# Dot Product and Span Summary

Dot Product $\mathbf{u}\cdot\mathbf{v}=\|u\|\|v\|\cos\theta$ $\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{\|u\|\|v\|}$

Span $\text{span}\{\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}\}=\{c_1\mathbf{u_1}+c_2\mathbf{u_2}+\dots+c_k\mathbf{u_k}\mid c_1,c_2,\dots,c_k\in\mathbb{R}\}=\text{set of all linear combinations of }$ $\{\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}\}.$

Subspaces $V\subseteq\mathbb{R}^n$ is a subspace of $\mathbb{R}^n$ if
1) $V=\text{span}\{\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}\}$ for some vectors $\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}$.
2) $V$ satisfies the closure properties:

(i) for all $\mathbf{u},\mathbf{v}\in V$, we must have $\mathbf{u}+\mathbf{v}\in V$.

(ii) for all $\mathbf{u}\in V$ and $c\in\mathbb{R}$, we must have $c\mathbf{u}\in V$.

3) $V$ is the solution set of a homogeneous system.

(Sufficient to check either one of Condition 1, 2, 3.)

Remark:
For $V$ to be a subspace, zero vector $\mathbf{0}$ must be in $V$. (Since for $\mathbf{u}\in V$, $0\in\mathbb{R}$, we have $0\mathbf{u}\in V$.)

Linear Independence and Dependence $\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}$ are linearly independent if the system $\displaystyle c_1\mathbf{u_1}+c_2\mathbf{u_2}+\dots+c_k\mathbf{u_k}=0$ has only the trivial solution, i.e. $c_1=c_2=\dots=c_k=0$.

If the system has non-trivial solutions, i.e. at least one $c_i$ not zero, then $\mathbf{u_1},\mathbf{u_2},\dots,\mathbf{u_k}$ are linearly dependent. ## Author: mathtuition88

https://mathtuition88.com/

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