Exercises – Long Day#
Exercise 1: Eigenvalues and Eigenvectors#
We are given the matrix
Determine whether the following vectors are eigenvectors of matrix \(\mathbf A\). For those that are, compute the corresponding eigenvalue.
\(\left[\begin{array}{c} 1 \\ 2\end{array}\right]\)
\(\left[\begin{array}{c} 1 \\ 1\end{array}\right]\)
Hint
The definition of an eigenvector can be found in Definition 12.1.1.
Hint
For each of the two vectors \({\mathbf v}\), investigate whether \({\mathbf A}\cdot {\mathbf v}\) can be written as a scalar multiple of \({\mathbf v}\).
Answer
The vector \(\left[\begin{array}{c} 1 \\ 2\end{array}\right]\) is not an eigenvector of \({\mathbf A}\).
The vector \(\left[\begin{array}{c} 1 \\ 1\end{array}\right]\) is an eigenvector of \({\mathbf A}\). The corresponding eigenvalue is \(5\).
Exercise 2: Real Eigenvalues and Eigenvectors#
As in the previous Exercise we consider the matrix
Question a#
Determine the characteristic polynomial of \(\mathbf A\,\) and use it to find the eigenvalues of \(\mathbf A\,\). Determine their algebraic multiplicities as well.
Hint
The characteristic polynomial of an \(n \times n\) matrix is defined as \(\mathrm{det}({\mathbf A}-Z\cdot {\mathbf I}_n)\). You can find this in the textbook after Theorem 12.1.1.
Answer
The characteristic polynomial is \(Z^2+Z-30\). The eigenvalues are the roots of this polynomial: \(5\,\) and \(-6\), and hence we have \(Z^2+Z-30=(Z-5)\cdot(Z+6)\). We see that both eigenvalues have an algebraic multiplicity of \(1\).
Question b#
Determine the eigenspace \(E_{-6}\) corresponding to the eigenvalue \(-6\). What is the geometric multiplicity of the eigenvalue \(-6\)?
Hint
If \(\lambda\) is an eigenvalue of an \(n \times n\) matrix \(\mathbf A\), then \(E_\lambda\) equals \(\mathrm{ker}({\mathbf A}-Z\cdot {\mathbf I}_n)\) according to Lemma 12.2.3 in the textbook.
Hint
In order to determine \(E_{-6}\) you can use the approach from Example 12.2.1 in the textbook. The geometric multiplicity of an eigenvalue is defined in Definition 12.2.1.
Answer
The matrix
has the reduced row-echelon form
Hence it applies that
This shows that \(\left\{\left[\begin{array}{c} -10 \\ 1\end{array}\right]\right\}\) is a basis for \(E_{-6}\). From this we conclude that \(\mathrm{dim}(E_{-6})=1\), which is the same as saying that the eigenvalue \(-6\) has a geometric multiplicity of \(1\).
Question c#
The eigenspace corresponding to the eigenvalue \(5\) is denoted by \(E_{5}\). Determine \(\mathrm{dim}(E_5)\) by use of your answer to Question a along with Theorem 12.2.4 in the textbook. Then use your results from Exercise 1 to provide a basis for \(E_{5}\).
Hint
From Question a we know that \(\mathrm{am}(5)=1\). Then Theorem 12.2.4 in the textbook implies that \(\mathrm{dim}(E_5)=1\).
Answer
From Question a we know that \(\mathrm{am}(5)=1\), and according to Definition 12.2.1 we have that \(\mathrm{dim}(E_5)=\mathrm{gm(5)}\). Hence, \(E_5\) has a dimension of \(1\). A basis for \(E_5\) will thus contain one eigenvector. We can for example use the vector \(\left[\begin{array}{c} 1 \\ 1\end{array}\right]\in E_5\) that we found in Exercise 1 to form a basis for \(E_{5}\).
