# Schedule

• Vector spaces and Subspaces
• Column space of A : Solving Ax=b
• Nullspace of A

# Vector space requirements – 向量空间满足条件

v+w and c*v are in the space. 满足相加和数乘.
All combs cv+dw are in the space. 满足相加和数乘的线性组合.

To subspaces S and T, intersection S \cap T is a subspace.

1. 加法(Sum)封闭
2. 数乘(Scale of Multiplication)封闭

# Column space of a matrix A – A的列空间

A=\begin{bmatrix}1&1&2\2&1&3\3&1&4\4&1&5\end{bmatrix}

Column space of A is subspace of R^4, which is all linear combs of columns of A.

A的列空间与线性方程式组联系起来.

Does Ax=b have a solution for every b ?

Which b allow this system to be solved ?

Ax=\begin{bmatrix}1&1&2\2&1&3\3&1&4\4&1&5\end{bmatrix}*\begin{bmatrix}x_1\x_2\x_3\end{bmatrix}=\begin{bmatrix}b_1\b_2\b_3\b_4\end{bmatrix}=b

Can solve Ax=b exactly when b is in C(A).

# Null space – 零空间

Null space of A is N(A) contains all solutions x=\begin{bmatrix}x_1\x_2\x_3\end{bmatrix} to the equation Ax=0.
Ax=\begin{bmatrix}1&1&2\2&1&3\3&1&4\4&1&5\end{bmatrix}*\begin{bmatrix}x_1\x_2\x_3\end{bmatrix}=\begin{bmatrix}0\0\0\0\end{bmatrix}=0

The solutions to Ax=0 always give a subspace.

Ax=\begin{bmatrix}1&1&2\2&1&3\3&1&4\4&1&5\end{bmatrix}*\begin{bmatrix}x_1\x_2\x_3\end{bmatrix}=\begin{bmatrix}1\2\3\4\end{bmatrix}=b

1. 可以从几个向量,通过线性组合的方式得到子空间;
Given a few vectors and fill it out, take combinations;
2. 可以从一个方程式组中,通过让x满足特定条件来得到子空间.
Given a system of equations the requirements that the x-s have to satisfy.