Problem 9.1
Describe geometrically the linear transformation \(T_A : \mathbb{R}^2 \rightarrow \mathbb{R}^2\) given by
\[A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\]
and then interpret the meanings of the eigenvalues and eigenvectors accordingly.
Solution. \(T_A\) is a reflection about the line \(y=x.\) Hence \(v\) is an eigenvector of \(T_A\) if and only if \(v\) is parallel to or orthogonal to the line \(y=x.\) \(T_A\) does not change the length of a vector, hence the eigenvalue is \(1\) or \(-1.\)
Problem 9.2
Find one or more eigenvalue/eigenvector pairs other than exponentials of the form \(Ce^{\lambda x}\) for the second derivative operator \(D^2 : C^{\infty} (\mathbb{R}) \rightarrow C^{\infty}(\mathbb{R}).\)
Solution. Consider the functions \(y= \sin kx,\) \(y=\cos kx,\) \(y=\sin kx + \cos kx \) for \(k\ne 0.\) The eigenvalue corresponding to these functions is \(-k^2.\)
Problem 9.3
Show that if \(T:V\rightarrow V\) is a linear transformation which is not injective, then \(0\) is an eigenvalue of \(T.\)
Solution. Since \(T\) is not one-to-one, \(\operatorname{Ker}(T) \ne \left\{ \mathbf{0} \right\},\) that is, there is a nonzero vector \(v\in\operatorname{Ker}(T).\) For this vector \(v,\) we have \(T(v) = \mathbf{0} = 0v.\) Hence \(0\) is an eigenvalue.
Problem 9.4
Find the characteristic polynomial, eigenvalues and corresponding eigenvectors for the matrix
\[A = \begin{pmatrix} 1 & 2 \\ 0 & 4 \end{pmatrix}.\]
Solution. \[\begin{align} tI - A &= \left[\begin{array}{cc} t-1 & -2 \\ 0 & t-4 \end{array}\right],\\[5pt] p_A(t) &= \det (tI-A) = (t-1)(t-4).\end{align}\] The equation \(p_A (t)=0\) has two solutions \(t=1\) and \(t=4.\)
Denote two eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = 4.\) The eigenvectors corresponding to these eigenvalues are \(v_1 = (1,\,0)\) and \(v_2 = (2,\,3),\) respectively.
Problem 9.5
For which real values of \(a\) does the matrix
\[A = \left( \begin{array}{rc} 2 & a \\ -1 & 1 \end{array} \right)\]
have real eigenvalues? State your answer as an inequality.
Solution. The characteristic polynomial of \(A\) is \[p_A (t) = \det\left[\begin{array}{cc} t-2 & -a \\ 1 & t-1 \end{array}\right] = t^2 - 3t + 2+a.\] Since \(t^2 - 3t + 2+a =0\) holds for some real number \(t,\) the discriminant has to be nonnegative, that is, \[D = 9 - 4(2+a) \ge 0.\] Therefore \[a \le \frac{1}{4}.\]
Problem 9.6
Compute the characteristic polynomial and eigenvalues for the matrix
\[A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 4 & 2 \\ 0 & 2 & 1 \end{pmatrix}.\]
Solution. It is straightforward. \[p_A (t) = \det\left[\begin{array}{ccc} t-1 & 0 & 0 \\ - & t-4 & -2 \\ 0 & -2 & t-1 \end{array}\right] = t(t-1)(t-5).\] Hence the eigenvalues are \[\lambda_1 = 0 ,\,\, \lambda_2 = 1 ,\,\, \lambda_3 = 5.\] The eigenvectors corresponding to these eigenvalues are \[ v_1 = \left[\begin{array}{r} 0 \\ 1 \\ -2 \end{array}\right] ,\,\, v_2 = \left[\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right] ,\,\, v_3 = \left[\begin{array}{r} 0 \\ 2 \\ 1 \end{array}\right] . \]
Problem 9.7
Show that the following matrix has no real eigenvalues. Interpret this geometrically.
\[A = \left(\begin{array}{rc} 0 & 1 \\ -1 & 0 \end{array}\right).\]
Solution. \[A = \left[\begin{array}{cr} \cos\frac{3}{2}\pi & -\sin\frac{3}{2}\pi \\ \sin\frac{3}{2}\pi & \cos\frac{3}{2}\pi \end{array}\right].\] Hence \(A\) represents a rotation by \(-90 ^\circ .\) No vector \(v\) other than zero vector satisfies \(T(v) = \lambda v\) for some scalar \(\lambda.\)
Problem 9.8
Let \(A\in M_n(\mathbb{R})\) where \(n\) is odd. Show that \(A\) has at least one real eigenvalue.
