Linearly Independent
A set of vectors \({\mathbf{v_1}, \cdots, \mathbf{v_p}}\) in \(\mathbb{R^n}\) is said to be linearly independent, if the vector equation
\(x\mathbf{v_1} + \cdots + x_p\mathbf{v_p} = 0\)
Has only the trivial solution.
trivial solution 만 있을 때, vector 앞에 있는 coefficient 가 다 0인 솔루션
- 어떤 vector set 에 LD 를 따지는 것은, \(A\mathbf{x} = 0\)을 푸는 것과 동일함
- free variable이 한 개라도 존재하면, Nontrivial solution → 해가 무수히 존재
Linearly Dependent
The set \({\mathbf{v_1}, \cdots, \mathbf{v_p}}\) in \(\mathbb{R^n}\) is said to be linearly dependent if there exist weights \(c_1, \cdots, c_p\), not all zero, such that
\(c_1\mathbf{v_1} + \cdots + c_p\mathbf{v_p} = 0\)
\(c_1 \sim c_p\) 중 최소 하나라도 nonzero
Linear Independence of Matrix Columns
\(A = \left[ a_1, \cdots, a_n \right] \\
A\mathbf{x} = 0\\
x_1\mathbf{a_1} + x_2\mathbf{a_2} + \cdots + x_n\mathbf{a_n} = 0\)
The columns of a matrix \(A\) are linearly independent if and only if the equations \(A\mathbf{x} = 0\) has only the trivial solution.
Example.
Determine if the columns of the following matrix are linearly independent.
모든 row에 pivot position이 있고, free variable이 없음 → only trivial solution : Linearly independent
Sets of One Vector
If a set contains only on vector, \(\mathbf{v}, then the set is linearly independent\)
Only when \(\mathbf{v} \ne 0 \\
\because \mbox{ if } \mathbf{v} = 0, x_1\mathbf{v} = x_10 = 0\)
Sets of Two Vectors
하나의 vector가 다른 v의 곱으로 표현된다면 linearly Dependent
A set \(\{\mathbf{v_1}, \mathbf{v_2}\}\) is linearly dependent
if at least one of the vectors is a multiple of the other \(\mathbf{v_1} = c\mathbf{v_2} \\ -\mathbf{v_1} + c\mathbf{v_2} = 0\)
The set is linearly independent, if and only if neither of the vectors is a multiple of the other.
\(x_1\mathbf{v_1} + x_2\mathbf{v_2} = 0\) → \(\mathbf{v_1} = -(x_2 / x_1) \mathbf{v_2}\) 는 맞지 않음 : 이 식 자체는 서로의 곱으로 표현되는데, 이는 즉 서로의 multiple 형태인 셈 → 불가능
-
A multiple of the other → linearly dependent \(\mathbf{v_1} = \left[ \begin{matrix}3\\1\\\end{matrix}\right] \mathbf{v_2} = \left[ \begin{matrix}6\\2\\\end{matrix}\right] \\ \mathbf{v_2} = 2\mathbf{v_1} \\2\mathbf{v_1} - \mathbf{v_2} = 0\)
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Neither → linearly independent \(\mathbf{v_1} = \left[ \begin{matrix}3\\3\\\end{matrix}\right] \mathbf{v_2} = \left[ \begin{matrix}6\\2\\\end{matrix}\right] \\ x_1\mathbf{v_1} + x_2\mathbf{v_2} = 0\\ \mathbf{x} = (0,0)\)
📖 Theorem 7. Characterisation of Linearly Dependent Sets
An indexed set \(S = \{\mathbf{v_1}, \cdots, \mathbf{v_p} \}\) of two or more vectors Is linearly dependent
(Non Trivial solution → 최소 하나의 벡터가 나머지 벡터의 linear combination으로 표현되는 경우)
If and only if at least one of the vectors in \(S\) is a linear combination of the others.
In fact, if \(S\) is linearly dependent and \(\mathbf{v_1} \ne 0\), then some \(\mathbf{v_j}\) (with j > 1) is a linear combination of the preceding vectors, \(\{v_1, \cdots, v_j-1\}\)
\(j\) : the largest subscript for which \(c_j \ne 0\)
\(c_1\mathbf{v_1} + \cdots + c_p\mathbf{v_p} = 0\)
\({\mathbf{u}, \mathbf{v}, \mathbf{w}}\) in \(\mathbb{R^3}\)
With \(\mathbf{u} \mbox{ and } \mathbf{v}\) linearly independent (서로가 스칼라곱의 형태로 표현되지 않을 때 )
\(\mathbf{w}\) is in \(/mbox{span} \{\mathbf{u}, \mathbf{v}\}\)
If and only if the set \({\mathbf{u}, \mathbf{v}, \mathbf{w}}\) is linearly dependent.
\(\mathbf{w} = c\mathbf{u} + d\mathbf{v}, -\mathbf{w} + c\mathbf{u} + d\mathbf{v} = 0\\
\mathbf{u} \ne 0 \mathbf{u} \ne c\mathbf{v}\)
📖 Theorem 8.
If a set contains more vectors than there are entries in each vector, then the set is linearly dependent.
When n rows p columns, \(p > n\) must be a free variable!
\(A\mathbf{x} = 0\) has a nontrivial solution.
📖 Theorem 8.
If a set contains the zero vector, then the set is linearly dependent.
\(1\mathbf{v_1} + 0\mathbf{v_2} + \cdots + 0\mathbf{v_p} = 0\)