LA学习笔记

Lecture 9 线性相关性 基 维数

关于这部分是我在B战学习MIT的Linear Algebra的学习笔记，在这里记录以用于之后的复习

其实整体学习下来就会发现这些概念都是一个接一个的引发出来，同时前面的知识有时候和新学习的知识属于同类知识哈哈哈

In this lecture, I learn some knowledges as follows:

Independent or Dependent

Vector $v_1, \dots, v_n$ are columns of A, where,

\begin{equation} \begin{cases} \textbf{Independent} \space if \space N(A) \space is \space zero-vector.\\ \textbf{Dependent} \space if \space Ac =0 \space for \space some \space non-zero-vector \space in \space N(A). \end{cases} \end{equation}

In other words, if there is no solution for Ac = 0 except for zero-vector, the columns of A are Independent.

Well… There is a phenomenon I find during the class. Lets set a matrix $A_{2 \times 3}$.

\begin{equation} \begin{Bmatrix} v_1 & v_2 & v_3 \end{Bmatrix} \end{equation}

where,

\begin{equation} v_1=\begin{Bmatrix} x_1 \\ x_2 \\ x_3 \end{Bmatrix} \end{equation}

All of the vectors $v_i$ is like the formulation above on.

If $v_1 $ and $v_2$ are independent, we always find that $v_3$ is the linear combination of $v_1$ and $v_2$. Therefore, the rank of A which is formulated as R(A) is going to be 2.

That is ,the combs of independent vectors can become any vector in this columns space which has the same dimension and these independent vectors can **span **a space,literally,the column space.

So the nation of “span” and “dimension” will introduce in the next two sections.

Vectors span a space

It means the space consists of the linear combs of these vectors.

Basis for a space

Its important to know that a space can have zillions of basis, but they are all a sequence of vectors with two properties down below:

• They are independent.
• They can span the space.

For example, a space $\in \R^3$, one of its infinite basis is

\begin{equation} v_1 = \begin{Bmatrix} 1 \\ 0 \\ 0 \end{Bmatrix}, v_2 = \begin{Bmatrix} 0 \\ 1 \\ 0 \end{Bmatrix}, v_3 = \begin{Bmatrix} 0 \\ 0 \\ 1 \end{Bmatrix} \end{equation}

Here we can find the nation of Dimension.

Tips: for n column vectors of A give basis if the $n \times n$ matrix A is invertible.

Dimension for a column space C(A)

Every basis of the space has the same number of vectors. **Here!!!**That number is the nation of dimension.

Here gives an example:

Matrix A is

\begin{equation} A = \begin{Bmatrix} 1 & 2 & 3 & 1 \\ 1 & 1 & 2 & 1\\ 1 & 2 & 3 & 1 \end{Bmatrix} \end{equation}

• First Q: the basis of A?

I want pick $[1, 1, 1,]^T$ and $[2, 1, 2]^T$ as the basis of C(A). NOT FOR A!!!

• Second Q: the N(A)?

If we pick the basis as above on, N(A) contains the solutions for Ac = 0. N(A) consists of $[1, 1, -1, 0]^T$ and $[-1, 0, 0, 1]^T$.

• Third Q: the R(A)?

R(A) = 2 = the number of pivot column(guess they are $[1, 1, 1,]^T$ and $[2, 1, 2]^T$) = the dimension of column space C(A).