ブロック対角行列の固有空間と広義固有空間
ブロック対角行列の固有空間と広義固有空間
このとき、ブロック対角行列\(\diag\left(A_{1},A_{2},\cdots,A_{r}\right)\)の固有空間\(W\left(\lambda\right)\)は
\[ W\left(\lambda\right)=\prod_{k\in\left\{ 1,2,\cdots,r\right\} }W_{k}\left(\lambda\right) \] となる。
このとき、ブロック対角行列\(\diag\left(A_{1},A_{2},\cdots,A_{r}\right)\)の広義固有空間\(\widetilde{W}\left(\lambda\right)\)は
\[ \widetilde{W}\left(\lambda\right)=\prod_{k\in\left\{ 1,2,\cdots,r\right\} }\widetilde{W}_{k}\left(\lambda\right) \] となる。
(1)
\(k\in\left\{ 1,2,\cdots,r\right\} \)として、\(n_{k}\)次正方行列\(A_{k}\)があり、この固有空間を\(W_{k}\left(\lambda\right)=\ker\left(A_{k}-\lambda I\right)\)とする。このとき、ブロック対角行列\(\diag\left(A_{1},A_{2},\cdots,A_{r}\right)\)の固有空間\(W\left(\lambda\right)\)は
\[ W\left(\lambda\right)=\prod_{k\in\left\{ 1,2,\cdots,r\right\} }W_{k}\left(\lambda\right) \] となる。
(2)
\(k\in\left\{ 1,2,\cdots,r\right\} \)として、\(n_{k}\)次正方行列\(A_{k}\)があり、この広義固有空間を\(\widetilde{W}_{k}\left(\lambda\right)=\ker\left(A_{k}-\lambda I\right)^{n_{k}}\)とする。このとき、ブロック対角行列\(\diag\left(A_{1},A_{2},\cdots,A_{r}\right)\)の広義固有空間\(\widetilde{W}\left(\lambda\right)\)は
\[ \widetilde{W}\left(\lambda\right)=\prod_{k\in\left\{ 1,2,\cdots,r\right\} }\widetilde{W}_{k}\left(\lambda\right) \] となる。
(1)
ブロック対角行列\[ A=\diag\left(\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right),\left(2\right)\right) \] は
\[ A_{1}=\left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right) \] \[ A_{2}=\left(2\right) \] とおくと、
\[ A=\diag\left(A_{1},A_{2}\right) \] となり、\(A_{1}\)の固有値は\(1,1\)となり、\(A_{2}\)の固有値は\(2\)となる。
このとき、\(A_{1}\)の固有空間広義\(W_{1}\left(1\right)\)と\(W_{1}\left(2\right)\)と固有空間\(\widetilde{W}_{1}\left(1\right)\)と\(\widetilde{W}_{1}\left(2\right)\)は
\begin{align*} W_{1}\left(1\right) & =\ker\left(A_{1}-I_{2}\right)\\ & =\ker\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)\\ & =\left\{ \left(x,0\right)^{T};x,y\in\mathbb{C}\right\} \end{align*} \begin{align*} W_{1}\left(2\right) & =\ker\left(A_{1}-2I_{2}\right)\\ & =\ker\left(\begin{array}{cc} -1 & 1\\ 0 & -1 \end{array}\right)\\ & =\left(0,0\right)^{T} \end{align*} \begin{align*} \widetilde{W}_{1}\left(1\right) & =\ker\left(A_{1}-I_{2}\right)^{2}\\ & =\ker\left(\begin{array}{cc} 0 & 1\\ 0 & 0 \end{array}\right)^{2}\\ & =\ker O\\ & =\left\{ \left(x,y\right)^{T};x,y\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}_{1}\left(2\right) & =\ker\left(A_{1}-2I_{2}\right)^{2}\\ & =\ker\left(\begin{array}{cc} -1 & 1\\ 0 & -1 \end{array}\right)^{2}\\ & =\ker\left(\begin{array}{cc} 1 & -2\\ 0 & 1 \end{array}\right)\\ & =\left(0,0\right)^{T} \end{align*} となり、\(A_{2}\)の固有空間広義\(W_{2}\left(1\right)\)と\(W_{2}\left(2\right)\)と固有空間\(\widetilde{W}_{2}\left(1\right)\)と\(\widetilde{W}_{2}\left(2\right)\)は
