible to find 3 linearly independent eigenvectors. Therefore a basis of the wanted type does not exist. And therefore the wanted diagonal matrix cannot be created ($\mB$ cannot be diagonalized).
$\,\mF\,$ maps the blue object on the red one. Find by moving the column vectors $\,\mathbf s1\,$ and $\,\mathbf s2\,$ a diagonal matrix that maps the blue object to a red object on the dashed position.
Also consider the maps that correspond to $\,\,\begin{matr}{rr} 3 &0 \newline 0 & -2 \end{matr}\,\,$ and $\,\,\begin{matr}{rr} 1 &0 \newline 0 & 2 \end{matr}\,\,.$
Explain that in general it holds true that the diagonal elements in diagonal matrices are eigenvalues of $\,\mathbf F\,$ with $\,\mathbf i\,$ and $\,\mathbf j\,$, respectively, as corresponding eigenvectors. What do the eigenvalues have to do with expansion or contraction in the direction of $\,\mathbf x1\,$ and $\,\mathbf x2\,$, respectively?
Move $\,\mathbf x1\,$ and $\,\mathbf x2\,$ such that $\,(\mathbf x1,\mathbf x2)\,$ becomes a new basis consisting of eigenvectors of $f$ , and state the corresponding eigenvalues. Note, the eigenvectors should be as short as possible while keeping their coordinates integers.
Which coordinates does the point $\,(6,1)\,$ have in the new $\,(O,\mathbf x1,\mathbf x2)$-coordinate system?
The blue object is fixed in the $\,(O,\mathbf x1,\mathbf x2)$-coordinate system.
Set the mapping matrix to $\,\,\mF=\begin{matr}{rr} 1 &-2 \newline -1 & 0 \end{matr}\,\,$ by moving the column vectors $\,\mathbf s1\,$ and $\,\mathbf s2\,.$
Find by moving $\,\mathbf x1\,$ and $\,\mathbf x2\,$ a new basis $\,(\mathbf x1,\mathbf x2)\,$ consisting of eigenvectors of $\mF$, and determine the corresponding eigenvalues. State the mapping matrix with respect to the basis $\,(\mathbf x1,\mathbf x2)\,.$ How do you see the relation between the blue and the red object?
Repeat the investigation in the preceding question with the mapping matrix that is given in the GeoGebra sheet Eigenvalue6.
Formulate a consolidated hypothesis about what eigenvalues and their corresponding eigenvectors say about the linear map they stem from.
Exercise 5: Eigenvalues in Functional Spaces
Consider the linear map $\,f:C^{\infty}(\reel)\rightarrow C^{\infty}(\reel)\,$ given by
$$ f(x(t))=x'(t)-x(t)\,.$$
A
Explain that for every $\,k \in \reel\,$ the function $\,\e^{k\cdot t}\,$ (where $\,t\in \reel\,$) is an eigenvector of $\,f\,,$ and state the corresponding eigenvalue.
answer
The eigenvalue of $\e^{k \cdot t}$ is $k-1$.
B
Explain that the four functions $\,\e^{k\cdot t}\,$ where $\,k\in\lbrace-1,0,1,2\rbrace\,$ are linearly independent.
hint
See Corollary 13.9 in eNote 13.
Let $\,U\,$ denote the subspace in $\,C^{\infty}(\reel)\,$ that has the basis $\,v=(\e^{-t},\,1,\,\e^t,\,\e^{2\cdot t}\,)\,.$
C
Show that the image space $\,f(U)\,$ is a subspace of $\,U\,,$ and determine the mapping matrix $\,\matind vFv\,$ of the map $f:U\rightarrow U\,$ with respect to basis $\,v\,.$
hint
$\,f(U)\,$ is spanned by the images of the basis vectors.
$v$ is an eigenbasis. What does this mean for the appearance of the mapping matrix?
$$x(t)= 3e^{-t}-2+e^{2t}+ce^t, \,\,\, c \in {\mathbb R}$$
E
Compare the result from the previous question with the output from Maple’s dsolve. Why are there not in $\,C^{\infty}(\reel)\,$ more solutions to the equation
Exercise 6: Diagonalization of a Matrix. Simulated manually
A linear map $\,f: \reel^3\rightarrow\reel^3\,$ has, with respect to the standard basis for $\,\reel^3\,$, the mapping matrix
\begin{equation}
\mA=\begin{matr}{rrr} 1 & 0 & 0 \newline 1 & 1 & 1 \newline 1 & 0 & 2 \end{matr}.
\end{equation}
A
State a basis $\,v\,$ for $\,\reel^3\,$ with respect to which the mapping matrix of $\,f\,$ becomes a diagonal matrix, and state the corresponding change-of-basis matrix $\,\matind eMv\,$ that changes from $v$-coordinates to $e$-coordinates.
answer
$\,\mA\,$ has the eigenvalues $1$ and $2$ where $\mathrm{am}(1)=2$ and $\mathrm{am}(2)=1$. The corresponding eigenvector spaces are
The eigenvalues are $1$, $0$ and $-1$ with $\mathrm{gm}(1) = \mathrm{am}(1) = 1$, $\mathrm{gm}(0) =1 < 2 = \mathrm{am}(0)$ and $\mathrm{gm}(-1) = \mathrm{am}(-1) = 1$.
Investigate whether a regular matrix $\,\mV\,$ and a diagonal matrix $\,\mathbf{\Lambda}\,$ exist, such that
$$\mathbf{\Lambda}=\mV^{-1}\cdot\mB\cdot\mV\,.$$
Exercise 8: Eigenvectors Linear Independence. Advanced
Assume that $\,\lambda_1\,$ and $\,\lambda_2\,$ are two different eigenvalues of a matrix $\,\mA\,.$ Then vectors $\,\mv_1\neq 0\,$ and $\,\mv_2\neq 0\,$ exist such that
\begin{equation}
\mA\cdot\mv_1=\lambda_1\cdot\mv_1\;\mathrm{and}\;\mA\cdot\mv_2=\lambda_2\cdot\mv_2,\quad\mathrm{where}\;\lambda_1\neq\lambda_2\,.
\end{equation}
A
Show that the eigenvectors $\,\mv_1\,$ and $\,\mv_2\,$ are linearly independent.
hint
The arguments can be given as a proof by contradiction: Assume that the opposite of what you want to prove is true and prove that this leads to a contradiction.
hint
The proof by contradiction can be carried out by assuming that $\mv_1$ and $\mv_2$ are linearly dependent. Then a $k\neq 0$ exists, such that $\mv_2=k\cdot\mv_1$.
hint
Now try to compute $\mA\cdot\mv_2$ while you look for a contradiction.
B
Compare the result obtained here with Corollary 13.9 in eNote 13 about the linear independence of eigenvectors.
