1 Jordan ブロックの冪 命題 1.1. K K K J m ( λ ) = λ I + N J_m(\lambda)=\lambda I+N J m ( λ ) = λ I + N m m m N m = 0 N^m=0 N m = 0 n ≥ 0 n\geq 0 n ≥ 0
J m ( λ ) n = ∑ k = 0 min ( n , m − 1 ) ( n k ) λ n − k N k J_m(\lambda)^n
=\sum_{k=0}^{\min(n,m-1)}
\binom{n}{k}\lambda^{\,n-k}N^k J m ( λ ) n = k = 0 ∑ m i n ( n , m − 1 ) ( k n ) λ n − k N k が成り立つ。
証明. λ I \lambda I λ I
( λ I + N ) n = ∑ k = 0 n ( n k ) λ n − k N k (\lambda I+N)^n=\sum_{k=0}^n\binom{n}{k}\lambda^{\,n-k}N^k ( λ I + N ) n = k = 0 ∑ n ( k n ) λ n − k N k を得る。k ≥ m k\geq m k ≥ m N k = 0 N^k=0 N k = 0 0 ≤ k ≤ min ( n , m − 1 ) 0\leq k\leq\min(n,m-1) 0 ≤ k ≤ min ( n , m − 1 ) ▨
証明. 表現行列を取る操作は積を保つので、[ T n ] B = [ T ] B n [T^n]_{\mathcal B}=[T]_{\mathcal B}^n [ T n ] B = [ T ] B n ▨
例 1.3. 係数体をR \R R
A = ( 2 1 0 2 ) = 2 I + N , x n + 1 = A x n , x 0 = ( 1 1 ) A=\begin{pmatrix}2&1\\0&2\end{pmatrix}=2I+N,\qquad
x_{n+1}=Ax_n,\qquad x_0=\binom11 A = ( 2 0 1 2 ) = 2 I + N , x n + 1 = A x n , x 0 = ( 1 1 ) とする。このとき、N 2 = 0 N^2=0 N 2 = 0 n ≥ 0 n\geq0 n ≥ 0
A n = 2 n I + n 2 n − 1 N = ( 2 n n 2 n − 1 0 2 n ) . A^n=2^nI+n2^{n-1}N
=\begin{pmatrix}2^n&n2^{n-1}\\0&2^n\end{pmatrix}. A n = 2 n I + n 2 n − 1 N = ( 2 n 0 n 2 n − 1 2 n ) . 帰納法からx n = A n x 0 x_n=A^nx_0 x n = A n x 0
x n = ( 2 n + n 2 n − 1 2 n ) x_n=\binom{2^n+n2^{n-1}}{2^n} x n = ( 2 n 2 n + n 2 n − 1 ) を得る。大きさ2 2 2 1 1 1
2 指数関数と微分方程式 K = R K=\R K = R C \C C T T T
exp ( T ) = ∑ n = 0 ∞ T n n ! \exp(T)=\sum_{n=0}^{\infty}\frac{T^n}{n!} exp ( T ) = n = 0 ∑ ∞ n ! T n と定める。基底を一つ選んで行列の任意のノルムを用いれば、この級数は絶対収束する。有限次元では異なるノルムにより同じ収束が定められるため、定義は選んだノルムに依存しない。
命題 2.1. J m ( λ ) = λ I + N J_m(\lambda)=\lambda I+N J m ( λ ) = λ I + N t ∈ R t\in\R t ∈ R t t t
exp ( t J m ( λ ) ) = e λ t ∑ k = 0 m − 1 t k k ! N k \exp(tJ_m(\lambda))
=e^{\lambda t}\sum_{k=0}^{m-1}\frac{t^k}{k!}N^k exp ( t J m ( λ )) = e λ t k = 0 ∑ m − 1 k ! t k N k が成り立つ。
証明. 可換する自己準同型S S S R R R
exp ( S ) exp ( R ) = ∑ a , b ≥ 0 S a R b a ! b ! = ∑ n ≥ 0 ( S + R ) n n ! = exp ( S + R ) \exp(S)\exp(R)
=\sum_{a,b\geq0}\frac{S^aR^b}{a!b!}
=\sum_{n\geq0}\frac{(S+R)^n}{n!}
=\exp(S+R) exp ( S ) exp ( R ) = a , b ≥ 0 ∑ a ! b ! S a R b = n ≥ 0 ∑ n ! ( S + R ) n = exp ( S + R ) を得る。S = t λ I S=t\lambda I S = t λ I R = t N R=tN R = tN