\begin{align*} W_{2}\left(1\right) & =\ker\left(A_{2}-I_{1}\right)\\ & =\ker\left(1\right)\\ & =\boldsymbol{0}_{V}\\ & =\left(0\right)^{T} \end{align*} \begin{align*} W_{2}\left(2\right) & =\ker\left(A_{2}-2I_{1}\right)\\ & =\ker\left(0\right)\\ & =\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}_{2}\left(1\right) & =\ker\left(A_{2}-I_{1}\right)^{1}\\ & =\ker\left(1\right)^{1}\\ & =\boldsymbol{0}_{V}\\ & =\left(0\right)^{T} \end{align*} \begin{align*} \widetilde{W}_{2}\left(2\right) & =\ker\left(A_{2}-2I_{1}\right)^{1}\\ & =\ker\left(0\right)^{1}\\ & =\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \end{align*} となるので、\(A\)の固有空間広義\(W\left(1\right)\)と\(W\left(2\right)\)広義固有空間\(\widetilde{W}\left(1\right)\)と\(\widetilde{W}\left(2\right)\)は
\begin{align*} W\left(1\right) & =W_{1}\left(1\right)\times W_{2}\left(1\right)\\ & =\left\{ \left(x,0\right)^{T};x\in\mathbb{C}\right\} \times\left(0\right)^{T}\\ & =\left\{ \left(x,0,0\right)^{T};x\in\mathbb{C}\right\} \end{align*} \begin{align*} W\left(2\right) & =W_{1}\left(2\right)\times W_{2}\left(2\right)\\ & =\left(0,0\right)^{T}\times\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \\ & =\left\{ \left(0,0,z\right)^{T};z\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}\left(1\right) & =\widetilde{W}_{1}\left(1\right)\times\widetilde{W}_{2}\left(1\right)\\ & =\left\{ \left(x,y\right)^{T};x,y\in\mathbb{C}\right\} \times\left(0\right)^{T}\\ & =\left\{ \left(x,y,0\right)^{T};x,y\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}\left(2\right) & =\widetilde{W}_{1}\left(2\right)\times\widetilde{W}_{2}\left(2\right)\\ & =\left(0,0\right)^{T}\times\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \\ & =\left\{ \left(0,0,z\right)^{T};z\in\mathbb{C}\right\} \end{align*} となる。
(2)
ブロック対角行列\[ A=\diag\left(\left(\begin{array}{cc} 1 & 1\\ 0 & 2 \end{array}\right),\left(1\right)\right) \] は
\[ A_{1}=\left(\begin{array}{cc} 1 & 1\\ 0 & 2 \end{array}\right) \] \[ A_{2}=\left(1\right) \] とおくと、
\[ A=\diag\left(A_{1},A_{2}\right) \] となり、\(A_{1}\)の固有値は\(1,2\)となり、\(A_{2}\)の固有値は\(1\)となる。
このとき、\(A_{1}\)の固有空間広義\(W_{1}\left(1\right)\)と\(W_{1}\left(2\right)\)と固有空間\(\widetilde{W}_{1}\left(1\right)\)と\(\widetilde{W}_{1}\left(2\right)\)は
\begin{align*} W_{1}\left(1\right) & =\ker\left(A_{1}-I_{2}\right)\\ & =\ker\left(\begin{array}{cc} 0 & 1\\ 0 & 1 \end{array}\right)\\ & =\left\{ \left(x,0\right)^{T};x\in\mathbb{C}\right\} \end{align*} \begin{align*} W_{1}\left(2\right) & =\ker\left(A_{1}-2I_{2}\right)\\ & =\ker\left(\begin{array}{cc} -1 & 1\\ 0 & 0 \end{array}\right)\\ & =\left\{ \left(x,x\right)^{T};x\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}_{1}\left(1\right) & =\ker\left(A_{1}-I_{2}\right)^{2}\\ & =\ker\left(\begin{array}{cc} 0 & 1\\ 0 & 1 \end{array}\right)^{2}\\ & =\ker\left(\begin{array}{cc} 0 & 1\\ 0 & 1 \end{array}\right)\\ & =\left\{ \left(x,0\right)^{T};x\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}_{1}\left(2\right) & =\ker\left(A_{1}-2I_{2}\right)^{2}\\ & =\ker\left(\begin{array}{cc} -1 & 1\\ 0 & 0 \end{array}\right)^{2}\\ & =\ker\left(\begin{array}{cc} 1 & -1\\ 0 & 0 \end{array}\right)\\ & =\left\{ \left(x,x\right)^{T};x\in\mathbb{C}\right\} \end{align*} となり、\(A_{2}\)の固有空間広義\(W_{2}\left(1\right)\)と\(W_{2}\left(2\right)\)と固有空間\(\widetilde{W}_{2}\left(1\right)\)と\(\widetilde{W}_{2}\left(2\right)\)は
\begin{align*} W_{2}\left(1\right) & =\ker\left(A_{2}-I_{1}\right)\\ & =\ker\left(0\right)^{1}\\ & =\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \end{align*} \begin{align*} W_{2}\left(2\right) & =\ker\left(A_{2}-2I_{1}\right)\\ & =\ker\left(-1\right)\\ & =\boldsymbol{0} \end{align*} \begin{align*} \widetilde{W}_{2}\left(1\right) & =\ker\left(A_{2}-I_{1}\right)^{1}\\ & =\ker\left(0\right)^{1}\\ & =\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}_{2}\left(2\right) & =\ker\left(A_{2}-2I_{1}\right)^{1}\\ & =\ker\left(-1\right)^{1}\\ & =\boldsymbol{0} \end{align*} となるので、\(A\)の固有空間広義\(W\left(1\right)\)と\(W\left(2\right)\)広義固有空間\(\widetilde{W}\left(1\right)\)と\(\widetilde{W}\left(2\right)\)は
\begin{align*} W\left(1\right) & =W_{1}\left(1\right)\times W_{2}\left(1\right)\\ & =\left\{ \left(x,0\right)^{T};x\in\mathbb{C}\right\} \times\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \\ & =\left\{ \left(x,0,z\right)^{T};x,z\in\mathbb{C}\right\} \end{align*} \begin{align*} W\left(2\right) & =W_{1}\left(2\right)\times W_{2}\left(2\right)\\ & =\left\{ \left(x,x\right)^{T};x\in\mathbb{C}\right\} \times\boldsymbol{0}\\ & =\left\{ \left(x,x,0\right)^{T};x\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}\left(1\right) & =\widetilde{W}_{1}\left(1\right)\times\widetilde{W}_{2}\left(1\right)\\ & =\left\{ \left(x,0\right)^{T};x\in\mathbb{C}\right\} \times\left\{ \left(z\right)^{T};z\in\mathbb{C}\right\} \\ & =\left\{ \left(x,0,z\right)^{T};x,z\in\mathbb{C}\right\} \end{align*} \begin{align*} \widetilde{W}\left(2\right) & =\widetilde{W}_{1}\left(2\right)\times\widetilde{W}_{2}\left(2\right)\\ & =\left\{ \left(x,x\right)^{T};x\in\mathbb{C}\right\} \times\boldsymbol{0}\\ & =\left\{ \left(x,x,0\right)^{T};x\in\mathbb{C}\right\} \end{align*} となる。
(1)
\(\sum_{k\in\left\{ 1,2,\cdots,r\right\} }n_{k}=n\)とする。ここで、\(V\)を\(n\times1\)行列、\(V_{k}\)を\(n_{k}\times1\)行列として、任意の\(\boldsymbol{v}\in V\)について、ある\(\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in\prod_{k\in\left\{ 1,2,\cdots,r\right\} }V_{r}\)が存在し、\(\boldsymbol{v}=\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in\prod_{k\in\left\{ 1,2,\cdots,r\right\} }V_{r}\)と表すことができる。
