Chapter 8 — Diagonalization

Solution

Suppose \( Tv = \lambda v \) for some \( v \in V \), \( \lambda \in \F \). Applying \(T\) repeatedly, \( T^2 v = \lambda T(v) = \lambda^2 v \), and \( T^n(v) = \lambda^n v = 0 \) (nilpotent), so either \( \lambda = 0 \) or \( v = 0 \) — every eigenvalue of \(T\) is \(0\). A diagonalizable \(T\) would then be similar to \( \diag(0,\dots,0) = O \), forcing \( T = 0 \) and contradicting \( T \neq 0 \). So \(T\) cannot be diagonalized.

For \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \), \( A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O_{2\times 2} \), so \(A\) is nilpotent and, by the above, cannot be diagonalized. (Nilpotent operators/matrices cannot be diagonalized.)

Solution

Suppose \( Tv = \lambda v \), \( v \neq 0 \). Applying \(T\), \( T^2 v = \lambda^2 v \), and repeating, \( T^k v = \lambda^k v = 0 \). For \( v \neq 0 \), \( \lambda = 0 \). Thus \(0\) is the only eigenvalue of \(T\).

Solution

If \( \lambda \) is an eigenvalue, \( Tv = \lambda v \) for some \( v \neq 0 \), so \( (T - \lambda I)v = 0 \), i.e. \( v \in \Null(T - \lambda I) \). Since \( v \neq 0 \), \( \Null(T - \lambda I) \neq \{0\} \), so \( T - \lambda I \) is not injective (LADR 3.16: injectivity \( \iff \) null space \( = \{0\} \)). Conversely the same chain runs backwards. For \( [T] = A \in \C^{n\times n} \) (finite-dimensional), \( T - \lambda I \) is invertible \( \iff \) injective \( \iff \) surjective (LADR 3.65), so \( \lambda \) is an eigenvalue \( \iff A - \lambda I \) is not invertible \( \iff \det(A - \lambda I) = 0 \).

Solution

With \( (A - tI_n)_{ij} = b_{ij} = \begin{cases} a_{ii} - t & i = j \\ a_{ij} & i \neq j \end{cases} \), the Leibniz formula gives \( \det(A - tI_n) = \sum_{\sigma\in\Pi_n}\sign(\sigma)\prod_{j=1}^{n}b_{\sigma(j),j} \). Splitting off the identity permutation \( \sigma_I \) (with \( \sign(\sigma_I) = 1 \)), \[ \det(A - tI_n) = \underbrace{\sum_{\sigma\in\Pi_n\setminus\{\sigma_I\}}\sign(\sigma)\prod_{j=1}^{n}a_{\sigma(j),j}}_{\text{lower-degree terms}} + \prod_{j=1}^{n}(a_{jj} - t), \] where \( \prod_{j=1}^{n}(a_{jj} - t) \) is a degree-\(n\) polynomial (roots \( a_{11}, \dots, a_{nn} \)). Collecting terms, \[ \det(A - tI_n) = (-1)^n t^n + c_1 t^{n-1} + \dots + c_{n-1}t + c_n \] for some constants. Thus \( \det(A - tI_n) \) is a polynomial of degree \(n\).

Solution

\(A\) and \(B\) represent \(T\) in different bases, so \( A \sim B \) (Ex 37), i.e. \( A = PBP^{-1} \) for some invertible \(P\). Since \( P(tI)P^{-1} = tI \), \[\begin{aligned} f_A(\lambda) &= \det(A - \lambda I) = \det(PBP^{-1} - \lambda PIP^{-1}) \\ &= \det\!\big(P(B - \lambda I)P^{-1}\big) \\ &= \det(P)\det(B - \lambda I)\det(P^{-1}) \\ &= \det(B - \lambda I) = f_B(\lambda). \end{aligned}\]

Solution

By induction. Base case \( i = 2 \): from \( a_1 v_1 + a_2 v_2 = 0 \) (①), apply \(T\): \( a_1\lambda_1 v_1 + a_2\lambda_2 v_2 = 0 \) (②). Compute \( \lambda_1\cdot① - ② \): \( a_2 v_2(\lambda_2 - \lambda_1) = 0 \); since \( \lambda_1 \neq \lambda_2 \), \( a_2 = 0 \), and then \( a_1 = 0 \). So \( v_1, v_2 \) are LI.

Inductive step. Assume \( v_1, \dots, v_{i-1} \) are LI. From \( a_1 v_1 + \dots + a_i v_i = 0 \) (③), apply \( (T - \lambda_i I) \): \[ a_1(\lambda_1 - \lambda_i)v_1 + \dots + a_{i-1}(\lambda_{i-1} - \lambda_i)v_{i-1} = 0. \] Since the \( \lambda_j \) are distinct and \( v_1, \dots, v_{i-1} \) are LI, \( a_1 = \dots = a_{i-1} = 0 \); putting back in ③, \( a_i v_i = 0 \Rightarrow a_i = 0 \). So \( v_1, \dots, v_i \) are LI, and by induction \( v_1, \dots, v_k \) are LI.

