Eigenvalues and eigenvectors¶
Exercises · Vectors and matrices · with worked solutions · Lecture notes (PDF) · Chapter 4.5: Eigenvalues and eigenvectors
Exercise 1
Compute the eigenvalues of the matrix
and one eigenvector associated with each eigenvalue.
Solution
The characteristic equation is
hence \(\lambda_1=2\) and \(\lambda_2=5\).
For \(\lambda_1=2\):
For \(\lambda_2=5\):
Check: \(\boldsymbol A\boldsymbol v_1=(4-2,\;2-6)'=(2,-4)'=2\,\boldsymbol v_1\) and \(\boldsymbol A\boldsymbol v_2=(5,5)'=5\,\boldsymbol v_2\).
Exercise 2
Compute the eigenvalues and one eigenvector for each eigenvalue of the symmetric matrix
Is \(\boldsymbol Q\) positive definite?
Solution
The characteristic equation is
hence \(\lambda_1=1\) and \(\lambda_2=6\).
For \(\lambda_1=1\):
For \(\lambda_2=6\):
Since \(\boldsymbol Q\) is symmetric and both eigenvalues are strictly positive, \(\boldsymbol Q\) is positive definite. (Note also that \(\boldsymbol v_1'\boldsymbol v_2=2-2=0\).)
Exercise 3
Without computing any determinant, find the eigenvalues of
Then find one eigenvector for each eigenvalue and check the trace and the determinant.
Solution
\(\boldsymbol U\) is upper triangular, hence its eigenvalues are its diagonal entries:
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\(\lambda_1=3\): \((\boldsymbol U-3\boldsymbol I)\boldsymbol v=\boldsymbol 0\) gives \(v_2-2v_3=0\), \(-4v_2+4v_3=0\), \(-v_3=0\), hence \(v_2=v_3=0\) and \(\boldsymbol v_1=(1,0,0)'\).
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\(\lambda_2=-1\):
\[ (\boldsymbol U+\boldsymbol I)\boldsymbol v= \begin{pmatrix} 4 & 1 & -2\\[0.5ex] 0 & 0 & 4\\[0.5ex] 0 & 0 & 3 \end{pmatrix} \boldsymbol v=\boldsymbol 0 \Longrightarrow v_3=0,\ \ 4v_1+v_2=0, \qquad \boldsymbol v_2=\begin{pmatrix}1\\[0.5ex]-4\\[0.5ex]0\end{pmatrix}. \] -
\(\lambda_3=2\):
\[ (\boldsymbol U-2\boldsymbol I)\boldsymbol v= \begin{pmatrix} 1 & 1 & -2\\[0.5ex] 0 & -3 & 4\\[0.5ex] 0 & 0 & 0 \end{pmatrix} \boldsymbol v=\boldsymbol 0 \Longrightarrow v_2=\tfrac{4}{3}v_3,\ \ v_1=-v_2+2v_3=\tfrac{2}{3}v_3. \]Choosing \(v_3=3\): \(\boldsymbol v_3=(2,4,3)'\).
Check: \(\operatorname{tr}(\boldsymbol U)=3-1+2=4=\lambda_1+\lambda_2+\lambda_3\) and \(\det(\boldsymbol U)=3\cdot(-1)\cdot 2=-6=\lambda_1\,\lambda_2\,\lambda_3\).
Exercise 4
Compute the characteristic polynomial, the eigenvalues and one eigenvector for each eigenvalue of
Classify \(\boldsymbol Q\) (definite, semidefinite or indefinite).
Solution
Expanding along the first row:
The eigenvalues are \(\lambda_1=-1\), \(\lambda_2=4\), \(\lambda_3=-3\).
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\(\lambda_1=-1\): \((\boldsymbol Q+\boldsymbol I)\boldsymbol v=\boldsymbol 0\) with \(\boldsymbol Q+\boldsymbol I=\begin{pmatrix}0 & 1 & 0\\ 1 & 3 & 3\\ 0 & 3 & 0\end{pmatrix}\) gives \(v_2=0\) and \(v_1+3v_3=0\): \(\boldsymbol v_1=(3,0,-1)'\).
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\(\lambda_2=4\): \(\boldsymbol Q-4\boldsymbol I=\begin{pmatrix}-5 & 1 & 0\\ 1 & -2 & 3\\ 0 & 3 & -5\end{pmatrix}\) gives \(v_2=5v_1\), \(v_3=\frac{3}{5}v_2=3v_1\) (and \(v_1-10v_1+9v_1=0\)): \(\boldsymbol v_2=(1,5,3)'\).
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\(\lambda_3=-3\): \(\boldsymbol Q+3\boldsymbol I=\begin{pmatrix}2 & 1 & 0\\ 1 & 5 & 3\\ 0 & 3 & 2\end{pmatrix}\) gives \(v_2=-2v_1\), \(v_3=-\frac{3}{2}v_2=3v_1\) (and \(v_1-10v_1+9v_1=0\)): \(\boldsymbol v_3=(1,-2,3)'\).
Check: \(\operatorname{tr}(\boldsymbol Q)=-1+2-1=0=-1+4-3\) and \(\det(\boldsymbol Q)=-1\cdot(-2-9)-1\cdot(-1-0)=12=(-1)\cdot 4\cdot(-3)\).
\(\boldsymbol Q\) is symmetric with a positive and a negative eigenvalue, hence it is indefinite.