exp ( t J m ( λ ) ) = exp ( t λ I ) exp ( t N ) = e λ t I ∑ k = 0 m − 1 t k k ! N k \exp(tJ_m(\lambda))
=\exp(t\lambda I)\exp(tN)
=e^{\lambda t}I\sum_{k=0}^{m-1}\frac{t^k}{k!}N^k exp ( t J m ( λ )) = exp ( t λ I ) exp ( tN ) = e λ t I k = 0 ∑ m − 1 k ! t k N k となる。最後の等式ではN k = 0 ( k ≥ m ) N^k=0\ (k\geq m) N k = 0 ( k ≥ m ) ▨
例 2.2. J = ( λ 1 0 λ ) = λ I + N J=\begin{pmatrix}\lambda&1\\0&\lambda\end{pmatrix}=\lambda I+N J = ( λ 0 1 λ ) = λ I + N
e t J = e λ t ( 1 t 0 1 ) . e^{tJ}=e^{\lambda t}
\begin{pmatrix}1&t\\0&1\end{pmatrix}. e t J = e λ t ( 1 0 t 1 ) . したがってx ′ ( t ) = J x ( t ) x'(t)=Jx(t) x ′ ( t ) = J x ( t ) x ( 0 ) = ( a , b ) T x(0)=(a,b)^{\mathsf T} x ( 0 ) = ( a , b ) T
x ( t ) = e λ t ( a + t b b ) x(t)=e^{\lambda t}\binom{a+tb}{b} x ( t ) = e λ t ( b a + t b ) である。実際に右辺を微分するとx ′ ( t ) = J x ( t ) x'(t)=Jx(t) x ′ ( t ) = J x ( t ) t = 0 t=0 t = 0
例 2.3.
J = ( 2 1 0 0 2 1 0 0 2 ) = 2 I + N J=\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}=2I+N J = 2 0 0 1 2 0 0 1 2 = 2 I + N ではN 3 = 0 N^3=0 N 3 = 0
e t J = e 2 t ( I + t N + t 2 2 N 2 ) = e 2 t ( 1 t t 2 / 2 0 1 t 0 0 1 ) . e^{tJ}
=e^{2t}\left(I+tN+\frac{t^2}{2}N^2\right)
=e^{2t}\begin{pmatrix}1&t&t^2/2\\0&1&t\\0&0&1\end{pmatrix}. e t J = e 2 t ( I + tN + 2 t 2 N 2 ) = e 2 t 1 0 0 t 1 0 t 2 /2 t 1 . 右辺を微分するとJ e t J J e^{tJ} J e t J t = 0 t=0 t = 0
定理 2.4. A ∈ M n ( C ) A\in M_n(\C) A ∈ M n ( C ) x 0 ∈ C n x_0\in\C^n x 0 ∈ C n
x ′ ( t ) = A x ( t ) , x ( 0 ) = x 0 x'(t)=Ax(t),\qquad x(0)=x_0 x ′ ( t ) = A x ( t ) , x ( 0 ) = x 0 の解はx ( t ) = e t A x 0 x(t)=e^{tA}x_0 x ( t ) = e t A x 0 A A A x 0 x_0 x 0
証明. 有限区間上では指数級数と項別微分した級数が一様収束するので、
d d t e t A = ∑ n = 1 ∞ n t n − 1 A n n ! = A ∑ n = 0 ∞ t n A n n ! = A e t A \frac{d}{dt}e^{tA}
=\sum_{n=1}^{\infty}\frac{nt^{n-1}A^n}{n!}
=A\sum_{n=0}^{\infty}\frac{t^nA^n}{n!}
=Ae^{tA} d t d e t A = n = 1 ∑ ∞ n ! n t n − 1 A n = A n = 0 ∑ ∞ n ! t n A n = A e t A である。またe 0 A = I e^{0A}=I e 0 A = I e t A x 0 e^{tA}x_0 e t A x 0
y y y e − t A e t A = I e^{-tA}e^{tA}=I e − t A e t A = I