Exercise 3: Eigenvalues and Eigenvectors in an Infinite-Dimensional Case#
Let \(C_\infty(\Bbb R)\) be the infinite-dimensional, real vector space mentioned in Example 10.4.5 in the textbook. It contains functions from \(\mathbb{R}\) to \(\mathbb{R}\) that can be differentiated an arbitrary number of times. A linear map \(L: C_\infty(\Bbb R) \to C_\infty(\Bbb R)\) is given by \(L(f)=f''\). In other words, \(L\) maps a function \(f \in C_\infty(\Bbb R)\) to its second-order derivative.
Question a#
Determine whether the following functions are eigenvectors of \(L\). For those that are, determine their corresponding eigenvalue.
\(f_1(t)=t^2\)
\(f_2(t)=\cos(t)\)
\(f_3(t)=\sin(t)\)
\(f_4(t)=\mathrm e^{4 t}\)
\(f_5(t)=t\mathrm e^t\)
Hint
Investigate for each of the given functions \(f_i\) for \(i=1,2,3,4,5\) whether \(L(f_i)\) can be written as a scalar multiple of \(f_i\). This is precisely the requirement for a vector being an eigenvector, as Definiton 12.1.1 tells.
Answer
The function \(f_1(t)=t^2\) is not an eigenvector of \(L\).
We see that \(\cos(t)''=(-\sin(t))'=-\cos(t)\). Hence the function \(f_2(t)=\cos(t)\) is an eigenvector of \(L\). The corresponding eigenvalue is \(-1\).
The function \(f_3(t)=\sin(t)\) is an eigenvector of \(L\). The corresponding eigenvalue is \(-1\).
The function \(f_4(t)=\mathrm e^{4t}\) is an eigenvector of \(L\). The corresponding eigenvalue is \(16\).
Using the chain rule of differentiation we see that \((t\mathrm e^t)''=(\mathrm e^t+t\mathrm e^t)'=2\mathrm e^t+t\mathrm e^t\). Hence the function \(f_5(t)=t\mathrm e^t\) is not an eigenvector of \(L\).
Exercise 4: Complex Eigenvalues and Eigenvectors#
We are given the matrix
Question a#
Set up the characteristic polynomial of \(\mathbf A\,\) and use it to compute the eigenvalues of \(\mathbf A\,\).
Hint
The characteristic polynomial of an \(n \times n\) matrix is defined as \(\mathrm{det}({\mathbf A}-Z\cdot {\mathbf I}_n)\). You can also find it in the textbook right after Theorem 12.1.1.
Answer
The characteristic polynomial is \(Z^2-6Z+10\). The eigenvalues are \(\,3+i\,\) and \(\,3-i\,.\)
Question b#
Determine the eigenspace \(E_{3+i}\) corresponding to the eigenvalue \(3+i\).
Hint
First, find the reduced row-echelon form of the matrix: $\(\mathbf A-(3+i)\mathbf I_2=\left[\begin{array}{cc} 2 & 2\\ -1 & 4\end{array}\right]-(3+i)\left[\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right]=\left[\begin{array}{cc} -1-i & 2\\ -1 & 1-i\end{array}\right] .\)$
Answer
\(E_{3+i}=\mathrm{Span}\left( \left[\begin{array}{cc} 1-i\\ 1\end{array}\right]\right)\,.\)
Question c#
State without further computations the eigenspace that belongs to the other eigenvalue.
Answer
When matrices have real coefficients, one eigenvector/eigenspace is found from the other as the complex conjugate. We thus see that \(E_{3-i}=\mathrm{Span}\left( \left[\begin{array}{cc} 1+i\\ 1\end{array}\right]\right)\,.\)
Exercise 5: A Nice Basis Change#
We consider the same matrix,
as in Exercise 4. Let \(\epsilon\) be the ordered standard basis for \(\mathbb{C}^2\). We define another ordered basis for \(\mathbb{C}^2\) as follows:
Question a#
Determine the change-of-basis matrix \({}_\epsilon[\mathrm{id}_{\mathbb{C}^2}]_\beta\). Then find the change-of-basis matrix \({}_\beta[\mathrm{id}_{\mathbb{C}^2}]_\epsilon\).