Solution. Since \[p_A (t) = \det(tI-A),\] \(p_A (t)\) is a polynomial of degree \(n,\) and the coefficient of \(t^n\) is \(1.\) Hence \[\lim_{t\rightarrow\infty} p_A(t) = \infty \,\,\text{and}\,\, \lim_{t\rightarrow -\infty} p_A (t) = - \infty.\] By the definitions of limits, \[p_A (t_0 ) < 0 \,\,\text{and}\,\, p_A (t_1) > 0\] for some \(t_0\) and \(t_1\). Furthermore \(t_0 \ne t_1.\) Since \(p_A(t)\) is a continuous function, by intermediate value theorem, \(p_A(\lambda)=0\) for some \(\lambda\) between \(t_0\) and \(t_1.\) \(\lambda\) is a real eigenvalue of \(A.\)
Problem 9.9
For any \(A\in M_n(\mathbb{R}),\) show that the number of imaginary roots of the characteristic polynomial is even. This gives an alternative approach to the previous problem.
Solution. Let \(p_A(t)\) be the characteristic polynomial of \(A.\) If \(\lambda = a+bi,\) \(a\in\mathbb{R},\) \(b\in\mathbb{R}^\times\) and \(p_A (\lambda)=0,\) then \(\overline{\lambda} \ne \lambda\) and \(p_A (\overline{\lambda})=0.\) Note that if \(A\) were a complex matrix, then we could not derive this result.
Problem 9.10
Let \(\lambda \in K\) be an eigenvalue of \(A\in M_n (K).\) Show that \(\lambda ^r\) is an eigenvalues of \(A^r ,\) the \(r\)th power of \(A,\) \(r\ge 0.\)
Solution. We introduce two solutions.
First, this is YC Lee's solution: Let \(v\) be a eigenvector corresponding to \(\lambda.\) Then \[A^r v = A^{r-1} (\lambda v) = A^{r-2} (\lambda^2 v) = \cdots = \lambda^r v.\] Hence \(\lambda^r\) is an eigenvalue of \(A^r.\)
Next, this is I Seul Bee's solution: Consider \[\lambda^r I - A^r = (\lambda I-A) (\lambda^{r-1} I + \lambda^{r-2} A + \lambda^{r-3} A^2 + \cdots + A^{r-1} ).\] Hence \[p_{A^r} (\lambda^r ) = \det(\lambda I-A) \det(\ast) = 0 \times \det(\ast) = 0.\] Therefore \(\lambda^r\) is an eigenvalue of \(A^r.\)
Problem 9.11
Let \(A\in M_n (K)\) be such that \(A^r\) is the zero matrix for some \(r \ge 1.\) (In this case, \(A^r\) represents \(A\) to the power \(r.\)) Show that all of \(A\text{'s}\) eigenvalues are \(0.\)
Solution. If \(\lambda \ne 0\) and \(\lambda\) is an eigenvalue of \(A,\) then \(\lambda^r\) is an eigenvalue of \(A^r.\) But the eigenvalue of \(O,\) the zero matrix, is only \(0.\) Hence \(\lambda\) must be zero.
Problem 9.12
Show that the eigenvalues of a triangular matrix are precisely the diagonal entries.
