Scalar product of two vectors expressed via orthonormal base

Question

Let $\left\{ v_{1}, \dots, v_{n} \right\}$ form an orthonormal base and let elements $$x = \sum_{i=1}^{n} (v_{i}^\intercal x) v_{i}$$ $$y = \sum_{i=1}^{n} (v_{i}^\intercal y) v_{i}$$

Show that the scalar product $$x^\intercal y = \sum_{i=1}^{n} (v_{i}^\intercal x) (v_{i}^\intercal y)$$

Now, $x^\intercal y = \left( \sum_{i=1}^{n} v_{i}^\intercal x v_{i} \right)^\intercal \left( \sum_{i=1}^{n} v_{i}^\intercal y v_{i} \right) = \sum_{i=1}^{n} v_{i}^\intercal x^\intercal v_{i} v_{i}^\intercal y v_{i} = ...?$ I expect $v_i^\intercal v_i = 1$ to give the desired result but I don't see how to get there.

Answer 1

Many thanks to @paulinho who did the n=3 case. Below generalization to $n$ orthonormal vectors.

$$x^\intercal y = \left( \sum_{i=1}^{n} (v_{i}^\intercal x) v_{i} \right)^\intercal \left( \sum_{i=1}^{n} (v_{i}^\intercal y) v_{i} \right) = \left( \sum_{i=1}^{n} v_{i}^\intercal ( x^\intercal v_{i} ) \right) \left( \sum_{i=1}^{n} (v_{i}^\intercal y) v_{i}\right)$$ $v_{i}^\intercal ( x^\intercal v_{i} ) = ( x^\intercal v_{i} ) v_{i}^\intercal $ because dot product is a scalar $$= \left( \sum_{i=1}^{n} ( x^\intercal v_{i} ) v_{i}^\intercal \right) \left( \sum_{i=1}^{n} v_{i} (v_{i}^\intercal y) \right)$$ decompose the second sum $$= \sum_{i=1}^{n} \left(( x^\intercal v_{i} ) v_{i}^\intercal \left( v_{i}( v_{i}^\intercal y ) + \sum_{\substack{j=1 \\ j \not = i}}^{n} v_{j}( v_{j}^\intercal y ) \right) \right)$$ $$=\left( \sum_{i=1}^{n} ( x^\intercal v_{i} ) v_{i}^\intercal v_{i}( v_{i}^\intercal y ) \right) + \left( \sum_{i=1}^{n} \sum_{\substack{j=1 \\ j \not = i}}^{n} ( x^\intercal v_{i} ) v_{i}^\intercal v_{j}( v_{j}^\intercal y ) \right)$$ by orthonormality $v_{i}^\intercal v_{i} = 1$ and $v_{i}^\intercal v_{j}=0, i \not = j$ hence $$=\sum_{i=1}^{n} ( x^\intercal v_{i} ) ( v_{i}^\intercal y )$$

Answer 2

Let's look at three dimensions, where we have $$\vec{x} = (\vec{v}^T_1 \vec{x}) \vec{v}_1 + (\vec{v}^T_2 \vec{x}) \vec{v}_2 + (\vec{v}^T_3 \vec{x}) \vec{v}_3, ~~ \vec{y} = (\vec{v}^T_1 \vec{y}) \vec{v}_1 + (\vec{v}^T_2 \vec{y}) \vec{v}_2 + (\vec{v}^T_3 \vec{y}) \vec{v}_3$$ We when we take the inner product of the two, we get \begin{align*} \vec{x}^T \vec{y} &= \left[(\vec{v}^T_1 \vec{x}) \vec{v}_1^T + (\vec{v}^T_2 \vec{x}) \vec{v}_2^T + (\vec{v}^T_3 \vec{x}) \vec{v}_3^T \right] \cdot \left[(\vec{v}^T_1 \vec{y}) \vec{v}_1 + (\vec{v}^T_2 \vec{y}) \vec{v}_2 + (\vec{v}^T_3 \vec{y}) \vec{v}_3\right] \\ \\ &= \color{blue} {(\vec{v}^T_1 \vec{x}) (\vec{v}^T_1 \vec{y})(\vec{v}_1^T \vec{v}_1)} + \color{red} {(\vec{v}^T_1 \vec{x}) (\vec{v}^T_2 \vec{y})(\vec{v}_1^T \vec{v}_2)} + \color{red} {(\vec{v}^T_1 \vec{x}) (\vec{v}^T_3 \vec{y})(\vec{v}_1^T \vec{v}_3)} \\ & ~~~~~~~+ \color{red} {(\vec{v}^T_2 \vec{x}) (\vec{v}^T_1 \vec{y})(\vec{v}_2^T \vec{v}_1)} + \color{blue} {(\vec{v}^T_2 \vec{x}) (\vec{v}^T_2 \vec{y})(\vec{v}_2^T \vec{v}_2)} + \color{red} {(\vec{v}^T_2 \vec{x}) (\vec{v}^T_3 \vec{y})(\vec{v}_2^T \vec{v}_3)} \\ & ~~~~~~~+ \color{red} {(\vec{v}^T_3 \vec{x}) (\vec{v}^T_1 \vec{y})(\vec{v}_3^T \vec{v}_1)} + \color{red} {(\vec{v}^T_3 \vec{x}) (\vec{v}^T_2 \vec{y})(\vec{v}_3^T \vec{v}_2)} + \color{blue} {(\vec{v}^T_3 \vec{x}) (\vec{v}^T_3 \vec{y})(\vec{v}_3^T \vec{v}_3)} \end{align*} Notice that all the red terms then drop out of the equation, because they include an inner product of two vectors of the orthonormal basis. The blue terms stay, and we can further get rid of the $\vec{v}_i^T \vec{v}_i$ term as it is one (because the vectors are unit). So indeed we get that $$\vec{x}^T \vec{y} = (\vec{v}^T_1 \vec{x}) (\vec{v}^T_1 \vec{y}) + (\vec{v}^T_2 \vec{x}) (\vec{v}^T_2 \vec{y}) + (\vec{v}^T_3 \vec{x}) (\vec{v}^T_3 \vec{y}) = \sum_{i = 1}^3 (\vec{v}_i^T \vec{x}) (\vec{v}_i^T \vec{y})$$ Can you generalize this to when the basis consists of $n$ orthonormal vectors? You just need to keep in mind all terms with an inner product of two different basis vectors drop out.