d d t ( e − t A y ( t ) ) = − A e − t A y ( t ) + e − t A A y ( t ) = 0 \frac{d}{dt}\bigl(e^{-tA}y(t)\bigr)
=-Ae^{-tA}y(t)+e^{-tA}Ay(t)=0 d t d ( e − t A y ( t ) ) = − A e − t A y ( t ) + e − t A A y ( t ) = 0 となる。したがって、すべてのt t t e − t A y ( t ) = y ( 0 ) = x 0 e^{-tA}y(t)=y(0)=x_0 e − t A y ( t ) = y ( 0 ) = x 0 e t A e^{tA} e t A y ( t ) = e t A x 0 y(t)=e^{tA}x_0 y ( t ) = e t A x 0 A , x 0 A,x_0 A , x 0 ▨
3 Jordan 標準形による行列関数 複素自己準同型T T T λ 1 , … , λ s \lambda_1,\ldots,\lambda_s λ 1 , … , λ s λ i \lambda_i λ i m i m_i m i §E3.30 系 2.3 により
m T ( t ) = ∏ i = 1 s ( t − λ i ) m i . m_T(t)=\prod_{i=1}^s(t-\lambda_i)^{m_i}. m T ( t ) = i = 1 ∏ s ( t − λ i ) m i . 命題 3.1. V V V T ∈ End C ( V ) T\in\operatorname{End}_{\C}(V) T ∈ End C ( V ) λ 1 , … , λ s ∈ C \lambda_1,\ldots,\lambda_s\in\C λ 1 , … , λ s ∈ C m 1 , … , m s m_1,\ldots,m_s m 1 , … , m s
m T ( t ) = ∏ i = 1 s ( t − λ i ) m i m_T(t)=\prod_{i=1}^s(t-\lambda_i)^{m_i} m T ( t ) = i = 1 ∏ s ( t − λ i ) m i と仮定する。このとき、複素数の族c i , k c_{i,k} c i , k 1 ≤ i ≤ s 1\leq i\leq s 1 ≤ i ≤ s 0 ≤ k < m i 0\leq k<m_i 0 ≤ k < m i
p ( k ) ( λ i ) = c i , k ( 1 ≤ i ≤ s , 0 ≤ k < m i ) p^{(k)}(\lambda_i)=c_{i,k}
\qquad(1\leq i\leq s,\ 0\leq k<m_i) p ( k ) ( λ i ) = c i , k ( 1 ≤ i ≤ s , 0 ≤ k < m i ) を満たす多項式p ∈ C [ t ] p\in\C[t] p ∈ C [ t ] m T m_T m T
p ( T ) p(T) p ( T ) は条件を満たすp p p
関数f f f λ i \lambda_i λ i m i − 1 m_i-1 m i − 1 c i , k = f ( k ) ( λ i ) c_{i,k}=f^{(k)}(\lambda_i) c i , k = f ( k ) ( λ i ) f ( T ) : = p ( T ) f(T):=p(T) f ( T ) := p ( T ) f ( T ) f(T) f ( T )
証明. q i ( t ) = ∑ k = 0 m i − 1 c i , k k ! ( t − λ i ) k q_i(t)=\sum_{k=0}^{m_i-1}\dfrac{c_{i,k}}{k!}(t-\lambda_i)^k q i ( t ) = ∑ k = 0 m i − 1 k ! c i , k ( t − λ i ) k q i ( k ) ( λ i ) = c i , k q_i^{(k)}(\lambda_i)=c_{i,k} q i ( k ) ( λ i ) = c i , k 0 ≤ k < m i 0\leq k<m_i 0 ≤ k < m i a i = ( t − λ i ) m i a_i=(t-\lambda_i)^{m_i} a i = ( t − λ i ) m i A i = ∏ j ≠ i a j A_i=\prod_{j\ne i}a_j A i = ∏ j = i a j gcd ( A i , a i ) = 1 \gcd(A_i,a_i)=1 g cd( A i , a i ) = 1 u i A i + v i a i = 1 u_iA_i+v_ia_i=1 u i A i + v i a i = 1 u i , v i u_i,v_i u i , v i