Solution. Let \(A = (a_{ij})_{n\times n}\) be an upper triangular matrix. Then \(a_{ij} =0\) for \(i > j.\) Since \(tI-A\) is also an upper triangular matrix, the characteristic polynomial of \(A\) is \[p_A (t) = \det(tI-A) = \prod_{i=1}^n (t-a_{ii}).\] Therefore the eigenvalues of \(A\) are exactly \(a_{11},\) \(a_{22},\) \(\cdots,\) \(a_{nn},\) the diagonal entries of \(A.\)
Problem 9.13
Find an example of a real \(2\times 2\) matrix \(A\) which is not diagonalizable as an endomorphism of \(\mathbb{R}^2,\) but is diagonalizable as an endomorphism of \(\mathbb{C}^2.\)
Solution. Take \[A = \left[\begin{array}{rc} 1 & 1 \\ -1 & 1 \end{array}\right].\] Then the characteristic polynomial of \(A\) is \[p_A (t) = (t-1)^2 +1.\] \(p_A(t)\) cannot assume zero unless \(t\) is an imaginary number.
Problem 9.14
Find all of the eigenvalues of the matrix
\[A=\left(\begin{array}{rc} 0 & 2 \\ -2 & 4 \end{array}\right).\]
Solution. Since \(p_A(t) = (t-2)^2,\) the eigenvalue of \(A\) is only \(2.\)
Problem 9.15
Show that the matrix of the previous problem is not diagonalizable over either the real or complex fields.
Solution. Since \(v=(1,\,1)\) is the only eigenvector of \(A,\) the eigenvector of \(A\) does not constitute an eigenbasis. Hence \(A\) is not diagonalizable.
Problem 9.16
Find the eigenvalues of the matrix
\[A = \begin{pmatrix} 2 & 1 \\ 0 & 5 \end{pmatrix}.\]
Solution. Since \(p_A(t) = (t-2)(t-5),\) the eigenvalues are \(\lambda_1 = 2\) and \(\lambda_2 = 5 ,\) and the eigenvectors corresponding to these eigenvalues are \(v_1 = (1,\,0)\) and \(v_2 = (1,\,3).\)
Problem 9.17
For the matrix \(A\) of the previous problem, find an invertible matrix \(P\) such that \(P^{-1} AP\) is a diagonal matrix.
Solution. \[\left[\begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array}\right] = \left[\begin{array}{cc} 1 & 1 \\ 0 & 3 \end{array}\right]^{-1} \left[\begin{array}{cc} 2 & 1 \\ 0 & 5 \end{array}\right] \left[\begin{array}{cc} 1 & 1 \\ 0 & 3 \end{array}\right]. \]
Problem 9.18
Let a sequence \(\left\{ a_n \right\}\) be defined recursively by
\[a_0 = 1 ,\,\, a_1 = 1 \,\,\text{and}\,\, a_{n+2} = 5a_{n+1} - 6a_n \,\,\text{for}\,\, n\ge 0.\]
Find the explicit formula for the general term \(a_n.\)
Solution. The recursive formula is expressible as \[\left[\begin{array}{c} a_n \\ a_{n-1} \end{array}\right] = \left[\begin{array}{cr} 5 & -6 \\ 1 & 0 \end{array}\right] \left[\begin{array}{c} a_{n-1} \\ a_{n-2} \end{array}\right].\] Hence \[\left[\begin{array}{c} a_n \\ a_{n-1} \end{array}\right] = \left[\begin{array}{cr} 5 & -6 \\ 1 & 0 \end{array}\right]^{n-1} \left[\begin{array}{c} 1 \\ 1 \end{array}\right].\] Take \[A = \left[\begin{array}{cr} 5 & -6 \\ 1 & 0 \end{array}\right].\] The eigenvalues of \(A\) are \(\lambda_1 = 2\) and \(\lambda_2 = 3,\) and the eigenvectors corresponding to these eigenvalues are \[v_1 = \left[\begin{array}{c} 2 \\ 1\end{array}\right] \,\,\text{and}\,\,v_2 = \left[\begin{array}{c} 3 \\ 1\end{array}\right].\] The diagonalization of \(A\) is \[ \left[\begin{array}{cr} 5 & -6 \\ 1 & 0 \end{array}\right] = \left[\begin{array}{cr} 2 & 3 \\ 1 & 1 \end{array}\right] \left[\begin{array}{cr} 2 & 0 \\ 0 & 3 \end{array}\right] \left[\begin{array}{cr} 2 & 3 \\ 1 & 1 \end{array}\right]^{-1}. \] Hence we have \[ \left[\begin{array}{c} a_n \\ a_{n-1} \end{array}\right] = \left[\begin{array}{cr} 2 & 3 \\ 1 & 1 \end{array}\right] \left[\begin{array}{cr} 2^{n-1} & 0 \\ 0 & 3^{n-1} \end{array}\right] \left[\begin{array}{cr} 2 & 3 \\ 1 & 1 \end{array}\right]^{-1} \left[\begin{array}{c} 1 \\ 1 \end{array}\right]. \] An evaluation gives us the formula \[a_n = 2^{n+1} - 3^{n}\] for \(n\ge 0.\)
