[Linear Algebra] Lecture 9

학습 목표

• Linear Independence
• Spanning a Space
• Basis and demension

Suppose \(A\) is \(m\) by \(n\) with \(m<n\).

Then there are nonzero solutions to \(Ax= 0\) (more unknown variables than equations)

• Reason: There will be free variables!

ex) 2 by 3 matrix \(\begin{bmatrix} - & - & - \\ - & - & - \end{bmatrix}\)

• unknown variables = 3, equations = 2

- Independence

Vectors \(x_1,x_2, \dots x_n\) are independent if no combination gives zero vector \(c_1 x_1 + c_2x_2 + \dots + c_n x_n \ne 0\) (except the zero combination, all \(c_i = 0\))

ex) \(v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) \(v_2 = \begin{bmatrix} 2 \\ 2 \end{bmatrix}\)라면,

• \(v_2 = 2 v_1\), dependent(double)
• \(\therefore 2v_1 - v_2 = 0\)

ex) \(v_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\), \(v_2 = 0\) 이라면,

• dependent, 0 이 속해 있으니까
• \(0 v_1 + 6 v_2 = 0\)

ex) \(A = \begin{bmatrix} 2 & 1 & 2.5 \\ 1 & 2 & -1 \end{bmatrix}\) \(\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}\)

• 이게 성립할 수 있다면 dependent임

Repeat when \(v_1, \dots v_n\) are columns of \(A\).

They are independent if nullspace of \(A\) is zero vector : rank = \(n\).

• independent : \(N(A) = \{ 0 \}\), no free variables

They are dependent if \(Ac=0\) for some nonzero \(c\).

• that is, rank \(<n\)(pivot 도 \(<n\))
• 왜냐하면 combination of columns 가 zero column이라서.
• free variables 도 있다.(free variable = n-r 임. r은 rank를 의미하고.)

- Spanning a Space

Vectors \(v_1, \dots , v_n\) span a space means: The space consists of all combinations of those vectors.

• vector들이 space를 span한다? : 공간이 그 벡터들의 모든 조합으로 구성되어 있다.

Basics for a space is a sequence of vectors \(v_1, v_2, \dots , v_l\) with 2 properties.

1. They are independent.
2. They span the space.

Example:

Space is \(\mathbb{R}^3\)

One basis is \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\) \(\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\) \(\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}\)

Another basis is \(\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\) \(\begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix}\) \(\begin{bmatrix} 3 \\ 3 \\ 8 \end{bmatrix}\)

\(n\) vectors gives basis if the \(n \times n\) matrix with those columns is invertible.

Another basis \(\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}\) \(\begin{bmatrix} 2 \\ 2 \\ 5 \end{bmatrix}\)

Given a space: columns

Every basis for the space have the same number of vectors

\(\to\) Definition: the dimension of the space

• = the number of baasis
• = the number of vectors

Space is \(C(A)\) = \(\begin{bmatrix} 1 & 2 & 3 & 1 \\ 1 & 1 & 2 & 1 \\ 1 & 2 & 3 & 1 \end{bmatrix}\)

• We can put the basis on 1,2 columns.
• We can’t put the basis on 3,4 columns.
• That is, Rank \(= 2\)

\(N(A)\) \(\to\) \(\begin{bmatrix} -1 \\ -1 \\ 1 \\ 0 \end{bmatrix}\), \(\begin{bmatrix} -1 \\ 0 \\ 0 \\ 1\end{bmatrix}\)

• \(2 =\) rank(\(A\)) = #¹ pivot columns = dimension of \(C(A)\)²

¹ The number of

² Column Space

dimension, column space알면 basis 알 수 있음?

• dimension 수랑 basis 수랑 같다고 했잖아

dim \(C(A)=r\)

Do these two special solutions form a basis for the null space?

= Does the null space consist of all combinations of those the guys?

• Yes!

The null space is two dimensions.

The dimension of null space is the number of free variables.

dim\(N(A)\) = # free variables = \(n-r\)