MIT18.06 Linear Algebra (4)

lec4

基础性质

1

$$⁍$$

proof.

$$(A B)\left(B^{-1} A^{-1}\right)=A\left(B B^{-1}\right) A^{-1}=A A^{-1}=I$$

3

$$(A B)^{\top}=B^{\top} A^{\top}$$

2

$$\left(A^{\top}\right)^{-1}=\left(A^{-1}\right)^{\top}$$

proof.

$$\begin{aligned}A A^{-1} &=I \\\left(A A^{-1}\right)^{\top} &=I^{\top} \\\left(A^{-1}\right)^{\top} A^{\top} &=I^{\top}=I\end{aligned}$$

$A=LU$ 分解

$A=LU$ without row exchange.

$$A=\begin{array}{c}   \begin{bmatrix}   1 & 0 & 0 \\x & 1 & 0 \\x & x & 1\end{bmatrix} &\begin{bmatrix}   x & x & x \\0 & x & x \\0 & 0 & x\end{bmatrix} \\\text{L(lower)} & \text{U(upper)} \end{array}$$

when #pivot = 0, need row exchange.

or:

$$A=\left[\begin{array}{lll} 1 & 0 & 0 \\ x & 1 & 0 \\ x & x & 1 \end{array}\right]\left[\begin{array}{lll} x & 0 & 0 \\ 0 & x & 0 \\ 0 & 0 & x \end{array}\right]\left[\begin{array}{lll} 1 & x & x \\ 0 & 1 & x \\ 0 & 0 & 1 \end{array}\right]$$

for $n \times n$ matrix, $\exists n!$ P’s.

置换矩阵（Permutation matrix）

when n = 3, 6 permutation matrices:

$$\begin{array}{l}{\left[\begin{array}{lll}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\1 & 0 & 0 \\0 & 0 & 1\end{array}\right]\left[\begin{array}{lll}0 & 0 & 1 \\0 & 1 & 0 \\1 & 0 & 0\end{array}\right]} \\{\left[\begin{array}{lll}1 & 0 & 0 \\0 & 0 & 1 \\0 & 1 & 0\end{array}\right]\left[\begin{array}{lll}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0\end{array}\right]\left[\begin{array}{lll}0 & 0 & 1 \\1 & 0 & 0\\0 & 1 & 0\end{array}\right]}\end{array}$$

$$P^{-1}=P^{T}$$

e.g.

$$\begin{array}{c}   \begin{bmatrix} 0 &0  &1 \\  1& 0 &0 \\0  & 0 &1\end{bmatrix} \\ A\end{array} \begin{array}{c}   \begin{bmatrix} 0 &1  &0 \\  0& 0 &1 \\ 1& 0 &0\end{bmatrix} \\ A\end{array}=\begin{array}{c}   \begin{bmatrix} 1 &0  &0 \\  0& 1 &0 \\0  & 0 &1\end{bmatrix} \\ C\end{array}$$

proof.

$$\begin{array}{l}\because A_{i j}=1 \Rightarrow \text { row i of } C=\text { row j of } B \\\therefore \text { row j of } B=\text { col j of } A \\\therefore C_{i i}=A_{i j}=1 \\C=I\end{array}$$