matrix & arithmetic (2)

矩阵乘法

设

$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 & 2 & 3 \\ -1 & -2 & 4 \\ 0 & 5 & 1 \end{pmatrix}$

求 $3AB-2A$ 及 $A^TB$

$3AB-2A = \begin{pmatrix} 0 & 15 & 24 \\ 0 & -15 & 18 \\ 6 & 27 & 0 \end{pmatrix} - \begin{pmatrix} 2 & 2 & 2 \\ 2 & 2 & -2 \\ 2 & -2 & 2 \end{pmatrix} = \begin{pmatrix} -2 & 13 & 22 \\ -2 & -17 & 20 \\ 4 & 29 & -2 \end{pmatrix}$ $A^TB = AB = \begin{pmatrix} 0 & 5 & 8 \\ 0 & -5 & 6 \\ 2 & 9 & 0 \end{pmatrix}$

举反例证明以下命题是错的

如果 $A^2=\omicron$ 则 $A=\omicron$ $A = \begin{pmatrix} 1 & 1 \\ -1 & -1 \end{pmatrix}$
如果 $A^2=A$ 则 $A=\omicron$ 或 $A=E$ $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$
如果 $AX=AY(A \neq \omicron)$ 则 $X=Y$ $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} X = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix} Y = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$

矩阵的转置

A、B是n阶矩阵，A是对称矩阵，则 $B^TAB$ 也是对称矩阵 $(B^TAB)^T = (AB)^TB = B^TAB$
A、B是n阶对称矩阵，AB是对称矩阵的充分必要条件是AB=BA $AB = (AB)^T = B^TA^T = BA$

求矩阵的逆矩阵：

$A = \begin{pmatrix} 1 & 2 & -1 \\ 3 & 4 & -2 \\ 5 & -4 & 1 \end{pmatrix}$ $det(A) = \begin{vmatrix} 1 & 0 & 0 \\ 3 & -2 & -1 \\ 5 & -16 & 6 \end{vmatrix} = 2$ $A^{-1} = \frac{A*}{|A|} = \frac{1}{2} \begin{pmatrix} -4 & -2 & 0 \\ 13 & 6 & 1 \\ -22 & -14 & -2 \end{pmatrix} = \begin{pmatrix} -2 & \frac{13}{2} & 0 \\ -1 & 3 & \frac{1}{2} \\ 0 & -7 & -1 \end{pmatrix}$

求矩阵X

$\begin{pmatrix} 1 & 4 \\ -1 & 2 \end{pmatrix}X\begin{pmatrix} 2 & 0 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 1 \\ 0 & -1 \end{pmatrix}$ $AXB = C => A^{-1}AXBB^{-1} = A^{-1}CB^{-1}C => X = A^{-1}CB^{-1}$ $A^{-1} = \frac{1}{6} \begin{pmatrix} 2 & 1 \\ -4 & 1 \end{pmatrix}, B^{-1} = \frac{1}{2} \begin{pmatrix} 1 & 4 \\ -1 & 2 \end{pmatrix}$ $X = \begin{pmatrix} 1 & 1 \\ \frac{1}{4} & 0 \end{pmatrix}$

推导

设A为3阶矩阵， $|A|=\frac{1}{2}$ ，求 $|(2A)^{-1} - 5A^* |$

$|(2A)^{-1} - 5A^* | = |\frac{1}{2}A^{-1} - \frac{5}{2}A^{-1} | = |-2A^{-1} |$ $|A^{-1}|=2 => |-2A^{-1} |= -16$

设 $A = \begin{pmatrix} 0 & 3 & 3 \\ 1 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix}$ ， AB = A + 2B，求B

$(A - 2E)B = A => B = (A-2E)^{-1} A$ $det(A-2E) = 2, (A-2E)* =\begin{pmatrix} -1 & 3 & 3 \\ -1 & 1 & 3 \\ 1 & 1 & -1 \end{pmatrix}$ $B = \frac{1}{2}\begin{pmatrix} -1 & 3 & 3 \\ -1 & 1 & 3 \\ 1 & 1 & -1 \end{pmatrix} \begin{pmatrix} 0 & 3 & 3 \\ 1 & 1 & 0 \\ -1 & 2 & 3 \end{pmatrix} = \begin{pmatrix} 0 & 3 & 3 \\ -1 & 2 & 3 \\ 1 & 1 & 0 \end{pmatrix}$

