[Linear Algebra] Lecture 6

Author

SEOYEON CHOI

Published

February 25, 2024

Import

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

학습목표

  • Vector Spaces and Subspaces
  • Column space of \(A\) : solving \(Ax=b\)
  • Nullspace of \(A\)

사전 학습

What is the vector space?

  • It is a bunch of vectors that where I allow I can add the two vectors in the space.
    • answer is that stayin in the space.

Vector space requirment

  • \(v+w\) and \(cv\) are in the space
  • all columns \(cv+dw\) are in the space.
    • the sum and the scale multiplication which combines into linear combinations
    • c랑 d에 어느 값을 곱해도(multiply) space 에 머무르는(stay)공간 = vector space
# 임의의 평면 방정식: ax + by + cz + d = 0 (원점을 지나므로 d = 0)
a, b, c = 1, 2, -3

# 임의의 선 방정식: x = t, y = 2t, z = 3t
t = np.linspace(-5, 5, 100)  # 매개변수 t 값 생성
x_line = t
y_line = 2 * t
z_line = 3 * t

# 그래프 생성
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# 원점을 지나는 평면 그리기
x_plane, y_plane = np.meshgrid(np.linspace(-5, 5, 10), np.linspace(-5, 5, 10))
z_plane = (-a * x_plane - b * y_plane) / c
ax.plot_surface(x_plane, y_plane, z_plane, alpha=0.5)

# 원점을 지나는 선 그리기
ax.plot(x_line, y_line, z_line, label='Line', color='r')

# 축 레이블 설정
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')

# 범례 추가
ax.legend()

# 그래프 출력
plt.show()

blue: plane through \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) is subspace of \(R^3\)

red: line is a subplace (through \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)) \(\to\) 모든 vector를 포함하지 않음.

- 2 subspaces : \(P\) and \(L\)

\(P \cup L\) = all vectors in \(P\) or \(L\) or both

  • This is not a subspace.

\(P \cap L\) = all vectors in both \(P\) and \(L\)

  • This is a subspace
  • \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\) 이 자체만으로 subspace 가능, 아무거나 multiply해도 0이잖아.

- Subspaced \(S\) and \(T\), interaction \(S \cap T\) is a subspace.

vector \(d\),\(W\)가 있을때,

  • \(\{d,w \} \in S\), \(\{d,w \} \in T\)라면,
  • \(\{d,w \}\cup S = S\), 왜냐하면 \(S\)에 \(\{d,w \}\)가 속하니까.
  • \(\{d,w \}\cup T = T\), 왜냐하면 \(T\)에 \(\{d,w \}\)가 속하니까.
  • \(\{7d,7w \}\in S\), multiply해도 속함.
  • \(\{7d,7w \}\in T\), multiply해도 속함.
  • original \(\{d,w \}\)가 \(S,T\)에 속하고 \(S,T\)는 subspace이다.

- Column Space of \(A\)

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

\(A\)는 \(4 \times 3\) matrix니까 column space of \(A\) is subspace of \(R^4\).

Column Space of \(A = C(A)\) \(\to\) All linear combinations of columns

  • Does \(Ax=b\) have a solution for every \(b\)?
    • NO
  • Which right hand side \(b\) ok?
    • \(Ax=b\) \(\to\) 4 equations and 3 unknowns
    • We can’t always do it. there are a lots of vectors \(b\). that are not combinations of these columns.

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

  • 즉, 몇 개의 vector \(b\)의 solve한 결과가 존재한다.
    • that’s the bunch of right-hand side.
  • Which \(b\)’s allow this system to be solved?
    • when \(b\)’s are all zero.
    • \(Ax = OI\)

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4\end{bmatrix}\)

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \\ 1\end{bmatrix}\)

  • can solve \(Ax=b\) exactly when the right hand side \(b\) is a vector in the column space \(C(A)\)
  • can solve \(Ax=b\) when \(b\) is a combination of the columns, when it’s in the column space.

Why?

  • because the column space by it’s definition contains all the combinations.
  • If \(b\) is a combination of the columns, then that combination tells me what \(x\) should be.
  • If \(b\) is not a combination of the columns, then there is no \(x\)(\(Ax=b\) 못 푼다.)

Can I throw away columns and have the same column space>

  • which one do you suggest I throw away?
    • 3 column : 1 column + 2 column = 3column이니까 관계 있어 보여서.
    • 1 column 해도 이 관계가 무너지긴 함.
  • throw away 한다면, the column space of this matrix \(A\) as a two dimensional subspace of \(R^4\) 되겠지..?

- Nullspace of \(A\)

  • all solutions \(x = \begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix}\) in \(R^3\) to \(Ax = 0\).

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}\)

Whatever that matrix is, its nullspace contains zero.

  • because \(A\) times zero vector sure gives the zero right-hand side.

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\)

Nullspace \(N(A)\) contains \(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\), \(c\begin{bmatrix} 1 \\ 1 \\ -1\end{bmatrix}\)

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} c \\ c \\ -c \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0\end{bmatrix}\)

이것은 아래와 같은 line으로 나타나며, line in \(R^3\)이다.

# 선의 방정식: x = x0 + at, y = y0 + bt, z = z0 + ct
# 이 예시에서는 (0,0,0)을 지나고 (1,1,1)을 향하는 선을 그립니다.
x0, y0, z0 = 0, 0, 0
a, b, c = 1, 1, 1
t = np.linspace(-10, 10, 100)  # t 값의 범위 설정

# 선 위의 점들
x_line = x0 + a * t
y_line = y0 + b * t
z_line = z0 + c * t

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# 선 그리기
ax.plot(x_line, y_line, z_line, color='r')

ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')

plt.show()

How do I know that the null space is a vector space?

  • Check that the solutions to \(Ax = 0\)
    • always give a subspace
  • If \(Av=0\) and \(Aw=0\)a then \(A(v+w)=0\), \(Av+Aw=0\)
    • \(v\) is in the nullspace.
    • \(w\) is in the nullspace.
    • \(v+w\) is in the nullspace.
    • then \(A(12v) = 0\), \(12Av = 0\).

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

Do I get the subspace if I get the solutions to \(\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}\)?

Do the x-s the solutions from the vector space?

  • No
  • the solutions to this do not from a subspace.
  • zero vector is not a solution.
    • solution can’t be a vector space.

solution

but it is not a subspace

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4\end{bmatrix}\)

\(Ax = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & 4 \\ 4 & 1 & 5 \end{bmatrix} \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \\ 4\end{bmatrix}\)