p = ∑ i = 1 s q i u i A i p=\sum_{i=1}^s q_i u_iA_i p = i = 1 ∑ s q i u i A i と置く。a i a_i a i i i i a i a_i a i u i A i ≡ 1 u_iA_i\equiv1 u i A i ≡ 1 p ≡ q i ( m o d a i ) p\equiv q_i\pmod{a_i} p ≡ q i ( mod a i ) p ≡ q i ( m o d a i ) p\equiv q_i\pmod{a_i} p ≡ q i ( mod a i ) p p p q i q_i q i λ i \lambda_i λ i m i − 1 m_i-1 m i − 1
p p p p ′ p' p ′ p − p ′ p-p' p − p ′ a i a_i a i a i a_i a i m T m_T m T ( p − p ′ ) ( T ) = 0 (p-p')(T)=0 ( p − p ′ ) ( T ) = 0 p ( T ) = p ′ ( T ) p(T)=p'(T) p ( T ) = p ′ ( T ) ▨
命題 3.2. f f f λ \lambda λ m − 1 m-1 m − 1 J m ( λ ) = λ I + N J_m(\lambda)=\lambda I+N J m ( λ ) = λ I + N
f ( J m ( λ ) ) = ∑ k = 0 m − 1 f ( k ) ( λ ) k ! N k . f(J_m(\lambda))
=\sum_{k=0}^{m-1}\frac{f^{(k)}(\lambda)}{k!}N^k. f ( J m ( λ )) = k = 0 ∑ m − 1 k ! f ( k ) ( λ ) N k . 証明. 命題 3.1 の補間多項式をp p p
p ( λ I + N ) = ∑ k ≥ 0 p ( k ) ( λ ) k ! N k . p(\lambda I+N)
=\sum_{k\geq0}\frac{p^{(k)}(\lambda)}{k!}N^k. p ( λ I + N ) = k ≥ 0 ∑ k ! p ( k ) ( λ ) N k . N m = 0 N^m=0 N m = 0 k ≥ m k\geq m k ≥ m p ( k ) ( λ ) = f ( k ) ( λ ) p^{(k)}(\lambda)=f^{(k)}(\lambda) p ( k ) ( λ ) = f ( k ) ( λ ) ▨
4 対数と平方根 補題 4.1. m ∈ N ≥ 1 m\in\NN m ∈ N ≥ 1
N ∈ C m × m N\in\C^{m\times m} N ∈ C m × m L ( N ) = ∑ j = 1 m ( − 1 ) j + 1 j N j L(N)=\sum_{j=1}^{m}\frac{(-1)^{j+1}}{j}N^j L ( N ) = ∑ j = 1 m j ( − 1 ) j + 1 N j e L ( N ) = I + N e^{L(N)}=I+N e L ( N ) = I + N C ∈ R m × m C\in\R^{m\times m} C ∈ R m × m e B = C 2 e^B=C^2 e B = C 2 B ∈ R m × m B\in\R^{m\times m} B ∈ R m × m 証明. 主張 (1) を示す。N m = 0 N^m=0 N m = 0 m = 1 m=1 m = 1 N = 0 N=0 N = 0 L ( N ) = 0 L(N)=0 L ( N ) = 0 e L ( N ) = I = I + N e^{L(N)}=I=I+N e L ( N ) = I = I + N m ≥ 2 m\ge2 m ≥ 2
ℓ ( z ) = ∑ j = 1 m − 1 ( − 1 ) j + 1 j z j \ell(z)=\sum_{j=1}^{m-1}\frac{(-1)^{j+1}}{j}z^j ℓ ( z ) = j = 1 ∑ m − 1 j ( − 1 ) j + 1 z j を取る。∑ k = 0 m − 1 ℓ ( z ) k / k ! \sum_{k=0}^{m-1}\ell(z)^k/k! ∑ k = 0 m − 1 ℓ ( z ) k / k ! z m z^m z m E ( z ) = ∑ j = 0 m − 1 a j z j E(z)=\sum_{j=0}^{m-1}a_jz^j E ( z ) = ∑ j = 0 m − 1 a j z j ℓ ( 0 ) = 0 \ell(0)=0 ℓ ( 0 ) = 0
E ′ ( z ) ≡ ℓ ′ ( z ) ∑ k = 0 m − 2 ℓ ( z ) k k ! ≡ ℓ ′ ( z ) E ( z ) ( m o d z m − 1 ) E'(z)\equiv\ell'(z)\sum_{k=0}^{m-2}\frac{\ell(z)^k}{k!}
