Calculate matrix $A^T A$ with pairwise orthogonal vectors

Question

I have a matrix $A$, that contains pairwise orthogonal vectors with length $1$, and I should calculate $A^T A$. I defined that:

$ v_{1}, v_{2}, v_{n-1}, v_{n} ∈ R^n \ and \ A ∈ R^{m _x n} $

and if I understand it right that are the rules for v1, v2, vn-1, vn:

$ v_2 = v_2 * \frac{1}{||v_2||} \\ v_n = v_n \frac{1}{||v_n||} \\ v_{n-1} * v_n = 0 \\ v_1 * v_2 \ = 0 \\ v^2_2 = (v_2 * \frac{1}{||v_2||})^2 = v_2 \\ v^2_n = (v_n * \frac{1}{||v_n||})^2 = v_n $ $ A = \begin{pmatrix}v1&v2&v_{n-1}&v_n\end{pmatrix}$

$ A^T * A = \begin{pmatrix}v_1\\ v_2\\ v_{n-1}\\ vn\end{pmatrix} * \begin{pmatrix}v1&v2&v_{n-1}&v_n\end{pmatrix} = \begin{pmatrix}v^2_1&0&v_1v_{n-1}&v_1v_n\\ 0&v_2&v_2v_{n-1}&v_2v_n\\ v_1v_{n-1}&v_2v_{n-1}&v^2_{n-1}&0\\ v_1v_n&v_2v_n&0&v_n\end{pmatrix} $

Is that the right approach or did I miss something?

Answer 1

Pairwise orthogonal means that $v_i\cdot v_j=0$ if $i\neq j$. Whereas saying that each of them has norm $1$ means that $v_i\cdot v_i=1$. Thus in general $$v_i\cdot v_j=\delta_{ij}$$ Now, if $A=(v_i)_i$ then $$(A^\top A)_{ij}=v_i\cdot v_j=\delta_{ij}$$ so the product is actually the identity matrix.