Exercises — Lesson 10: Eigenvalues, Eigenvectors, and Diagonalization
Exercise 10.1. [Hand] Find all eigenvalues and eigenvectors of (a) \(\left(\begin{smallmatrix}3&0\\0&-2\end{smallmatrix}\right)\), (b) \(\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\) (reflection through \(x_2 = x_1\) — interpret the answer geometrically), (c) \(\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)\) over \(\mathbb{R}\) and then over \(\mathbb{C}\).
Solution.
(a) \(A = \begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}\), which acts as \(\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} 3x_1 \\ -2x_2 \end{pmatrix}\).
\[ \boxed{\; \begin{aligned} \lambda_1 &= 3 &&\Longleftrightarrow\quad v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\[6pt] \lambda_2 &= -2 &&\Longleftrightarrow\quad v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \end{aligned} \;} \]
(b) \(A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\) — reflection over \(x_2 = x_1\).
\[ \begin{aligned} (0 - \lambda)^2 - 1 &= 0 \\[6pt] \lambda^2 &= 1 \end{aligned} \]
\[ \boxed{\; \begin{aligned} \lambda_1 &= 1 &&\Longleftrightarrow\quad v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\[6pt] \lambda_2 &= -1 &&\Longleftrightarrow\quad v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \end{aligned} \;} \]
(c) \(A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\) — rotation by \(90^\circ\) counterclockwise.
\[ \begin{aligned} (0 - \lambda)^2 + 1 &= 0 \\[6pt] \lambda^2 &= -1 \\[6pt] i &= \sqrt{-1} \end{aligned} \]
Over \(\mathbb{R}\): no real eigenvalues. Over \(\mathbb{C}\):
\[ \boxed{\; \begin{aligned} \lambda_1 &= i &&\Longleftrightarrow\quad v_1 = \begin{pmatrix} 1 \\ -i \end{pmatrix} \\[6pt] \lambda_2 &= -i &&\Longleftrightarrow\quad v_2 = \begin{pmatrix} 1 \\ i \end{pmatrix} \end{aligned} \;} \]
\[ \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ -i \end{pmatrix} = \begin{pmatrix} i \\ 1 \end{pmatrix} \quad \checkmark \]
Exercise 10.2. [Hand] Diagonalize \(A = \left(\begin{smallmatrix}1&2\\2&1\end{smallmatrix}\right)\) and use Proposition 10.6 to find a closed form for \(A^k(1,0)\).
Solution. Diagonalize \(A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}\).
\[ \begin{aligned} (1 - \lambda)^2 - 4 &= 0 \\[6pt] 1 - 2\lambda + \lambda^2 - 4 &= 0 \\[6pt] \lambda^2 - 2\lambda - 3 &= 0 \\[6pt] (\lambda - 3)(\lambda + 1) &= 0 \end{aligned} \qquad \begin{aligned} \lambda_1 &= -1 \\[6pt] \lambda_2 &= 3 \end{aligned} \]
Eigenvectors, by row reduction of \((A - \lambda I)v = 0\):
\[ \left[\begin{array}{cc|c} 2 & 2 & 0 \\ 2 & 2 & 0 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{cc|c} 2 & 2 & 0 \\ 0 & 0 & 0 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{cc|c} 1 & 1 & 0 \\ 0 & 0 & 0 \end{array}\right] \]
\[ \left[\begin{array}{cc|c} -2 & 2 & 0 \\ 2 & -2 & 0 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{cc|c} 0 & 0 & 0 \\ 2 & -2 & 0 \end{array}\right] \;\longrightarrow\; \left[\begin{array}{cc|c} 1 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right] \]
\[ \begin{aligned} \lambda_1 &= -1 &&\Longleftrightarrow\quad v_1 = \begin{pmatrix} 1 \\ -1 \end{pmatrix} \\[6pt] \lambda_2 &= 3 &&\Longleftrightarrow\quad v_2 = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \end{aligned} \]