\equiv\ell'(z)E(z)\pmod {z^{m-1}} E ′ ( z ) ≡ ℓ ′ ( z ) k = 0 ∑ m − 2 k ! ℓ ( z ) k ≡ ℓ ′ ( z ) E ( z ) ( mod z m − 1 ) となる。また、有限等比和から
( 1 + z ) ℓ ′ ( z ) = 1 − ( − z ) m − 1 ≡ 1 ( m o d z m − 1 ) (1+z)\ell'(z)=1-(-z)^{m-1}\equiv1\pmod {z^{m-1}} ( 1 + z ) ℓ ′ ( z ) = 1 − ( − z ) m − 1 ≡ 1 ( mod z m − 1 ) であるため、( 1 + z ) E ′ ( z ) ≡ E ( z ) ( m o d z m − 1 ) (1+z)E'(z)\equiv E(z)\pmod {z^{m-1}} ( 1 + z ) E ′ ( z ) ≡ E ( z ) ( mod z m − 1 ) a 0 = E ( 0 ) = 1 a_0=E(0)=1 a 0 = E ( 0 ) = 1 j j j
( j + 1 ) a j + 1 = ( 1 − j ) a j ( 0 ≤ j ≤ m − 2 ) (j+1)a_{j+1}=(1-j)a_j\qquad(0\le j\le m-2) ( j + 1 ) a j + 1 = ( 1 − j ) a j ( 0 ≤ j ≤ m − 2 ) を得る。したがってa 1 = 1 a_1=1 a 1 = 1 2 ≤ j ≤ m − 1 2\le j\le m-1 2 ≤ j ≤ m − 1 a j = 0 a_j=0 a j = 0 L ( N ) = ℓ ( N ) L(N)=\ell(N) L ( N ) = ℓ ( N ) N N N L ( N ) m = 0 L(N)^m=0 L ( N ) m = 0 e L ( N ) e^{L(N)} e L ( N ) N m = 0 N^m=0 N m = 0 z m z^m z m N N N
e L ( N ) = ∑ k = 0 m − 1 ℓ ( N ) k k ! = E ( N ) = I + N e^{L(N)}=\sum_{k=0}^{m-1}\frac{\ell(N)^k}{k!}=E(N)=I+N e L ( N ) = k = 0 ∑ m − 1 k ! ℓ ( N ) k = E ( N ) = I + N である。
主張 (2) を示す。C C C C m × m \C^{m\times m} C m × m λ 1 , … , λ r \lambda_1,\ldots,\lambda_r λ 1 , … , λ r C C C λ j ≠ 0 \lambda_j\ne0 λ j = 0 G j = G λ j ( C ) G_j=G_{\lambda_j}(C) G j = G λ j ( C ) N j = λ j − 1 ( C − λ j I ) ∣ G j N_j=\lambda_j^{-1}(C-\lambda_jI)|_{G_j} N j = λ j − 1 ( C − λ j I ) ∣ G j §E3.26 命題 1.3 と§E3.26 定理 2.1 によりN j N_j N j C m = G 1 ⊕ ⋯ ⊕ G r \C^m=G_1\oplus\cdots\oplus G_r C m = G 1 ⊕ ⋯ ⊕ G r G j G_j G j C = λ j ( I + N j ) C=\lambda_j(I+N_j) C = λ j ( I + N j ) C C C μ j ∈ C \mu_j\in\C μ j ∈ C e μ j = λ j 2 e^{\mu_j}=\lambda_j^2 e μ j = λ j 2 λ j ∈ R \lambda_j\in\R λ j ∈ R μ j = ln ( λ j 2 ) ∈ R \mu_j=\ln(\lambda_j^2)\in\R μ j = ln ( λ j 2 ) ∈ R { λ j , λ k = λ j ‾ } \{\lambda_j,\lambda_k=\overline{\lambda_j}\} { λ j , λ k = λ j } λ j = r j ( cos θ j + i sin θ j ) \lambda_j=r_j(\cos\theta_j+i\sin\theta_j) λ j = r j ( cos θ j + i sin θ j ) r j > 0 r_j>0 r j > 0 θ j ∈ R \theta_j\in\R θ j ∈ R μ j = 2 ln r j + 2 i θ j \mu_j=2\ln r_j+2i\theta_j μ j = 2 ln r j + 2 i θ j μ k = μ j ‾ \mu_k=\overline{\mu_j} μ k = μ j B ∈ C m × m B\in\C^{m\times m} B ∈ C m × m G j G_j G j