このとき、
\begin{align*} W\left(\lambda\right) & =\ker\left(A-\lambda I_{n}\right)\\ & =\left\{ \boldsymbol{v}\in V;\left(A-\lambda I_{n}\right)\boldsymbol{v}=\boldsymbol{0}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\left(\diag\left(A_{1},A_{2},\cdots,A_{r}\right)-\lambda\left(I_{n_{1}},I_{n_{2}},\cdots,I_{n_{r}}\right)\right)\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\left(\diag\left(A_{1}-\lambda I_{n_{1}},A_{2}-\lambda I_{n_{2}},\cdots,A_{r}-\lambda I_{n_{r}}\right)\right)\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\diag\left(\left(A_{1}-\lambda I_{n_{1}}\right)\boldsymbol{v}_{1},\left(A_{2}-\lambda I_{n_{2}}\right)\boldsymbol{v}_{2},\cdots,\left(A_{r}-\lambda I_{n_{r}}\right)\boldsymbol{v}_{r}\right)=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\left(A_{1}-\lambda I_{n_{1}}\right)\boldsymbol{v}_{1}=\boldsymbol{0},\left(A_{2}-\lambda I_{n_{2}}\right)\boldsymbol{v}_{2}=\boldsymbol{0},\cdots,\left(A_{r}-\lambda I_{n_{r}}\right)\boldsymbol{v}_{r}=\boldsymbol{0}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\forall k\in\left\{ 1,2,\cdots,r\right\} ,\left(A_{k}-\lambda I_{n_{k}}\right)\boldsymbol{v}_{k}=\boldsymbol{0}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\forall k\in\left\{ 1,2,\cdots,r\right\} ,\boldsymbol{v}_{k}\in\ker\left(A_{k}-\lambda I_{n_{k}}\right)\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\forall k\in\left\{ 1,2,\cdots,r\right\} ,\boldsymbol{v}_{k}\in W_{k}\left(\lambda\right)\right\} \\ & =\prod_{k\in\left\{ 1,2,\cdots,r\right\} }W_{k}\left(\lambda\right) \end{align*} となる。
従って、題意は成り立つ。
(2)
\(\sum_{k\in\left\{ 1,2,\cdots,r\right\} }n_{k}=n\)とする。ここで、\(V\)を\(n\times1\)行列、\(V_{k}\)を\(n_{k}\times1\)行列として、任意の\(\boldsymbol{v}\in V\)について、ある\(\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in\prod_{k\in\left\{ 1,2,\cdots,r\right\} }V_{r}\)が存在し、\(\boldsymbol{v}=\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in\prod_{k\in\left\{ 1,2,\cdots,r\right\} }V_{r}\)と表すことができる。
このとき、
\begin{align*} \widetilde{W}\left(\lambda\right) & =\ker\left(A-\lambda I_{n}\right)^{n}\\ & =\left\{ \boldsymbol{v}\in V;\left(A-\lambda I_{n}\right)^{n}\boldsymbol{v}=\boldsymbol{0}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\left(\diag\left(A_{1},A_{2},\cdots,A_{r}\right)-\lambda\left(I_{n_{1}},I_{n_{2}},\cdots,I_{n_{r}}\right)\right)^{n}\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\left(\diag\left(A_{1}-\lambda I_{n_{1}},A_{2}-\lambda I_{n_{2}},\cdots,A_{r}-\lambda I_{n_{r}}\right)\right)^{n}\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\diag\left(\left(A_{1}-\lambda I_{n_{1}}\right)^{n},\left(A_{2}-\lambda I_{n_{2}}\right)^{n},\cdots,\left(A_{r}-\lambda