Linearly Independence
A set of vectors $\begin{Bmatrix}\mathbf{v}_1 & \mathbf{v}_2&\cdots& \mathbf{v}_p\end{Bmatrix}$ in $\mathbb{R}^n$ is sasid to be linearly independent
if the vector equation $x_1\mathbf{v}_1+ x_2\mathbf{v}_2+\cdots + x_p\mathbf{v}_p=\mathbf{0} $ has only the trivial solution
→ coefficient가 모두 0인 솔루션만 존재
Linearly Dependence
A set of vectors $\begin{Bmatrix}\mathbf{v}_1 & \mathbf{v}_2&\cdots& \mathbf{v}_p\end{Bmatrix}$ in $\mathbb{R}^n$ is sasid to be linearly dependent
if there exist weights $c_1, c_2, \cdots , c_p$, not al zero, such that $c_1\mathbf{v}_1+ c_2\mathbf{v}_2+\cdots + c_1\mathbf{v}_p=\mathbf{0}$
→ $c_1, c_2, \cdots, c_p$의 solution set 중 하나라도 nonzero가 있으면(=nontrivial solution) linearly independent
그러면 $set\begin{Bmatrix}\mathbf{v}_1 & \mathbf{v}_2 & \cdots & \mathbf{v}_p \end{Bmatrix}$ in $\mathbb{R}^n$이 inearly independent 할 때, $\mathbf{v}_j=a_1\mathbf{v}_1+ \cdots + a_p\mathbf{v}_p$가 항상 성립할까?
→ No($\mathbf{v}_j$의 coefficeint가 0일 수 있기 때문)
Ex. Determine whether the set of vectors are linearly dependent
$$\mathbf{v}_1=\begin{bmatrix}1\\2\\3\end{bmatrix},\;\mathbf{v}_2=\begin{bmatrix}4\\5\\6
\end{bmatrix}, \; \mathbf{v}_3=\begin{bmatrix}2\\1\\0\end{bmatrix}$$
$$\begin{bmatrix}1&4&2&0
\\ 2&5&1&0
\\ 3&6&0&0
\end{bmatrix} \sim
\begin{bmatrix}1&0&-2&0
\\ 0&1&1&0
\\ 0&0&0&0
\end{bmatrix} $$
$\left\{\begin{matrix} x_1=2x_3
\\ x_2=-x_3
\\ x_3\;is\;free
\end{matrix}\right.$ → $\mathbf{X}=\mathbf{x}_3\begin{bmatrix}2\\-1\\1\end{bmatrix}$
→ linearly dependent
Linear Independence of Matrix Columns
$$A=\begin{bmatrix}\mathbf{a}_1&\mathbf{a}_2&\cdots&\mathbf{a}_n\end{bmatrix}$$
$$A\mathbf{X}=\mathbf{0},\;x_1\mathbf{a}_1+x_1\mathbf{a}_1+\cdots+x_n\mathbf{a}_n=0$$
The columns of a matrix $A$ are linearly independent
if and only if the equations $A\mathbf{X}=\mathbf{0}$ has only the trivial solution
Ex. Determine if the columns of the following matrix are linearly independent.
$$A=\begin{bmatrix}0&1&4
\\ 1&2&-1
\\ 5&8&0
\end{bmatrix}$$
$$\sim \begin{bmatrix} 1&2&-1&0
\\ 0&1&4&0
\\ 0&0&13&0
\end{bmatrix} \rightarrow \mathbf{X}=\begin{bmatrix}0\\0\\0\end{bmatrix}$$
→ only trivial solution
→ linearly independent
Sets of one vector
If a set contains only one vector, $\mathbf{v}$,
then the set is linearly independent only when $\mathbf{v}\neq\mathbf{0}$
pf)
if $\mathbf{v}=\mathbf{0}$
$x_1\mathbf{v}=\mathbf{0}$
→ $x_1$이 어떤 값이 오더라도 $\mathbf{0}$ (nontrivial solution)
Sets of Two vectors
A $set\begin{Bmatrix}\mathbf{v}_1&\mathbf{v}_2\end{Bmatrix}$ is linearly dependent
if at least of the vectors is a multiple of the other.
$$\mathbf{v}_1=c\mathbf{v}_2\;\rightarrow\;\mathbf{v}_1-c\mathbf{v}_2=0$$
$c=0$이더라도 $\mathbf{v}_1$의 coefficient의 값은 0이 아니다.
$\rightarrow$ nontrivial solution
$\rightarrow$ linearly dependent
The set is linearly independent, if and only if neither of the vectors is a multiple of the other.
pf)
$$x_1\mathbf{v}_1+x_2\mathbf{v}_2=0$$
$$x_1\mathbf{v}_1=-x_2\mathbf{v}_2$$
$$\mathbf{v}_1=-x_1/x_2\mathbf{v}_2$$
$\rightarrow$ 그러나, 가정을 두 벡터가 scalar곱의 관계가 없다에서 시작했기 때문에 위 식이 성립하지 않는다.
