# Interpretation of a proof on inner product space.

Let $$\mathcal{B} = \{v_1,…,v_n\}$$ be a basis for the finite dimensional vector space $$V$$ ($$\dim V = n < +\infty$$) such that $$\langle v_i,v_j\rangle =0$$ if $$i \neq j$$. If $$\mathcal{B}$$ is an orthogonal basis and $$x \in V$$ then

\begin{align*} x = \sum_{i=1}^n \frac{\langle x,v_i\rangle}{\|v_i\|^2} v_i \end{align*}

Proof

Since $$\displaystyle x = \sum_{i=1}^n c_i v_i$$ for some $$c_1,\ldots,c_n \in\textbf{F}$$, one has that

$$\langle x,v_j\rangle = \Bigg\langle\sum_{i=1}^{n} c_i v_i, v_j\Bigg\rangle =\sum_{i=1}^{n} c_i \langle v_i, v_j\rangle = c_j \|v_j\|^2$$

Thus we have that $$\langle x,v_j\rangle = c_j\|v_j\|^2$$, so $$\displaystyle c_j = \frac{\langle x,v_j\rangle}{\|v_j\|^2}$$

My question is: what is the interpretation of $$\langle x,v_j\rangle$$ on an inner product space?

• Did you learn dot products in physics or your calculus class? – Ted Shifrin Mar 8 at 2:09
• @TedShifrin yes, but I am not sure the geometric interpretation. – shine Mar 8 at 2:12
• The 3blue1brown series on Youtube has an excellent exposition on how to interpret the dot product geometrically in real Euclidean space. – Elliot Herrington Mar 8 at 3:17

You can think of $$\langle x,y\rangle/\|y\|$$ as the signed length of the projection of the vector $$x$$ on the direction of $$y/\|y\|$$ (provided that $$y\neq 0$$): a plus sign corresponds to the same direction and a minus sign corresponds to opposite directions.
Such interpretation comes from $$\textbf{R}^{3}$$, where the standard inner product satisfies $$\langle x,y\rangle = \|x\|\|y\|\cos(\theta)$$. Indeed, given $$x = (x_{1},x_{2},x_{3})$$, $$y = (y_{1},y_{2},y_{3})$$ and $$z = x - y = (x_{1} - y_{1}, x_{2} - y_{2}, x_{3} - y_{3})$$, the application of the cosine's law $$\|z\|^{2} = \|x\|^{2} + \|y\|^{2} - 2\|x\|\|y\|\cos(\theta)$$ results into \begin{align*} & (x_{1}-y_{1})^{2} + (x_{2} - y_{2})^{2} + (x_{3}-y_{3})^{2} = x^{2}_{1} + x^{2}_{2} + x^{2}_{3} + y^{2}_{1} + y^{2}_{2} + y^{2}_{3} - 2\|x\|\|y\|\cos(\theta) \Longleftrightarrow\\\\ & -2(x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}) = -2\|x\|\|y\|\cos(\theta) \Longleftrightarrow x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3} = \|x\|\|y\|\cos(\theta) \end{align*}
where $$\langle x,y\rangle = x_{1}y_{1} + x_{2}y_{2} + x_{3}y_{3}$$ stands for the standard inner product on $$\textbf{R}^{3}$$. In fact, this is how we define angles between vectors within euclidean spaces. More precisely, given $$x\neq 0$$ and $$y\neq 0$$, one has \begin{align*} \cos(\theta) = \frac{\langle x,y\rangle}{\|x\|\|y\|} \end{align*} which is between $$-1$$ and $$1$$ due to the Cauchy-Schwarz inequality $$|\langle x,y\rangle|\leq\|x\|\|y\|$$. In order to prove it, you can approach it by considering the expression $$\|x + \lambda y\|^{2}\geq 0$$, which is quadratic in the variable $$\lambda$$.
• Nitpicking: $\|x+\lambda y\|^2\ge 0$ is quadratic in $\lambda.$............+1 – DanielWainfleet Mar 8 at 4:29
In an inner-product space over $$\Bbb R,$$ if $$x\ne 0\ne v$$ and if $$x$$ is not a scalar multiple of $$v$$ then $$=\|x\|\cdot \|v\|\cdot \cos A$$ where $$A$$ is the angle at $$0$$ in the $$\triangle$$ with vertices $$0,x,v.$$ The inner-product vector-subspace generated by a linearly independent pair $$\{x,v\}$$ is isomorphic to $$\Bbb R^2.$$ If also $$\|v\|=1$$ then $$||=\|y\|$$ where $$y$$ is the foot of the perpendicular from $$x$$ to the line through $$0,v.$$
BTW. I found that helpful to remember the Linear Correlation Co-efficient formula from Statistics: Given $$x=$$ and $$y=,$$ first we "normalize" $$x$$ and $$y$$ by taking $$x'=$$ and $$y'=$$ where $$m_x=n^{-1}\sum_{j=1}^nx_j$$ and $$m_y=n^{-1}\sum_{j=1}^ny_j$$ are the means (averages). The correlation co-efficient is $$\cos \angle x'0y',$$ which, if $$x'\ne 0\ne y',$$ is $$\frac {}{\|x'\|\cdot \|y'\|}.$$