Problem 9.19
Find new coordinates \(x ' ,\) \(y ' \) so that the following quadratic forms can be written as \(\lambda_1 (x ' )^2 + \lambda_2 ( y ' )^2 .\) Express the following forms in the form \(\lambda_1 (x ' )^2 + \lambda_2 ( y ' )^2 .\)
(1) \(x^2 + 4xy + y^2 \)
(2) \(2x^2 + 2xy + 2y^2 \)
(3) \(x^2 -12xy -4y^2 \)
(4) \(3x^2 + 2xy + 3y^2 \)
(5) \(x^2 - 2xy + y^2 \)
Solution. (1) The form \(x^2 + 4xy + y^2\) is expressible as \[ \left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} 1 & 2 \\ 2 & 1 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] =: \left[\begin{array}{cc} x & y \end{array}\right]A\left[\begin{array}{c} x \\ y \end{array}\right] .\] The eigenvalues of \(A\) are \[\lambda_1 = 3 ,\,\, \lambda_2 = -1\] and the corresponding eigenvectors are \[ \left[\begin{array}{c} 1 \\ 1\end{array}\right] ,\,\, \left[\begin{array}{r} -1 \\ 1 \end{array}\right]. \] Normalizing these eigenvectors, we have an orthonormal eigenbasis: \[ \left[\begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right] ,\,\, \left[\begin{array}{r} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right]. \] Hence the new coordinate is \[\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{cr} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right] \left[\begin{array}{c} x' \\ y' \end{array}\right].\] As a consequence, we have \[x^2 + 4xy + y^2 = 3 ( x ' )^2 - ( y ' )^2 .\]
(2) The form \(2x^2 + 2xy + 2y^2\) is expressible as \[ \left[\begin{array}{cc} x & y \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right] =: \left[\begin{array}{cc} x & y \end{array}\right]A\left[\begin{array}{c} x \\ y \end{array}\right] .\] The eigenvalues of \(A\) are \[\lambda_1 = 3 ,\,\, \lambda_2 = 1\] and the corresponding eigenvectors are \[ \left[\begin{array}{c} 1 \\ 1\end{array}\right] ,\,\, \left[\begin{array}{r} -1 \\ 1 \end{array}\right]. \] Normalizing these eigenvectors, we have an orthonormal eigenbasis: \[ \left[\begin{array}{c} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right] ,\,\, \left[\begin{array}{r} -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \end{array}\right]. \] Hence the new coordinate is \[\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{cr} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right] \left[\begin{array}{c} x' \\ y' \end{array}\right].\] As a consequence, we have \[2x^2 + 2xy + 2y^2 = 3 ( x ' )^2 + ( y ' )^2 .\]
(3) The new coordinate is \[\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{rc} -\frac{3}{\sqrt{13}} & \frac{2}{\sqrt{13}} \\ \frac{2}{\sqrt{13}} & \frac{3}{\sqrt{13}} \end{array}\right] \left[\begin{array}{c} x' \\ y' \end{array}\right].\] As a consequence, we have \[x^2 - 12xy - 4y^2 = 5 ( x ' )^2 - 8 ( y ' )^2 .\]
(4) The new coordinate is \[\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{cr} \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right] \left[\begin{array}{c} x' \\ y' \end{array}\right].\] As a consequence, we have \[3x^2 + 2xy +3y^2 = 4 ( x ' )^2 +2 ( y ' )^2 .\]
(5) The new coordinate is \[\left[\begin{array}{c} x \\ y \end{array}\right] = \left[\begin{array}{rc} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{array}\right] \left[\begin{array}{c} x' \\ y' \end{array}\right].\] As a consequence, we have \[x^2 - 2xy + y^2 = 2 ( x ' )^2 .\]