I_{n_{r}}\right)^{n}\right)\left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\diag\left(\left(A_{1}-\lambda I_{n_{1}}\right)^{n}\boldsymbol{v}_{1},\left(A_{2}-\lambda I_{n_{2}}\right)^{n}\boldsymbol{v}_{2},\cdots,\left(A_{r}-\lambda I_{n_{r}}\right)^{n}\boldsymbol{v}_{r}\right)=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\diag\left(\left(A_{1}-\lambda I_{n_{1}}\right)^{n_{1}}\boldsymbol{v}_{1},\left(A_{2}-\lambda I_{n_{2}}\right)^{n_{2}}\boldsymbol{v}_{2},\cdots,\left(A_{r}-\lambda I_{n_{r}}\right)^{n_{r}}\boldsymbol{v}_{r}\right)=\left(\boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0}\right)^{T}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\left(A_{1}-\lambda I_{n_{1}}\right)^{n_{1}}\boldsymbol{v}_{1}=\boldsymbol{0},\left(A_{2}-\lambda I_{n_{2}}\right)^{n_{2}}\boldsymbol{v}_{2}=\boldsymbol{0},\cdots,\left(A_{r}-\lambda I_{n_{r}}\right)^{n_{r}}\boldsymbol{v}_{r}=\boldsymbol{0}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\forall k\in\left\{ 1,2,\cdots,r\right\} ,\left(A_{k}-\lambda I_{n_{k}}\right)^{n_{k}}\boldsymbol{v}_{k}=\boldsymbol{0}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\forall k\in\left\{ 1,2,\cdots,r\right\} ,\boldsymbol{v}_{k}\in\ker\left(A_{k}-\lambda I_{n_{k}}\right)^{n_{k}}\right\} \\ & =\left\{ \left(\boldsymbol{v}_{1},\boldsymbol{v}_{2},\cdots,\boldsymbol{v}_{r}\right)^{T}\in V;\forall k\in\left\{ 1,2,\cdots,r\right\} ,\boldsymbol{v}_{k}\in\widetilde{W}_{k}\left(\lambda\right)\right\} \\ & =\prod_{k\in\left\{ 1,2,\cdots,r\right\} }\widetilde{W}_{k}\left(\lambda\right) \end{align*} となる。
従って、題意は成り立つ。
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ブロック対角行列の固有多項式と固有値
\[
p_{A}\left(\lambda\right)=\prod_{k\in\left\{ 1,2,\cdots,r\right\} }p_{A_{k}}\left(\lambda\right)
\]
ブロック対角行列の逆行列
\[
\left(\begin{array}{cccc}
A_{1,1} & O & \cdots & O\\
O & A_{2,2} & \ddots & O\\
\vdots & \ddots & \ddots & \vdots\\
O & O & \cdots & A_{p,p}
\end{array}\right)^{-1}=\left(\begin{array}{cccc}
A_{1,1}^{-1} & O & \cdots & O\\
O & A_{2,2}^{-1} & \ddots & O\\
\vdots & \ddots & \ddots & \vdots\\
O & O & \cdots & A_{p,p}^{-1}
\end{array}\right)
\]
対称ブロック分けのトーレス
\[
\tr\left(\begin{array}{cccc}
A_{1,1} & A_{1,2} & \cdots & A_{1,p}\\
A_{2,1} & A_{2,2} & \ddots & A_{2,p}\\
\vdots & \ddots & \ddots & \vdots\\
A_{p,1} & A_{p,2} & \cdots & A_{p,p}
\end{array}\right)=\sum_{k=1}^{p}\tr\left(A_{k,k}\right)
\]
ブロック対角行列の和・積・べき乗
\[
\left(\begin{array}{cccc}
A_{11} & O & \cdots & O\\
O & A_{22} & \ddots & O\\
\vdots & \ddots & \ddots & \vdots\\
O & O & \cdots & A_{pp}
\end{array}\right)^{k}=\left(\begin{array}{cccc}
A_{11}^{k} & O & \cdots & O\\
O & A_{22}^{k} & \ddots & O\\
\vdots & \ddots & \ddots & \vdots\\
O & O & \cdots & A_{pp}^{k}
\end{array}\right)
\]