MIT学习笔记-Linear Algebra

画了两个图，分别表示col和row形式。row就是用正常的方程组写，然后解代表直线（二维）的交点。而col是vector相加的形式，即linear combination（这表明矩阵的col代表vector）。

Can I solve Ax=b for every b?如果有解的话，就可以用消元法解出来。同样的问法 Do the linear combination of the columns fill three dimensional space?这意味着方程组的解b代表着线性空间的点。
matrix multiply 方法1、[2 5;1 3]【1 2】=1【1 2】+2【1 3】。
Ax is comb of columns of A数据量大的时候这么理解更方便。
Determinants行列式
step1:subtract 3 row1 from row2、step2:subtract 2 row2 from row3 **
[1 0 0 [1 0 0 [1 2 1
-3 1 0 * 3 8 2 = 0 2 -2
0 0 1] 0 4 1] 0 4 1]
E21(E为Elementary Matrics)
a important fact move the parentheses arbitrary 任意交换括号
Problem1：image Permutation Matrics想象[a b;c d]的置换矩阵（行置换）
Problem2：image Matrix’s [1 0 0;-3 1 0;0 0 1] Inverses

four way of matrix multiplication 1st:normal way, row*col
2ed: cols way; 3st: rows way; 4th:col * row(m by 1 times 1 by p equals m by p); 5th:Block way.

A-1（if this is exist） * A = I = A * A-1 左逆与右逆没有区别
Gauss-Jordan：[A I]->[I A-1]将【A I】同乘A-1即可理解

与EA=U相比，更简洁.例：E=【1 0 0；-2 1 0；10 -5 1】则L=【1 0 0；2 1 0；0 5 1】
If no row exchanges the multipliers go directly into L(应该有负号？)
**How many operations on n*n matrix A?**应以Matlab算法复杂度的角度思考。count 1/3n3 on A，count n2 on b（back substitution）。

(A-1)T=(AT)-1
number of permutation 3 by 3 matrix have 6P(Pemutation matrices)'s;n by n matrix have n! P’s
P-1=PT;PPT=I
R * RT is always symmetric

must satisfied multiply and addition

R2 = all 2-dim real vectors
R3 = all columns with 3 real components

subspaces of R2
1、all of R2
2、any line through [0 0]
3、zero vector only

A = [1 3;2 3;4 1]; for columns of A, in R3 , all their combinations from subspace called column space, C(A).

mit linear