$$|u{\cdot}v| \leq ||u||||v|| \tag{1}$$ I tried to derive the above.

Expand the LHS: $$|u{\cdot}v| = \left|\sum_{i=1}^n a_i + b_i\right| \tag{2}$$

Which can be rewritten as: $$|u{\cdot}v| = \left|\sum_{i=1}^n a_i + \sum_{i=1}^n b_i\right| \tag{3}$$

The RHS of $(1)$ can be written as: $$||u||||v|| = \sqrt{\sum_{i=1}^n a_i^2} \times \sqrt{\sum_{i=1}^n b_i^2} \tag{4}$$

Taking the square of $(4)$: $$||u||^2||v||^2 = \sum_{i=1}^n a_i^2 \times \sum_{i=1}^n b_i^2$$

From $(3)$: $$u{\cdot}v = \sum_{i=1}^n a_i + \sum_{i=1}^n b_i \tag{3.1}$$

Square $(3.1)$: $$(u{\cdot}v)^2 = \left(\sum_{i=1}^n {a_i}\right)^2 + 2\sum_{i=1}^n {a_i}{\cdot}\sum_{i=1}^n b_i + \left(\sum_{i=1}^n b_i\right)^2 \tag{5}$$

Now, $X^2 \lt Y^2 \implies |Y| \gt |X| \tag{*}$

Rewriting $(4)$: $$||u||^2||v||^2 = \sum_{i=1}^n a_i{\cdot}a_i \times \sum_{i=1}^n b_i{\cdot}b_i \tag{4.1}$$

Rewriting $(5)$: $$(u{\cdot}v)^2 = \sum_{i=1}^n {a_i}\sum_{i=1}^n {a_i} + 2\sum_{i=1}^n {a_i}{\cdot}\sum_{i=1}^n b_i + \sum_{i=1}^n b_i\sum_{i=1}^n b_i \tag{5.1}$$

It seems trivial that $(5.1) \le (4.1)$ which from $(*) \implies |u{\cdot}v| \geq ||u||||v||$ directly contradicting the theorem.

What am I doing wrong? Where did I misstep (and please properly derive the proof).
$(*)$ this is trivial and left as a proof for the reader. ;)


Equation 2 is wrong.

It should be

$$u\cdot v =\sum u_i v_i $$

  • $\begingroup$ The question was what am I doing wrong. I answered that. $\endgroup$ – marty cohen Jul 5 '17 at 13:45
  • $\begingroup$ very true, comment deleted (and answer upvoted). $\endgroup$ – Frits Veerman Jul 5 '17 at 14:12

In (2) it should be $$\mathbf{u}\cdot\mathbf{v}=a_1b_1+a_2b_2+...+a_nb_n$$ and you can finish your proof. I suggest this way: let $\lambda=\|\mathbf{v}\|^{-2}\mathbf{u}\cdot\mathbf{v}$ \begin{align} 0&\le\|\mathbf{u}-\lambda\mathbf{v}\|^2=(\mathbf{u}-\lambda\mathbf{v})\cdot(\mathbf{u}-\lambda\mathbf{v})\\ &=\|\mathbf{u}\|^2-2\lambda \mathbf{u}\cdot\mathbf{v}+\lambda^2\|\mathbf{v}\|^2\\ &=\|\mathbf{u}\|^2-\|\mathbf{v}\|^{-2}|\mathbf{u}\cdot\mathbf{v}|^2 \Rightarrow|\mathbf{u}\cdot\mathbf{v}|\ge\|\mathbf{u}\|\|\mathbf{v}\|. \end{align}

  • 1
    $\begingroup$ Proofs by Magic are not helpful. $\endgroup$ – marty cohen Jul 5 '17 at 13:47

$$u\cdot v = \sum_{i=1}^{n}a_ib_i \neq \sum_{i=1}^{n}a_i+b_i$$

Properly derive the proof:



$$x_i = \frac{|a_i|}{(\sum_{j=1}^{n}a_j^2)^{1/2}}, y_i = \frac{|b_i|}{(\sum_{j=1}^{n}b_j^2)^{1/2}}$$


$$\sum_{i=1}^{n}x_i^2 = \sum_{i=1}^{n}\frac{|a_i|^2}{(\sum_{j=1}^{n}a_j^2)} = 1$$

Similarly, $\sum_{i=1}^ny_i^2 = 1 \implies \sum_{i=1}^{n}x_i^2+\sum_{i=1}^{n}y_i^2=2$ and,

$$2 \geq \sum_{i=1}^{n}2x_iy_i \implies \sum_{i=1}^{n}\frac{|a_i||b_i|}{(\sum_{j=1}^{n}a_j^2)^{1/2}(\sum_{j=1}^{n}b_j^2)^{1/2}} \leq 1\implies |a.b| \leq \|a\|\|b\|$$


Just thought I would rederive a well-known identity which implies the inequality and the conditions for equality for real vectors.

Want to show $(\sum_i a_ib_i)^2 \le \sum_i a_i^2 \sum_i b_i^2 $.

$\begin{array}\\ S &=(\sum_i a_ib_i)^2\\ &=\sum_i a_ib_i\sum_j a_jb_j\\ &=\sum_i\sum_j a_ib_i a_jb_j\\ \end{array} $

$\begin{array}\\ T &=\sum_i a_i^2 \sum_i b_i^2\\ &=\sum_i \sum_ja_i^2 b_j^2\\ &=\sum_i \sum_ja_j^2 b_i^2 \qquad\text{exchanging }i \text{ and }j \text{ and order of summation}\\ \end{array} $

$\begin{array}\\ 2(T-S) &=\sum_i \sum_j(a_i^2 b_j^2+a_j^2 b_i^2 -2a_ib_i a_jb_j)\\ &=\sum_i \sum_j(a_i b_j-a_j b_i)^2\\ &> 0 \qquad\text{unless }a_i b_j=a_j b_i \text{ for all }i, j\\ \end{array} $


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