Suggest edit — dot product

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title: "dot product"
source: https://www.jemoka.com/posts/kbhdot_product/
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The dot product is a property of real vector spaces which is a simplified version of an inner product; specifically, it obviates the need to complex-conjugate anything because, well, \(\bar{n} = n, n \in \mathbb{R}\). The dot-product also yield a real number.
constituents \(x, y \in \mathbb{R}^{n}\) (NOTE the realness) where, \(x = (x_1, \dots, x_{n})\) and \(y = (y_1, &hellip;, y_{n})\) requirements As we are familiar with, element-wise product and sum
\begin{equation} x\cdot y = x_1y_1 + \dots + x_{n}y_{n} \end{equation}
additional information properties of the dot product For fixed \(y \in \mathbb{R}^{n}\), the dot product map that sends \(x\) to \(x \cdot y\) is linear (inheriting add. and homo. from algebra) \(x \cdot x = 0\) IFF \(x =0\) (no negs allowed (above), so every slot has to have a zero to multiply to 0) \(x \cdot x &gt; 0\) for all \(x \in \mathbb{R}^{n}\) (neg times neg is pos) \(x \cdot y = y \cdot x\) for reals; by inheriting from each element&rsquo;s field orthogonality test The dot product is an orthogonality test. If the dot product between the two vectors is \(0\), they are definitely orthogonal.
geometric interpretation of the dot product Well, we have some shape between two vectors; then, we can first write out the law of cosines. Then, we can see that, for two vectors from the same origin, we can say that the projection of vector \(\vec{A}\) onto \(\vec{B}\) is written as:
\begin{equation} |\vec{A}||\vec{B}|\cos \theta \end{equation}
where, \(\theta\) is the angle between the two vectors.