$\rightarrow$ linearly independent
Ex.
$$\mathbf{v}_1=\begin{bmatrix}3\\1\end{bmatrix},\;\mathbf{v}_2=\begin{bmatrix}6\\2\end{bmatrix}$$
$$\mathbf{v}_2=2\mathbf{v}_1$$
$\rightarrow$ linearly dependent
$$\mathbf{v}_1=\begin{bmatrix}3\\3\end{bmatrix}, \; \mathbf{v}_2=\begin{bmatrix}6\\2\end{bmatrix}$$
두 벡터간 관계가 존재하지 않음
$$x_1\mathbf{v}_1+ x_2\mathbf{v}_2=\mathbf{0}$$
$$\rightarrow\;\mathbf{X}=\begin{bmatrix}0\\0\end{bmatrix}$$
$\rightarrow$ trivial solution(only)
$\rightarrow$ linearly independent
Theorem 7. Charaterization of Linearly Dependent Sets
An indexed set $S=\begin{Bmatrix}\mathbf{v}_1& \mathbf{v}_2&\cdots& \mathbf{v}_p \end{Bmatrix}$ of two or two more vectors is inearly dependent
if and only if at least one of the vectors in $S$ is a linear combindation of the others.
Ex.
$$\mathbf{v}_1=a_2\mathbf{v}_2+ a_3\mathbf{v}_3 +\cdots+ a_p\mathbf{v}_p$$
$= -\mathbf{v}_1+ a_2\mathbf{v}_2+ a_3\mathbf{v}_3 +\cdots+ a_p\mathbf{v}_p =\mathbf{0}$
$\rightarrow$ nontrivial solution
$\rightarrow$ linearly dependent
In fact, if $S$ is linearly dependent and $\mathbf{v}_1\neq\mathbf{0}$,
then sure $\mathbf{v}_j(j>1)$ is a linear combination of the preceding vectors, $\mathbf{v}_1,\; \mathbf{v}_2,\; \mathbf{v}_{j-1} $
pf)
$$c_1\mathbf{v}_1+ c_2\mathbf{v}_2+\cdots+ c_p\mathbf{v}_p=\mathbf{0}$$
Let $j$: the longest subscript for which $c_j\neq0$
식 정리 $\rightarrow$ $c_1\mathbf{v}_1+\cdots+ c_j\mathbf{v}_j +\mathbf{0}\mathbf{v}_{j+}+\cdots+\mathbf{0}\mathbf{v}_p=\mathbf{0} $
1) $j=1$ $\rightarrow$ $c_1\mathbf{v}_1=\mathbf{0}$ $\rightarrow$ $c_1=0$ 성립 안됨(모순)
2) $j>1$ $\rightarrow$ $\mathbf{v}_j=(-c_1/c_j)\mathbf{v}_1+\cdots+(-c_{j-1}/c_j)\mathbf{v}_{j-1}$
$\begin{Bmatrix}\mathbf{u}& \mathbf{v}& \mathbf{w} \end{Bmatrix}$ in $\mathbb{R}^3$ with $\mathbf{u}$ and $\mathbf{v}$ are linearly independent
$\mathbf{w}$ is in $Span\begin{Bmatrix}\mathbf{u} & \mathbf{v} \end{Bmatrix}$
if and only if the $set\begin{Bmatrix}\mathbf{u}& \mathbf{v}& \mathbf{w}\end{Bmatrix}$ is linearly dependent.
Theorem 8.
If a set contains more vectors than there are entries in each vector,
then the set is linaerly independent.
Ex.
$$ \begin{bmatrix}2\\1\end{bmatrix},\begin{bmatrix}4\\-1\end{bmatrix},\begin{bmatrix}2\\-2\end{bmatrix}$$
$$\begin{bmatrix}2&4&2&0
\\1&-1&-2&0
\end{bmatrix} \rightarrow n<p$$
$\rightarrow$ linaerly dependent
Theorem 9.
If a set contains the zero vector, then the set is linearly dependent.
$$\mathbf{1}\mathbf{v}_1+ \mathbf{0}\mathbf{v}_2+\cdots+ \mathbf{0}\mathbf{v}_p=0 $$
$\rightarrow$ nontrivial solution
Ex.
$$\begin{bmatrix}1\\7\\0\end{bmatrix},\begin{bmatrix}2\\0\\9\end{bmatrix},
\begin{bmatrix}3\\1\\5\end{bmatrix}, \begin{bmatrix}4\\1\\8\end{bmatrix} \rightarrow n<p$$
$\rightarrow$ linearly dependent
$$\begin{bmatrix}2\\3\\5\end{bmatrix},\begin{bmatrix}0\\0\\0\end{bmatrix},
\begin{bmatrix}2\\1\\5\end{bmatrix}$$
$\rightarrow$ linearly dependent
$$\begin{bmatrix}-2\\4\\6\\10\end{bmatrix},\begin{bmatrix}-1\\2\\3\\1\end{bmatrix}$$
$$\mathbf{u} \neq c\mathbf{v}$$
$\rightarrow$ trivial solution
$\rightarrow$ linearly independent
※ Reference
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