A \boldsymbol{A} A 按行分塊:
A = [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a m 1 a m 2 ⋯ a m n ] = [ A 1 A 2 ⋮ A m ]
\boldsymbol{A} =
\begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n}\\
a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn}\\
\end{bmatrix} =
\begin{bmatrix}
\boldsymbol{A}_1\\
\boldsymbol{A}_2\\
\vdots\\
\boldsymbol{A}_m
\end{bmatrix}
A = ⎣ ⎢ ⎢ ⎢ ⎡ a 1 1 a 2 1 ⋮ a m 1 a 1 2 a 2 2 ⋮ a m 2 ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a m n ⎦ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎡ A 1 A 2 ⋮ A m ⎦ ⎥ ⎥ ⎥ ⎤
其中,A i = [ a i 1 , a i 2 , ⋯ , a i n ] ( i = 1 , 2 , ⋯ , m ) \boldsymbol{A}_i=\begin{bmatrix} a_{i1},a_{i2},\cdots,a_{in} \end{bmatrix}(i=1,2,\cdots,m) A i = [ a i 1 , a i 2 , ⋯ , a i n ] ( i = 1 , 2 , ⋯ , m ) 是 A \boldsymbol{A} A 的一個子塊。
B \boldsymbol{B} B 按列分塊:
B = [ b 11 b 12 ⋯ b 1 n b 21 b 22 ⋯ b 2 n ⋮ ⋮ ⋮ b n 1 b n 2 ⋯ b n n ] = [ B 1 , B 2 , ⋯ , B n ]
\boldsymbol{B} =
\begin{bmatrix}
b_{11} & b_{12} & \cdots & b_{1n}\\
b_{21} & b_{22} & \cdots & b_{2n}\\
\vdots & \vdots & & \vdots\\
b_{n1} & b_{n2} & \cdots & b_{nn}\\
\end{bmatrix} =
\begin{bmatrix}
\boldsymbol{B}_1,\boldsymbol{B}_2,\cdots,\boldsymbol{B}_n
\end{bmatrix}
B = ⎣ ⎢ ⎢ ⎢ ⎡ b 1 1 b 2 1 ⋮ b n 1 b 1 2 b 2 2 ⋮ b n 2 ⋯ ⋯ ⋯ b 1 n b 2 n ⋮ b n n ⎦ ⎥ ⎥ ⎥ ⎤ = [ B 1 , B 2 , ⋯ , B n ]
其中,B j = [ b 1 j , b 2 j , ⋯ , b m j ] T ( j = 1 , 2 , ⋯ , n ) \boldsymbol{B}_j=\begin{bmatrix} b_{1j},b_{2j},\cdots,b_{mj} \end{bmatrix}^T(j=1,2,\cdots,n) B j = [ b 1 j , b 2 j , ⋯ , b m j ] T ( j = 1 , 2 , ⋯ , n ) 是 B \boldsymbol{B} B 的一個子塊。
加法:同型且分法一致,則 [ A 1 A 2 A 3 A 4 ] + [ B 1 B 2 B 3 B 4 ] = [ A 1 + B 1 A 2 + B 2 A 3 + B 3 A 4 + B 4 ] \begin{bmatrix} \boldsymbol{A}_1 & \boldsymbol{A}_2 \\ \boldsymbol{A}_3 & \boldsymbol{A}_4 \end{bmatrix} + \begin{bmatrix} \boldsymbol{B}_1 & \boldsymbol{B}_2 \\ \boldsymbol{B}_3 & \boldsymbol{B}_4 \end{bmatrix} = \begin{bmatrix} \boldsymbol{A}_1+\boldsymbol{B}_1 & \boldsymbol{A}_2+\boldsymbol{B}_2 \\ \boldsymbol{A}_3+\boldsymbol{B}_3 & \boldsymbol{A}_4+\boldsymbol{B}_4 \end{bmatrix} [ A 1 A 3 A 2 A 4 ] + [ B 1 B 3 B 2 B 4 ] = [ A 1 + B 1 A 3 + B 3 A 2 + B 2 A 4 + B 4 ] 。