Hint
For the second part of this Question, use the third part of Lemma 11.3.6 in the textbook.
Question b#
Determine the matrix \({}_\beta [\mathrm{id}_{\mathbb{C}^2}]_\epsilon \cdot {\mathbf A} \cdot {}_\epsilon [\mathrm{id}_{\mathbb{C}^2}]_\beta\). The result will be a diagonal matrix. How could we have known that based on the answers to Exercise 4 without further calculations?
Hint
Note that the vectors in the ordered basis \(\beta\) are the eigenvectors of the matrix \(\mathbf A\). See your answers to Exercise 4.
Hint
The first column in the matrix \({}_\beta [\mathrm{id}_{\mathbb{C}^2}]_\epsilon \cdot {\mathbf A} \cdot {}_\epsilon [\mathrm{id}_{\mathbb{C}^2}]_\beta\) is identical to \({}_\beta [\mathrm{id}_{\mathbb{C}^2}]_\epsilon \cdot {\mathbf A} \cdot {}_\epsilon [\mathrm{id}_{\mathbb{C}^2}]_\beta\cdot \begin{bmatrix}1\\0\end{bmatrix}\). Now notice that the column vector \({}_\epsilon [\mathrm{id}_{\mathbb{C}^2}]_\beta\cdot \begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}1-i\\1\end{bmatrix}\) is an eigenvector of \(\mathbf A\) with an eigenvalue of \(3+i\). Can this be used to find \({}_\beta [\mathrm{id}_{\mathbb{C}^2}]_\epsilon \cdot {\mathbf A} \cdot {}_\epsilon [\mathrm{id}_{\mathbb{C}^2}]_\beta\cdot \begin{bmatrix}1\\0\end{bmatrix}\) without having to do calculations?
Question 6: Eigenvalues and Eigenvectors of a Real \(3 \times 3\) Matrix#
A linear map \(f: \mathbb{R}^3 \to \mathbb{R}^3\,\) between real vector spaces has with respect to the ordered standard basis for \(\mathbb{R}^3\,\) the following mapping matrix:
Question a#
Determine the characteristic polynomial and find the eigenvalues of \(f\,\). State the algebraic multiplicities of the eigenvalues.
Hint
The characteristic polynomial of a linear map can be determined by use of Definition 12.1.3 in the textbook. It might be helpful to choose a convenient row from which to expand the determinant that is mentioned in this definition without having to multiply through fully.
Answer
By expanding \(\mathrm{det}(\mathbf A-Z\mathbf I_3)\) from the third row the wanted characteristic polynomial is found to be:
The eigenvalues are \(2\) with \(\mathrm{gm}(2)=\mathrm{am}(2)=1\) and \(3\) with \(\mathrm{gm}(3)=\mathrm{am}(3)=2\).
Question a#
Determine the eigenspaces that belong to each of the eigenvalues of \(f\), and state the geometric multiplicities of the eigenvalues. Are the algebraic and geometric multiplicities of the eigenvalues identical?
Hint
The geometric multiplicity is the dimension of the eigenspace that belongs to \(\lambda\) (meaning, the subspace that is spanned by those eigenvectors of \(f\) that have the eigenvalue \(\lambda\)).
Hint
In order to determine the geometric multiplicities, the eigenspaces are to be found. Lemma 12.2.3 describes how the eigenspaces of a matrix look.
Answer
A possible ordered basis for the eigenspace \(E_2=\mathrm{ker}(\mathbf A-2\mathbf I_3)\) is
Hence, we see that \(\mathrm{gm}(2)=1\), which is identical to the algebraic multiplicity of the eigenvalue.