\[ P = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}, \qquad D = \begin{pmatrix} -1 & 0 \\ 0 & 3 \end{pmatrix}, \qquad \det P = 1 + 1 = 2, \qquad P^{-1} = \begin{pmatrix} 1/2 & -1/2 \\ 1/2 & 1/2 \end{pmatrix} \]
\[ A = PDP^{-1} = \begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1/2 & -1/2 \\ 1/2 & 1/2 \end{pmatrix}, \qquad A^k = PD^kP^{-1} \]
Decomposing the starting vector along the eigenbasis,
\[ \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 \\ -1 \end{pmatrix} + \frac{1}{2}\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \]
\[ \boxed{\; A^k \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \frac{1}{2}(-1)^k \begin{pmatrix} 1 \\ -1 \end{pmatrix} + \frac{1}{2}(3)^k \begin{pmatrix} 1 \\ 1 \end{pmatrix} \;} \]
Checking against direct multiplication:
\[ \underbrace{ \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} }_{k \,=\, 1} \qquad \underbrace{ \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \end{pmatrix} }_{k \,=\, 2} \qquad \underbrace{ \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} 5 \\ 4 \end{pmatrix} = \begin{pmatrix} 13 \\ 14 \end{pmatrix} }_{k \,=\, 3} \]
Exercise 10.3. [Proof, \(\star\)] Prove that if \(T\) is invertible with eigenvalue \(\lambda\) (necessarily \(\lambda \neq 0\) — why?), then \(\lambda^{-1}\) is an eigenvalue of \(T^{-1}\), with the same eigenvectors. Also show: \(\lambda^k\) is an eigenvalue of \(T^k\) for every \(k \ge 1\).
Proof. Let \(v \in V\) be an eigenvector corresponding to eigenvalue \(\lambda\). Then \(Tv = \lambda v\). So
\[ \begin{aligned} &\Longrightarrow\quad T^{-1}(Tv) = T^{-1}(\lambda v) \\[6pt] &\Longrightarrow\quad v = \lambda T^{-1} v \\[6pt] &\Longrightarrow\quad T^{-1} v = \lambda^{-1} v \qquad \square \end{aligned} \]
Note \(\lambda \neq 0\) because \(T\) is invertible, so \(\operatorname{null} T = \{0\}\).
Base: \(k = 1\). \(T^1 v = Tv = \lambda v\).
Inductive hypothesis. Assume \(T^{k-1} v = \lambda^{k-1} v\), for \(k \ge 2\), \(k \in \mathbb{N}\).
Inductive step.
\[ T^k v = T^{k-1}(Tv) = T^{k-1}(\lambda v) = \lambda T^{k-1} v = \lambda \lambda^{k-1} v = \lambda^k v \qquad \square \]
Exercise 10.4. [Proof] Show the shear \(S = \left(\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right)\) is not diagonalizable: find all eigenvalues, show the eigenspace is one-dimensional, and conclude no basis of eigenvectors exists.
Proof. Show \(S = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\) is not diagonalizable.
\[ \begin{aligned} (1 - \lambda)^2 &= 0 \\[6pt] \lambda &= 1 \end{aligned} \qquad\qquad \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad\Longrightarrow\quad x_2 = 0 \]
\[ \boxed{\; \lambda = 1 \quad\Longleftrightarrow\quad \operatorname{span}\!\left( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \right) \;} \]
\(S\) has only one eigenvalue whose eigenspace is \(\operatorname{span}\left( \begin{smallmatrix} 1 \\ 0 \end{smallmatrix} \right)\). Thus the dimension of the eigenspace is \(1\), but for a \(2 \times 2\) matrix to be diagonalizable, the dimension of the eigenspace must be \(2\), since you need two linearly independent eigenvectors to be able to diagonalize a \(2 \times 2\) matrix.
\(\square\)