$ \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\abs}[2][]{\left\lvert#2\right\rvert_{\text{#1}}} \newcommand{\ket}[1]{\left\lvert#1 \right.\rangle} \newcommand{\bra}[1]{\langle\left. #1\right\rvert} \newcommand{\braket}[1]{\langle #1 \rangle} \newcommand{\dd}{\text{d}} \newcommand{\dv}[2]{\frac{\dd #1}{\dd #2}} \newcommand{\pdv}[2]{\frac{\partial}{\partial #1}} $
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Let $V$ be a vector space over $\mathbb{F}$. A subset $S\subseteq V$ is linearly dependent if there are distinct $x_1, \dots, x_k\in S$ for some $k\geq 1$ and scalars $a_1, \dots, a_n\in\mathbb{F}$ where at least one $a_i$ is nonzero, such that $$\begin{align*} a_1x_1+\dots + x_kx_k=0 \end{align*}$$
If $S$ is not linearly dependent, then it is linearly independent. That is, the only solution to $$\begin{align*} a_1x_1+\dots + a_kx_k=0 \end{align*}$$ is $a_1=\dots = a_k=0$.

Concepts

If $S$ is linearly dependent, then we can express $x_i\in S$ as a linear combination of the rest of the elements. As well, then $$\begin{align*} \text{span }\{x_1,\dots, x_k\} = \text{span }\{x_2, \dots, x_k\} \end{align*}$$ Since we can rearrange the order of the set, that means any vector can be removed and the span remains the same.
If $0\in S$, then $S$ is linearly dependent.
If $S$ contains a linearly dependent set, then $S$ is linearly dependent
If $S$ is linearly independent, then so is every subset of $S$.

Used In

Linear independence is used in the definition of a basis.

Hypothesis

Coming soon

Results

Coming soon
Let $V$ be a vector space over $\mathbb{F}$. A subset $S\subseteq V$ is linearly dependent if there are distinct $x_1, \dots, x_k\in S$ for some $k\geq 1$ and scalars $a_1, \dots, a_n\in\mathbb{F}$ where at least one $a_i$ is nonzero, such that $$\begin{align*} a_1x_1+\dots + x_kx_k=0 \end{align*}$$
If $S$ is not linearly dependent, then it is linearly independent. That is, the only solution to $$\begin{align*} a_1x_1+\dots + a_kx_k=0 \end{align*}$$ is $a_1=\dots = a_k=0$.

Concepts

If $S$ is linearly dependent, then we can express $x_i\in S$ as a linear combination of the rest of the elements. As well, then $$\begin{align*} \text{span }\{x_1,\dots, x_k\} = \text{span }\{x_2, \dots, x_k\} \end{align*}$$ Since we can rearrange the order of the set, that means any vector can be removed and the span remains the same.
If $0\in S$, then $S$ is linearly dependent.
If $S$ contains a linearly dependent set, then $S$ is linearly dependent
If $S$ is linearly independent, then so is every subset of $S$.

Used In

Linear independence is used in the definition of a basis.

Hypothesis

Coming soon

Results

Coming soon
FullPage
definition
concepts
used in
hypothesis
results
FullPage
definition
concepts
used in
hypothesis
results
Let $V$ be a vector space over $\mathbb{F}$. A subset $S\subseteq V$ is linearly dependent if there are distinct $x_1, \dots, x_k\in S$ for some $k\geq 1$ and scalars $a_1, \dots, a_n\in\mathbb{F}$ where at least one $a_i$ is nonzero, such that $$\begin{align*} a_1x_1+\dots + x_kx_k=0 \end{align*}$$
If $S$ is not linearly dependent, then it is linearly independent. That is, the only solution to $$\begin{align*} a_1x_1+\dots + a_kx_k=0 \end{align*}$$ is $a_1=\dots = a_k=0$.

Concepts

If $S$ is linearly dependent, then we can express $x_i\in S$ as a linear combination of the rest of the elements. As well, then $$\begin{align*} \text{span }\{x_1,\dots, x_k\} = \text{span }\{x_2, \dots, x_k\} \end{align*}$$ Since we can rearrange the order of the set, that means any vector can be removed and the span remains the same.
If $0\in S$, then $S$ is linearly dependent.
If $S$ contains a linearly dependent set, then $S$ is linearly dependent
If $S$ is linearly independent, then so is every subset of $S$.

Used In

Linear independence is used in the definition of a basis.

Hypothesis

Coming soon

Results

Coming soon
Let $V$ be a vector space over $\mathbb{F}$. A subset $S\subseteq V$ is linearly dependent if there are distinct $x_1, \dots, x_k\in S$ for some $k\geq 1$ and scalars $a_1, \dots, a_n\in\mathbb{F}$ where at least one $a_i$ is nonzero, such that $$\begin{align*} a_1x_1+\dots + x_kx_k=0 \end{align*}$$
If $S$ is not linearly dependent, then it is linearly independent. That is, the only solution to $$\begin{align*} a_1x_1+\dots + a_kx_k=0 \end{align*}$$ is $a_1=\dots = a_k=0$.

Concepts

If $S$ is linearly dependent, then we can express $x_i\in S$ as a linear combination of the rest of the elements. As well, then $$\begin{align*} \text{span }\{x_1,\dots, x_k\} = \text{span }\{x_2, \dots, x_k\} \end{align*}$$ Since we can rearrange the order of the set, that means any vector can be removed and the span remains the same.
If $0\in S$, then $S$ is linearly dependent.
If $S$ contains a linearly dependent set, then $S$ is linearly dependent
If $S$ is linearly independent, then so is every subset of $S$.

Used In

Linear independence is used in the definition of a basis.

Hypothesis

Coming soon

Results

Coming soon
FullPage
definition
concepts
used in
hypothesis
results