Calculation of the squared Euclidean norm

7.2k Views Asked by At

Just a simple question. I'm reading an article and come across an equation that I cannot replicate as done in the original.

\begin{align} &\|\mathbf{x} - \mathbf{\alpha} \|^2 - \|\mathbf{x} - \mathbf{\beta}\|^2\\ &= \|\mathbf{x}\| \|\mathbf{x}\| - 2 \|\mathbf{\mathbf{\alpha}}\| \|\mathbf{x}\| + \|\mathbf{\alpha}\| \|\mathbf{\mathbf{\alpha}}\| - \|\mathbf{x}\| \|\mathbf{x}\| +2 \|\mathbf{\beta}\| \|\mathbf{x}\| - \|\mathbf{\beta}\| \|\mathbf{\beta}\|\\ &= \mathbf{\alpha}^T \mathbf{\alpha} - \beta^T \beta + 2(\sqrt{\mathbf{\beta}\cdot\mathbf{\beta}}-\sqrt{\mathbf{\alpha}\cdot\mathbf{\alpha}})\|\mathbf{x}\|\\ &= \mathbf{\alpha}^T \mathbf{\alpha} - \beta^T \beta + 2(\sqrt{\mathbf{\beta}\cdot\mathbf{\beta}}-\sqrt{\mathbf{\alpha}\cdot\mathbf{\alpha}})(\sqrt{\mathbf{x}\cdot\mathbf{x}}), \end{align} where I used the fact that $$\|a\|\|a\| = \sqrt{a\cdot a}\sqrt{a\cdot a} = \sqrt{a^T a}\sqrt{a^T a} = a^Ta.$$

However, the article gives $$2(\beta-\alpha)^T \mathbf{x} + \alpha^T \alpha - \beta^T\beta $$

1

There are 1 best solutions below

4
On BEST ANSWER

Your transition from the first line to the second is incorrect. We should have $$ \|x - \alpha\|^2 - \|x - \beta\|^2 = \\ (x - \alpha)^T(x - \alpha) - (x - \beta)^T(x - \beta) = \\ \|x\|^2 + \|\alpha\|^2 - \|x\|^2 - \|\beta\|^2 - x^T\alpha - \alpha^Tx + x^T\beta + \beta^Tx =\\ \|\alpha\|^2 - \|\beta\|^2 - 2\alpha^Tx + 2 \beta^Tx =\\ \alpha^T\alpha - \beta^T\beta + 2(\beta - \alpha)^Tx $$ which is the desired result.