设 $A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix}$ 且 $AB + E = A^2 + B$ ，求B

$B = (A - E)^{-1}(A^2 - E)$ $B = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 1 & 0 & 2 \\ 0 & 3 & 0 \\ 2 & 0 & 1 \end{pmatrix} = A + E$

设A = diag(1, -2, 1)， $A* BA = 2BA-8E$ ，求B

$B = 8(2A + 2E)^{-1} = 2A$

已知矩阵A的伴随矩阵A*为diag(1, 1, 1, 8)，且 $ABA^{-1} = BA^{-1} + 3E$ ，求B

$A = diag(2, 2, 2, \frac{1}{4}), B = (A - E)^{-1} * 3A => B = diag(6, 6, 6, -1)$

二项式

设
$P^{-1}AP = \Lambda$ ， $P = \begin{pmatrix} -1 & -4 \\ 1 & 1 \end{pmatrix}, \Lambda = \begin{pmatrix} -1 & 0 \\ 0 & 2 \end{pmatrix}$ ，求A^11

$A^11 = P \Lambda^{11} P^{-1}$ $= \begin{pmatrix} -1 & -4 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} -1 & 0 \\ 0 & 2^{11} \end{pmatrix}\frac{1}{3}\begin{pmatrix} 1 & 1 \\ 4 & -1 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 1+2^{13} & 4+2^{13} \\ -1-2^{1} & -4-2^{11} \end{pmatrix}$

设 $AP = P\Lambda$

$P = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & -2 \\ 1 & -1 & 1 \end{pmatrix} \Lambda = \begin{pmatrix} -1 & & \\ & 1 & \\ & & 5 \end{pmatrix}$

求 $\varphi(A) = A^8(5E - 6A + A^2)$

$\varphi(A) = P\lambda^8 (5E - 6\Lambda + \Lambda^2)P^{-1}$ $= \begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & -2 \\ 1 & -1 & 1 \end{pmatrix} \begin{pmatrix} 1 & & \\ & 1 & \\ & & 5^8 \end{pmatrix} \begin{pmatrix} 12 & & \\ & 0 & \\ & & 0 \end{pmatrix}\frac{1}{6} \begin{pmatrix} -2 & -2 & -2 \\ -3 & 0 & 3 \\ -1 & 2 & -1 \end{pmatrix}$ $\frac{1}{6} \begin{pmatrix} 12 & 0 & 0 \\ 12 & 0 & 0 \\ 12 & 0 & 0 \end{pmatrix} \begin{pmatrix} -2 & -2 & -2 \\ -3 & 0 & 3 \\ -1 & 2 & -1 \end{pmatrix} = 4 \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$

矩阵分块法

$\begin{pmatrix} 1 & 2 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & 0 & 3 & 1 \\ 0 & 1 & 2 & -1 \\ 0 & 0 & -2 & 3 \\ 0 & 0 & 0 & -3 \end{pmatrix}$

$\begin{pmatrix} a_1 & a_2 \\ \omicron & a_3 \end{pmatrix}\begin{pmatrix} b_1 & b_2 \\ \omicron & b_3 \end{pmatrix} = \begin{pmatrix} a_1b_1 & a_1b_2+a_2b_3 \\ \omicron & a_3b_3 \end{pmatrix} = \begin{pmatrix} 1 & 2 & 5 & 2 \\ 0 & 1 & 2 & -4 \\ 0 & 0 & -4 & 3 \\ 0 & 0 & 0 & -9 \end{pmatrix}$

$\begin{pmatrix} 3 & 4 & 0 & 0 \\ 4 & -3 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 2 & 2 \\ \end{pmatrix}$ ，求|A^8|及A^4

$A^4 = \begin{pmatrix} 5^4 & 0 & 0 & 0 \\ 0 & 5^4 & 0 & 0 \\ 0 & 0 & 2^4 & 0 \\ 0 & 0 & 2^6 & 2^4 \\ \end{pmatrix}, |A^8|=|A^4 * A^4|=5^8 * 5^8 * 2^8 * 2^8 = 10^{16}$