A possible ordered basis for the eigenspace \(E_3=\mathrm{ker}(\mathbf A-3\mathbf I_3)\) is
Hence, we see that \(\mathrm{gm}(3)=2, which again here is equal to the algebraic multiplicity of the eigenvalue.\)
Exercise 7: Eigenvalues and their Multiplicities of another Real \(3 \times 3\) Matrix#
We now consider the matrix
We are being informed that the characteristic polynomial of the matrix \(\mathbf B\) is \((1-Z)\cdot(Z+1)\cdot(Z-1)\).
Question a#
Find the eigenvalues of \(\mathbf B\) and state their algebraic multiplicities.
Answer
The eigenvalues are \(1\) with \(\mathrm{am}(1) = 2\) and \(-1\) with \(\mathrm{am}(-1) = 1\).
Question b#
State the geometric multiplicities of the eigenvalues. Are the algebraic and geometric multiplicities identical?
Hint
First, consider why Theorem 12.2.4 implies the following: if an eigenvalue has an algebraic multiplicity of \(1\) then its geometric multiplicity is also \(1\).
Answer
As explained in the hint, \(\mathrm{am}(\lambda)=1\) implies that \(\mathrm{gm}(\lambda)=1\). Thus we get without further calculations that \(\mathrm{gm}(-1)=\mathrm{am}(-1)=1\).
This trick cannot be used for the eigenvalue \(1\), since \(\mathrm{am}(1)=2\). We find that the eigenspace \(E_1\) is spanned by the vector
Hence we have \(\mathrm{gm}(1)=1\), which is strictly less than the algebraic multiplicity of the eigenvalue.
Exercise 8: The Characteristic Polynomial of a \(2 \times 2\) Matrix#
Let \(\mathbb{F}\) be a field and \(a,b,c,d\) elements from \(\mathbb{F}\). We consider the matrix
Question a#
Show that the matrix \(\mathbf A\) has the characteristic polynomial \(Z^2-(a+d)Z+\mathrm{det}({\mathbf A})\). Note: The expression \(a+d\) is called the trace of the matrix and is denoted by \(\mathrm{tr}({\mathbf A})\).
Question b#
Assume that \({\mathbf A}\) has eigenvalues \(\lambda_1\) and \(\lambda_2\). Use the previous question to realise that \(\lambda_1+\lambda_2=\mathrm{tr}({\mathbf A})\) and \(\lambda_1 \cdot \lambda_2=\mathrm{det}({\mathbf A})\).
Hint
If a \(2 \times 2\) matrix \({\mathbf A}\) has eigenvalues \(\lambda_1\) and \(\lambda_2\), then it holds true that the characteristic polynomial of the matrix is \((Z-\lambda_1)\cdot (Z-\lambda_2).\)
Answer
On the one hand we have \(p_{\mathbf A}(Z)=Z^2-\mathrm{tr}({\mathbf A})Z+\mathrm{det}({\mathbf A})\). On the other hand we have \(p_{\mathbf A}(Z)=(Z-\lambda_1)\cdot (Z-\lambda_2)=Z^2-(\lambda_1+\lambda_2)Z+\lambda_1\cdot \lambda_2\). Comparing the coefficients will provide what we are seeking.
Exercise 9: Linear Independency of two Eigenvectors#
We are given a matrix \({\mathbf A} \in \mathbb{C}^{2 \times 2}\) as well as two of its eigenvectors \({\mathbf v}_1,{\mathbf v}_2 \in \mathbb{C}^2\). Assume that the corresponding eigenvalues \(\lambda_1,\lambda_2\) are different. Show that \({\mathbf v}_1\) and \({\mathbf v}_2\) are linearly independent.
Hint
If \({\mathbf v}_1\) and \({\mathbf v}_2\) are linearly dependent, then two complex numbers \(c_1\) and \(c_2\), which are not both equal to zero, exist such that
Hint
From the previous hint it follows that if two vectors are linearly dependent, then one of them is a scalar multiple of the other. Is this possible if the two vectors are eigenvectors of the same matrix \({\mathbf A}\) but with different eigenvalues?