Prove $X = (\vec{x_1},...,\vec{x_n})^T$, then $X^T X = \sum \vec{x_i} \vec{x_i}^T$

Question

How would I prove the above identity - I don't want to write out each component and do the multiplication. Is there any other way?

Answer 1

$X^TX = (\vec x_1,\ldots,\vec x_n)\cdot \left(\begin{array}{c} \vec x_1^T\\\ldots \\ \vec x_n^T\end{array}\right) = \vec x_1\vec x_1^T +\ldots + \vec x_n \vec x_n^T$.

Answer 2

$ \def\bbR#1{{\mathbb R}^{#1}} \def\e{\varepsilon} \def\L{\left}\def\R{\right}\def\LR#1{\L(#1\R)} \def\m#1{\left[\begin{array}{c}#1\end{array}\right]} $Given the standard basis vectors $\e_k\in\bbR n$ and a matrix whose column vectors are $\,a_k\in\bbR m$
we can expand the matrix as a sum of vector pairs $$\eqalign{ A \;=\; \m{a_1&a_2&\ldots&a_n} \;=\; \LR{\sum_{i=1}^n a_i\,\e_i^T} \in\, \bbR{m\times n} \\ }$$ This expansion can be used to calculate the Gram matrix of $A$
$$\eqalign{ A^TA &= \sum_{i=1}^n \sum_{j=1}^n (\e_i a_i^T)\,(a_j\e_j^T) \;=\; \sum_{i=1}^n \sum_{j=1}^n (a_i^Ta_j)\,(\e_i\e_j^T) \\ &= \m{ a_1^Ta_1&a_1^Ta_2&\ldots&a_1^Ta_n \\ a_2^Ta_1&a_2^Ta_2&\ldots&a_2^Ta_n \\ \vdots&\vdots&\ddots&\vdots \\ a_n^Ta_1&a_n^Ta_2&\ldots&a_n^Ta_n \\ } \\ }$$ and the Gram matrix of $A^T$ as $$\eqalign{ AA^T &= \sum_{i=1}^n \sum_{j=1}^n (a_i\e_i^T)\,(\e_ja_j^T) \\ &= \sum_{i=1}^n \sum_{j=1}^n a_i \,\delta_{ij}\, a_j^T \\ &= \sum_{i=1}^n a_i a_i^T \\ }$$ where $\delta_{ij}$ is the Kronecker delta symbol.