B ∣ G j = μ j I + 2 L ( N j ) B|_{G_j}=\mu_jI+2L(N_j) B ∣ G j = μ j I + 2 L ( N j ) と定める。μ j I \mu_jI μ j I L ( N j ) L(N_j) L ( N j ) 主張 (1) と可換な行列の指数法則によりG j G_j G j
e B = e μ j e 2 L ( N j ) = λ j 2 ( e L ( N j ) ) 2 = λ j 2 ( I + N j ) 2 = C 2 e^B=e^{\mu_j}e^{2L(N_j)}=\lambda_j^2\bigl(e^{L(N_j)}\bigr)^2=\lambda_j^2(I+N_j)^2=C^2 e B = e μ j e 2 L ( N j ) = λ j 2 ( e L ( N j ) ) 2 = λ j 2 ( I + N j ) 2 = C 2 であり、e B = C 2 e^B=C^2 e B = C 2 C m \C^m C m
κ : C m → C m \kappa\colon\C^m\to\C^m κ : C m → C m C C C κ C = C κ \kappa C=C\kappa κ C = C κ κ ( C − λ I ) m v = ( C − λ ‾ I ) m κ v \kappa(C-\lambda I)^m\boldsymbol v=(C-\overline\lambda I)^m\kappa\boldsymbol v κ ( C − λ I ) m v = ( C − λ I ) m κ v κ ( G λ ( C ) ) = G λ ‾ ( C ) \kappa(G_{\lambda}(C))=G_{\overline\lambda}(C) κ ( G λ ( C )) = G λ ( C ) λ k = λ j ‾ \lambda_k=\overline{\lambda_j} λ k = λ j v ∈ G j \boldsymbol v\in G_j v ∈ G j N k κ v = λ j ‾ − 1 ( C − λ j ‾ I ) κ v = κ N j v N_k\kappa\boldsymbol v=\overline{\lambda_j}^{-1}(C-\overline{\lambda_j}I)\kappa\boldsymbol v=\kappa N_j\boldsymbol v N k κ v = λ j − 1 ( C − λ j I ) κ v = κ N j v L L L μ k = μ j ‾ \mu_k=\overline{\mu_j} μ k = μ j v ∈ G j \boldsymbol v\in G_j v ∈ G j
κ B v = μ j ‾ κ v + 2 L ( N k ) κ v = B κ v \kappa B\boldsymbol v=\overline{\mu_j}\,\kappa\boldsymbol v+2L(N_k)\kappa\boldsymbol v=B\kappa\boldsymbol v κ B v = μ j κ v + 2 L ( N k ) κ v = B κ v である（λ j ∈ R \lambda_j\in\R λ j ∈ R k = j k=j k = j κ \kappa κ κ B = B κ \kappa B=B\kappa κ B = B κ C m \C^m C m e i \boldsymbol e_i e i B e i = B κ e i = κ B e i B\boldsymbol e_i=B\kappa\boldsymbol e_i=\kappa B\boldsymbol e_i B e i = B κ e i = κ B e i B B B B ∈ R m × m B\in\R^{m\times m} B ∈ R m × m ▨
定理 4.2. 複素正方行列A A A
e L = A , R 2 = A e^L=A,\qquad R^2=A e L = A , R 2 = A を満たす複素行列L , R L,R L , R
証明. A A A J m ( λ ) = λ ( I + X ) J_m(\lambda)=\lambda(I+X) J m ( λ ) = λ ( I + X ) λ ≠ 0 \lambda\ne0 λ = 0 X = N / λ X=N/\lambda X = N / λ X m = 0 X^m=0 X m = 0
e μ = λ e^\mu=\lambda e μ = λ μ ∈ C \mu\in\C μ ∈ C
L λ , m = μ I + ∑ k = 1 m − 1 ( − 1 ) k + 1 k X k L_{\lambda,m}
