> [!NOTE] Theorem (Dot Product is Inner Product) > Let $n\geq 1.$ Let $\mathbb{R}^{n}$ denote the [[Real n-Space|real n-space]]. For all $v,w\in \mathbb{R}^{n},$ let $v\cdot w$ denote their [[Dot Product in Real n-Space|dot product]]. Then the following hold: > >(1) [[Commutativity|Commutativity]] of dot product: for all $\underline{v},\underline{w}\in \mathbb{R}^{n},$ $\underline{v} \cdot \underline{w}=\underline{w} \cdot \underline{v}.$ > >(2) [[Bilinearity|Bilinearity]] of dot product: for all $\lambda_{1},\lambda_{2}\in \mathbb{R},$ $\underline{v}_{1},\underline{v}_{2},\underline{w}\in \mathbb{R}^{n},$ $(\lambda_{1} \underline{v}_{1}+\lambda_{2}\underline{v}_{2})\cdot \underline{w}=\lambda_{1}(\underline{v}_{1}\cdot \underline{w})+\lambda_{2}(\underline{v}_{2}\cdot \underline{w})$ > >(3) Non-negative definite: for any $\underline{v}\in \mathbb{R}^{n}$, $\underline{v} \cdot \underline{v} \geq 0$, and furthermore $\underline{v} \cdot \underline{v}=0\iff \underline{v} = \underline{0}.$ **Proof**: Let $\underline{v}= (a_{1},\dots,a_{n})$ and $\underline{w}=(b_{1},\dots,b_{n}) \in \mathbb{R}^{n}.$ It follows from [[Associativity of Multiplication of Real Numbers]] that $(1)$ is true since $a_{i}c_{i}=c_{i}a_{i}$ for each $i=1,\dots,n.$ Let $\underline{v}_{1}=(a_{1},\dots,a_{n}),\;\underline{v}_{2}=(b_{1},\dots,b_{n})$ with $\underline{w}$ as before. The $i$-th component of $(\lambda_{1}\underline{v}_{1}+\lambda_{2}v_{2})\cdot \underline{w}$ is given by $(\lambda_{1}a_{i}+\lambda_{2}b_{i})c_{i}=\lambda_{1}a_{i}c_{i}+\lambda_{2}b_{i}c_{i}$which equals the $i$th component of $\lambda_{1}(\underline{v}_{1}\cdot \underline{w})+\lambda_{2}(\underline{v}_{2}\cdot \underline{w}).$ Thus $(2)$ is true. We have $\underline{v}\cdot \underline{v}=a_{1}^{2}+a_{2}^{2}+\dots+a_{n}^{2}$ is a sum of squares, so cannot be negative. Furthermore, the sum can only be zero if each $a_{i}^{2}=0$ and that only happens if $\underline{v}=\underline{0}$. Thus $(3)$ is true.