Exercise 5
The matrix
has the eigenvalue \(\lambda=3\) with multiplicity \(2\). Using the trace, find the third eigenvalue; then check the result with the determinant and find an eigenvector for it.
Solution
The eigenvalues, counted with multiplicity, are \(\lambda_1=\lambda_2=3\) and \(\lambda_3\). Since the trace is the sum of the eigenvalues:
Check with the determinant (Laplace expansion along the first row):
Eigenvector for \(\lambda_3=6\):
Adding the first two equations: \(-3v_1-3v_2=0\), i.e., \(v_2=-v_1\); then from the first equation \(v_3=2v_1+v_2=v_1\). Hence \(\boldsymbol v=(1,-1,1)'\); indeed \(\boldsymbol Q\boldsymbol v=(4+1+1,\,-1-4-1,\,1+1+4)'=(6,-6,6)'=6\,\boldsymbol v\).
Exercise 6
Using the eigenvalues, classify the following symmetric matrices:
Solution
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a) \((-3-\lambda)^2-1=\lambda^2+6\lambda+8=(\lambda+2)(\lambda+4)\): the eigenvalues are \(-2\) and \(-4\), both strictly negative, hence \(\boldsymbol Q_1\) is negative definite.
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b) \((1-\lambda)^2-9=\lambda^2-2\lambda-8=(\lambda-4)(\lambda+2)\): the eigenvalues are \(4\) and \(-2\), hence \(\boldsymbol Q_2\) is indefinite.
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c) \((2-\lambda)^2-4=\lambda^2-4\lambda=\lambda(\lambda-4)\): the eigenvalues are \(0\) and \(4\), both non-negative, hence \(\boldsymbol Q_3\) is positive semidefinite (but not positive definite, since \(0\) is an eigenvalue).
Exercise 7
Using Sylvester's criterion, prove that the matrix
is positive definite.
Solution
\(\boldsymbol Q\) is symmetric and its leading principal minors are
All the leading principal minors are strictly positive, hence \(\boldsymbol Q\) is positive definite. (Here Sylvester's criterion is convenient: the eigenvalues of \(\boldsymbol Q\) are \(2\) and \(2\pm\sqrt 2\), which are not integers.)
Exercise 8
Using Sylvester's criterion, classify
Solution
\(\boldsymbol Q\) is symmetric and its leading principal minors are
The leading principal minors alternate in sign starting with a negative one, hence \(\boldsymbol Q\) is negative definite. (Indeed, its eigenvalues are \(-1\), \(-2\) and \(-4\).)
Exercise 9
Using the principal minors, decide whether the following symmetric matrices are positive semidefinite:
Solution
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a) The leading principal minors are \(\Delta_1=2\), \(\Delta_2=4-4=0\), \(\Delta_3=\det(\boldsymbol Q_a)=0\): \(\boldsymbol Q_a\) is not positive definite. We check all principal minors:
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order 1: \(2,\ 2,\ 1\);
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order 2: \(\det\begin{pmatrix}2 & -2\\ -2 & 2\end{pmatrix}=0\), \(\det\begin{pmatrix}2 & 0\\ 0 & 1\end{pmatrix}=2\), \(\det\begin{pmatrix}2 & 0\\ 0 & 1\end{pmatrix}=2\);
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order 3: \(\det(\boldsymbol Q_a)=0\).
All are \(\ge 0\): \(\boldsymbol Q_a\) is positive semidefinite. Indeed \(\boldsymbol x'\boldsymbol Q_a\boldsymbol x=2(x_1-x_2)^2+x_3^2\ge 0\), and the eigenvalues are \(0,1,4\).
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b) The leading principal minors are \(\Delta_1=1\), \(\Delta_2=1-1=0\), \(\Delta_3=\det(\boldsymbol Q_b)=-1\cdot 0=0\): they are all \(\ge 0\), but this is not enough. The principal minor of order 1 given by the entry \(q_{33}=-1\) is negative, hence \(\boldsymbol Q_b\) is not positive semidefinite: for \(\boldsymbol x=(0,0,1)'\) we have \(\boldsymbol x'\boldsymbol Q_b\boldsymbol x=-1<0\). Since for \(\boldsymbol x=(1,0,0)'\) we have \(\boldsymbol x'\boldsymbol Q_b\boldsymbol x=1>0\), the matrix is indefinite (its eigenvalues are \(2,0,-1\)).
Exercise 10
Let \(k\in\R\) and
For which values of \(k\) is \(\boldsymbol Q(k)\) positive definite? For which values is it positive semidefinite?
Solution
\(\boldsymbol Q(k)\) is symmetric for every \(k\). Its leading principal minors are
By Sylvester's criterion, \(\boldsymbol Q(k)\) is positive definite if and only if
that is, if and only if \(\boldsymbol{k>1}\).
For positive semidefiniteness we need all principal minors \(\ge 0\): order 1: \(2,\ k,\ 2\); order 2: \(2k-1\), \(\det\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}=4\), \(2k-1\); order 3: \(4k-4\). They are all \(\ge 0\) if and only if \(k\ge 0\), \(k\ge\frac{1}{2}\) and \(k\ge 1\), i.e., if and only if \(\boldsymbol{k\ge 1}\). For \(k=1\) the matrix is positive semidefinite but not positive definite: its eigenvalues are \(0,2,3\).