【线性代数】MIT Linear Algebra Lecture 6: Column space and nullspace

Author｜ Rickyの水果摊

Time ｜ 2022.9.12

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Lecture 6: Column space and nullspace

Lecture Info

1. Instructor: Prof. Gilbert Strang

2. Course Number: 18.06

3. Topics: Linear Algebra

4. Official Lecture Resource: Resource Index of Linear Algebra

Excellent Notes on GitHub

There are some classic, excellent notes from other authors on GitHub, wihch I highly recommend you to star ⭐️ and read 📖

notes-linear-algebra (A systematic notes written in Chinese)

The-Art-of-Linear-Algebra (Focus on visualization of important concept of Linear Algebra)

Video Link

Lecture 6: Column space and nullspace (bilibili)

Lecture 6: Column space and nullspace (YouTube)

Key Points

1. column space
2. nullspace

Active Recall Questions

1. Suppose we have subspaces $P,Q$ , if $P\cap Q$ / $P\cup Q$ a subspace?
2. What is the mathematical meaning of $Ax$ ？
3. Does $Ax=b$ always have solutions for every b ? If the answer is no, which b’s allow $Ax=b$ to be solvable ? (From the perspective of $C(A)$)
4. What’s the definiton of the nullspace $N(A)$ ?
5. Suppose $A_{m\ast n}$ , why $N(A)$ is a subspace of ${\mathbb{R}}^n$ ? (Hint: rules that a subspace have to follow)
6. What are the similarities and differences between $C(A),N(A)$ ? (Hint: focus, dimension, the way they span, …)
7. Suppose all the solutions ($xs$) to $Ax=b$ ($b$ is not zero vector) form a new space $N(A{)}^{\prime }$, if $N(A{)}^{\prime }$ a subspace? (Hint: compare to the $N(A)$ that $Ax=0$ form)

Answer

1. $P\cap Q$ is a subspace, $P\cup Q$ is not a subspace

2. $Ax$ are all the linear combinations of columns of $A$, which form $C(A)$ (Some columns of A contribute nothing to $C(A)$, we will discuss them in the next lecture)

3. No. $Ax=b$ is solvable only when $b$ is in $C(A)$

4. The nullspace $N(A)$ consists of all solutions to $Ax=0$.

5. If we want to check if a space is a subspace, we need to check it from the definition of subspace

   First, we take any two vectors $x_1,x_2$ in the nullspace, they all satisfy $Ax=0$, we need to check if $x_3=c_1\ast x_1+c_2\ast x_2$ is still in the nullspace.

   If we can prove $A{x}_3=0$, then we know that $x_3$ is in $N(A)$, and here is the derivation: $A{x}_3=A(c_1\ast x_1+c_2\ast x_2)=c_1\ast A{x}_1+c_2\ast A{x}_2=c_1\ast 0+c_2\ast 0=0$

   That means $x_3$ is still in $N(A)$. Hence $N(A)$ is a subspace

6. Here are their similarites and differences:

   1. $C(A)$ focus on $A$, $N(A)$ focus on $x$.
   2. dim of $C(A)$ is $rank(A)$, dim of $N(A)$ is $(n-rank(A))$
   3. $C(A)$ is a subspace of ${\mathbb{R}}^m$, $N(A)$ is a subspace of ${\mathbb{R}}^n$
   4. $C(A)$ is formed by the pivot columns of $A$, $N(A)$ is formed by all the solutions to $Ax=0$ (We will learn how to determine pivot columns and solutions to $Ax=0$ in the next lecture)

7. This derivation is similar to Q5.

   First, we take any two vectors $x_1,x_2$ in $N(A{)}^{\prime }$, they all satisfy $Ax=b$, we need to check if $x_3=c_1\ast x_1+c_2\ast x_2$ is still in $N(A{)}^{\prime }$

   If we can prove $A{x}_3=b$, then we know that $x_3$ is in $N(A{)}^{\prime }$, however $A{x}_3=A(c_1\ast x_1+c_2\ast x_2)=c_1\ast A{x}_1+c_2\ast A{x}_2=c_1\ast b+c_2\ast b\ne b$

   That means $x_3$ is not in $N(A{)}^{\prime }$. Hence $N(A{)}^{\prime }$ is not a subspace