數乘:k [ A B C D ] = [ k A k B k C k D ] k\begin{bmatrix} \boldsymbol{A} & \boldsymbol{B} \\ \boldsymbol{C} & \boldsymbol{D} \\ \end{bmatrix} = \begin{bmatrix} k\boldsymbol{A} & k\boldsymbol{B} \\ k\boldsymbol{C} & k\boldsymbol{D} \\ \end{bmatrix} k [ A C B D ] = [ k A k C k B k D ] 。
乘法:[ A B C D ] [ X Y Z W ] = [ A X + B Z A Y + B W C X + D Z C Y + D W ] \begin{bmatrix} \boldsymbol{A} & \boldsymbol{B} \\ \boldsymbol{C} & \boldsymbol{D} \\ \end{bmatrix} \begin{bmatrix} \boldsymbol{X} & \boldsymbol{Y} \\ \boldsymbol{Z} & \boldsymbol{W} \\ \end{bmatrix} = \begin{bmatrix} \boldsymbol{A}\boldsymbol{X}+\boldsymbol{B}\boldsymbol{Z} & \boldsymbol{A}\boldsymbol{Y}+\boldsymbol{B}\boldsymbol{W} \\ \boldsymbol{C}\boldsymbol{X}+\boldsymbol{D}\boldsymbol{Z} & \boldsymbol{C}\boldsymbol{Y}+\boldsymbol{D}\boldsymbol{W} \\ \end{bmatrix} [ A C B D ] [ X Z Y W ] = [ A X + B Z C X + D Z A Y + B W C Y + D W ] ,要可乘、可加。
若 A , B \boldsymbol{A},\boldsymbol{B} A , B 分別爲 m , n m,n m , n 階方陣,則分塊對角矩陣的冪爲 [ A 0 0 B ] n = [ A n 0 0 B n ] \begin{bmatrix} \boldsymbol{A} & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{B} \\ \end{bmatrix}^n = \begin{bmatrix} \boldsymbol{A}^n & \boldsymbol{0} \\ \boldsymbol{0} & \boldsymbol{B}^n \end{bmatrix} [ A 0 0 B ] n = [ A n 0 0 B n ] 。
若 A m × n B n × s = 0 \boldsymbol{A}_{m \times n}\boldsymbol{B}_{n \times s}=\boldsymbol{0} A m × n B n × s = 0 ,將 B , 0 \boldsymbol{B},\boldsymbol{0} B , 0 按列分塊,有 A B = A [ β 1 , β 2 , ⋯ , β s ] = [ A β 1 , A β 2 , ⋯ , A β s ] = [ 0 , 0 , ⋯ , 0 ] = 0 \boldsymbol{A}\boldsymbol{B}=\boldsymbol{A}\begin{bmatrix} \boldsymbol{\beta}_1,\boldsymbol{\beta}_2,\cdots,\boldsymbol{\beta}_s \end{bmatrix}=\begin{bmatrix} \boldsymbol{\boldsymbol{A}\beta}_1,\boldsymbol{A}\boldsymbol{\beta}_2,\cdots,\boldsymbol{A}\boldsymbol{\beta}_s \end{bmatrix}=\begin{bmatrix} \boldsymbol{0},\boldsymbol{0},\cdots,\boldsymbol{0} \end{bmatrix}=\boldsymbol{0} A B = A [ β 1 , β 2 , ⋯ , β s ] = [ A β 1 , A β 2 , ⋯ , A β s ] = [ 0 , 0 , ⋯ , 0 ] = 0 ,則 A β i = 0 ( i = 1 , 2 , ⋯ , s ) \boldsymbol{A}\boldsymbol{\beta}_i=\boldsymbol{0}(i=1,2,\cdots,s) A β i = 0 ( i = 1 , 2 , ⋯ , s ) ,β i \boldsymbol{\beta}_i β i 是 A x = 0 \boldsymbol{A}\boldsymbol{x}=\boldsymbol{0} A x = 0 的解。