Solution

Let \( g = g(\lambda) \), \( a = a(\lambda) \), \( [T] = A \), \( \dim\Null(A - \lambda I) = g \). Let \( x_1, \dots, x_g \) be a basis of \( \Null(A - \lambda I) \) (so \( Ax_j = \lambda x_j \)), and extend to a basis \( x_1, \dots, x_g, x_{g+1}, \dots, x_n \) of \( \C^n \). Then \[ A\,[\,x_1 \cdots x_g \mid x_{g+1} \cdots x_n\,] = [\,x_1 \cdots x_n\,]\underbrace{\begin{bmatrix} \lambda I_g & B \\ O & C \end{bmatrix}}_{\Sigma}, \qquad AX = X\Sigma, \] so \( A = X\Sigma X^{-1} \) (\(X\) invertible as its columns are LI). Then \[\begin{aligned} f_A(t) &= \det(A - tI_n) = \det\!\big(X(\Sigma - tI_n)X^{-1}\big) \\ &= \det(\Sigma - tI_n) \\ &= \det\begin{bmatrix} (\lambda - t)I_g & B \\ O & C - tI_{n-g} \end{bmatrix} = (\lambda - t)^g\,h(t), \end{aligned}\] using \( \det\begin{bmatrix} A & B \\ O & C \end{bmatrix} = \det(A)\det(C) \). So \( f_A(t) = (-1)^g(t - \lambda)^g h(t) \), and by the definition of algebraic multiplicity \( g \le a \), i.e. \( a(\lambda) \ge g(\lambda) \).

For \( A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \): \( f_A(t) = t^2 = 0 \), so \( \lambda = 0 \) with \( a(\lambda) = 2 \). From \( (A - \lambda I)v = 0 \), \( v_2 = 0 \), so \( \begin{bmatrix} 1 \\ 0 \end{bmatrix} \) is the only eigenvector direction, \( \nullity(A - \lambda I) = 1 \), giving \( g(\lambda) = 1 < a(\lambda) = 2 \). An operator is defective if some eigenvalue has algebraic multiplicity strictly larger than its geometric multiplicity.

Solution

\( u_i \in \Null(T - \lambda_i I) \) means \( (T - \lambda_i I)u_i = 0 \), i.e. \( Tu_i = \lambda_i u_i \), so each non-zero \( u_i \) is an eigenvector for \( \lambda_i \). Eigenvectors for distinct eigenvalues are LI, so \( u_1 + \dots + u_k = 0 \Rightarrow u_1 = \dots = u_k = 0 \) (the converse is trivial).

Solution

From \( A = U\Lambda U\herm \) with \( U\herm = U^{-1} \), \( \Lambda = U\herm A U \), so \( A \sim \Lambda \) and \( f_A(t) = f_\Lambda(t) = \det(\Lambda - tI) \). For an upper triangular matrix \( \Lambda \) with diagonal \( \lambda_1, \dots, \lambda_n \), the Leibniz expansion leaves only the identity permutation, so \( \det(\Lambda) = \prod_{i=1}^{n}\lambda_i \) (determinant of an upper triangular matrix is the product of its diagonal). Hence \( \det(\Lambda - tI) = \prod_{i=1}^{n}(\lambda_i - t) \), whose roots are \( \lambda_1, \dots, \lambda_n \). So the diagonal of \( \Lambda \) gives the eigenvalues of \(A\).

Solution

Diagonal case. For \( A = \diag(\lambda_1, \dots, \lambda_n) \), \( f_A(t) = \prod_{i=1}^{n}(\lambda_i - t) \), so \( f_A(A) = \prod_{i=1}^{n}(\lambda_i I - A) \). Each \( \lambda_i I - A = \diag(\lambda_i - \lambda_1, \dots, \lambda_i - \lambda_n) \) is diagonal with a \(0\) in the \(i\)-th slot. Since the product of diagonal matrices multiplies entrywise, the \(j\)-th diagonal entry of \( \prod_i(\lambda_i I - A) \) contains the factor \( (\lambda_j - \lambda_j) = 0 \). Hence \( f_A(A) = O \) for diagonal \(A\).

Diagonalizable case. If \( A = PDP^{-1} \) with \(D\) diagonal, then \( f_A(t) = f_D(t) \) (similar matrices). Writing \( f_A(t) = (-1)^n t^n + a_{n-1}t^{n-1} + \dots + a_0 \), \[\begin{aligned} f_A(A) &= (-1)^n A^n + \dots + a_0 I \\ &= P\big[(-1)^n(P^{-1}AP)^n + \dots + a_0 I\big]P^{-1} \\ &= P\,f_D(D)\,P^{-1}, \end{aligned}\] using \( (P^{-1}AP)^m = P^{-1}A^m P \). Since \( f_D(D) = 0 \), \( f_A(A) = 0 \). Hence Cayley–Hamilton holds for diagonalizable matrices.

Cayley–Hamilton for all matrices

Using Lemma 2, Ex 91, and Ex 72 of Ch. 6 (if \( \|A_n - A\| \to 0 \) then \( \|p(A_n) - p(A)\| \to 0 \) for any polynomial \(p\)): the characteristic polynomial \( f_A(t) \) is continuous in both \(A\) and \(t\). For any \( A \in \C^{n\times n} \) take diagonalizable \( A_k \to A \). Since \( f_{A_k}(A_k) = 0 \) (Ex 91) and everything is continuous, \[ f_A(A) = \lim_{k\to\infty} f_{A_k}(A_k) = 0. \] Hence the Cayley–Hamilton theorem holds for any \( A \in \C^{n\times n} \).