=\mu I+\sum_{k=1}^{m-1}\frac{(-1)^{k+1}}{k}X^k L λ , m = μ I + k = 1 ∑ m − 1 k ( − 1 ) k + 1 X k と置く。X m = 0 X^m=0 X m = 0 補題 4.1 (1) と可換な行列の指数法則によりe L λ , m = e μ e L ( X ) = λ ( I + X ) = J m ( λ ) e^{L_{\lambda,m}}=e^{\mu}e^{L(X)}=\lambda(I+X)=J_m(\lambda) e L λ , m = e μ e L ( X ) = λ ( I + X ) = J m ( λ )
次にν 2 = λ \nu^2=\lambda ν 2 = λ ν ∈ C \nu\in\C ν ∈ C
R λ , m = ν ∑ k = 0 m − 1 ( 1 / 2 k ) X k R_{\lambda,m}
=\nu\sum_{k=0}^{m-1}\binom{1/2}{k}X^k R λ , m = ν k = 0 ∑ m − 1 ( k 1/2 ) X k と置く。形式冪級数の恒等式( ∑ k ≥ 0 ( 1 / 2 k ) z k ) 2 = 1 + z \bigl(\sum_{k\geq0}\binom{1/2}{k}z^k\bigr)^2=1+z ( ∑ k ≥ 0 ( k 1/2 ) z k ) 2 = 1 + z z m z^m z m R λ , m 2 = J m ( λ ) R_{\lambda,m}^2=J_m(\lambda) R λ , m 2 = J m ( λ )
全 Jordan ブロックについて構成したL λ , m L_{\lambda,m} L λ , m R λ , m R_{\lambda,m} R λ , m L L L R R R ▨
5 演習 問題 5.1 (零固有値をもつ行列の平方根).
N = J 2 ( 0 ) ∈ M 2 ( C ) N=J_2(0)\in M_2(\C) N = J 2 ( 0 ) ∈ M 2 ( C ) R 2 = N R^2=N R 2 = N R ∈ M 2 ( C ) R\in M_2(\C) R ∈ M 2 ( C ) E 13 ∈ M 3 ( C ) E_{13}\in M_3(\C) E 13 ∈ M 3 ( C ) ( 1 , 3 ) (1,3) ( 1 , 3 ) 1 1 1 0 0 0 R 2 = E 13 R^2=E_{13} R 2 = E 13 R ∈ M 3 ( C ) R\in M_3(\C) R ∈ M 3 ( C ) (1) と(2) から、定理 4.2 の平方根に関する主張で可逆性の仮定を外すことができないこと、および零固有値をもつ行列には平方根をもつものともたないものがあることを結論せよ。 解答. (1) を示す。R 2 = N R^2=N R 2 = N N 2 = 0 N^2=0 N 2 = 0 R 4 = 0 R^4=0 R 4 = 0 det ( R ) 4 = det ( R 4 ) = 0 \det(R)^4=\det(R^4)=0 det ( R ) 4 = det ( R 4 ) = 0 R R R ker R ≠ 0 \ker R\neq0 ker R = 0 R = 0 R=0 R = 0 N = 0 N=0 N = 0 dim ker R = 1 \dim\ker R=1 dim ker R = 1 W = im R W=\operatorname{im}R W = im R R ( W ) ⊆ W R(W)\subseteq W R ( W ) ⊆ W R ( W ) = W R(W)=W R ( W ) = W R k ( C 2 ) = W R^k(\C^2)=W R k ( C 2 ) = W k ≥ 1 k\geq1 k ≥ 1 R 4 = 0 R^4=0 R 4 = 0 R ( W ) R(W) R ( W ) W W W R ( W ) = 0 R(W)=0 R ( W ) = 0 R 2 = 0 R^2=0 R 2 = 0 R 2 = N ≠ 0 R^2=N\neq0 R 2 = N = 0 N N N
(2) ではR = J 3 ( 0 ) R=J_3(0) R = J 3 ( 0 ) R e 1 = 0 Re_1=0 R e 1 = 0 R e 2 = e 1 Re_2=e_1 R e 2 = e 1 R e 3 = e 2 Re_3=e_2 R e 3 = e 2 R 2 e 1 = R 2 e 2 = 0 R^2e_1=R^2e_2=0 R 2 e 1 = R 2 e 2 = 0 R 2 e 3 = e 1 R^2e_3=e_1 R 2 e 3 = e 1 R 2 = E 13 R^2=E_{13} R 2 = E 13
(3) を示す。N N N E 13 E_{13} E 13 0 0 0 N N N 定理 4.2 の平方根の主張は可逆性を外すと成り立たない。一方E 13 E_{13} E 13 0 0 0 ▨