若 A m × n B n × s = C m × s \boldsymbol{A}_{m \times n}\boldsymbol{B}_{n \times s}=\boldsymbol{C}_{m \times s} A m × n B n × s = C m × s ,
將 B , C \boldsymbol{B},\boldsymbol{C} B , C 按行分塊,有 [ a 11 a 12 ⋯ a 1 n a 21 a 22 ⋯ a 2 n ⋮ ⋮ ⋮ a m 1 a m 2 ⋯ a m n ] [ β 1 β 2 ⋮ β n ] = [ γ 1 γ 2 ⋮ γ n ] \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn}\\ \end{bmatrix}\begin{bmatrix} \boldsymbol{\beta}_1\\ \boldsymbol{\beta}_2\\ \vdots\\ \boldsymbol{\beta}_n \end{bmatrix}=\begin{bmatrix} \boldsymbol{\gamma}_1\\ \boldsymbol{\gamma}_2\\ \vdots\\ \boldsymbol{\gamma}_n \end{bmatrix} ⎣ ⎢ ⎢ ⎢ ⎡ a 1 1 a 2 1 ⋮ a m 1 a 1 2 a 2 2 ⋮ a m 2 ⋯ ⋯ ⋯ a 1 n a 2 n ⋮ a m n ⎦ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎡ β 1 β 2 ⋮ β n ⎦ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎡ γ 1 γ 2 ⋮ γ n ⎦ ⎥ ⎥ ⎥ ⎤ ,則 γ i = a i 1 β 1 + a i 2 β 2 + ⋯ + a i n β n ( i = 1 , 2 , ⋯ , m ) \boldsymbol{\gamma}_i=a_{i1}\boldsymbol{\beta}_1+a_{i2}\boldsymbol{\beta}_2+\cdots+a_{in}\boldsymbol{\beta}_n(i=1,2,\cdots,m) γ i = a i 1 β 1 + a i 2 β 2 + ⋯ + a i n β n ( i = 1 , 2 , ⋯ , m ) ,故 C \boldsymbol{C} C 的行向量是 B \boldsymbol{B} B 的行向量的線性組合;
將 A , C \boldsymbol{A},\boldsymbol{C} A , C 按列分塊,有 [ α 1 , α 2 , ⋯ , α n ] [ b 11 b 12 ⋯ b 1 s b 21 b 22 ⋯ b 2 s ⋮ ⋮ ⋮ b n 1 b n 2 ⋯ b n s ] = [ ξ 1 , ξ 2 , ⋯ , ξ n ] \begin{bmatrix} \boldsymbol{\alpha}_1,\boldsymbol{\alpha}_2,\cdots,\boldsymbol{\alpha}_n \end{bmatrix}\begin{bmatrix} b_{11} & b_{12} & \cdots & b_{1s}\\ b_{21} & b_{22} & \cdots & b_{2s}\\ \vdots & \vdots & & \vdots\\
b_{n1} & b_{n2} & \cdots & b_{ns}\\ \end{bmatrix} = \begin{bmatrix} \boldsymbol{\xi}_1,\boldsymbol{\xi}_2,\cdots,\boldsymbol{\xi}_n \end{bmatrix} [ α 1 , α 2 , ⋯ , α n ] ⎣ ⎢ ⎢ ⎢ ⎡ b 1 1 b 2 1 ⋮ b n 1 b 1 2 b 2 2 ⋮ b n 2 ⋯ ⋯ ⋯ b 1 s b 2 s ⋮ b n s ⎦ ⎥ ⎥ ⎥ ⎤ = [ ξ 1 , ξ 2 , ⋯ , ξ n ] ,則 ξ i = b 1 i α 1 + b 2 i α 2 + ⋯ + b n i α n ( i = 1 , 2 , ⋯ , s ) \boldsymbol{\xi}_i=b_{1i}\boldsymbol{\alpha}_1+b_{2i}\boldsymbol{\alpha}_2+\cdots+b_{ni}\boldsymbol{\alpha}_n(i=1,2,\cdots,s) ξ i = b 1 i α 1 + b 2 i α 2 + ⋯ + b n i α n ( i = 1 , 2 , ⋯ , s ) ,故 C \boldsymbol{C} C 的列向量是 A \boldsymbol{A} A 的列向量的線形組合。