Calculating $A^T A$ in matrix with orthogonal columns I have a matrix $A$ with three orthogonal columns, and I know that the length (2-norm) of each column is $4$.
The question is: what is $A^T A$?
Which properties should I use to solve this?
Thanks in advance.
 A: Write $A = [a_1, a_2, a_3]$, by condition,
$$A^TA = \begin{bmatrix}a_1^T \\ a_2^T \\  a_3^T  \end{bmatrix} \begin{bmatrix} a_1 & a_2 & a_3 \end{bmatrix} = \begin{bmatrix}a_1^Ta_1 & a_1^Ta_2 & a_1^Ta_3 \\ a_2^Ta_1 & a_2^Ta_2 & a_2^Ta_3 \\ a_3^Ta_1 & a_3^Ta_2 & a_3^Ta_3 \end{bmatrix} = \begin{bmatrix} 16 & 0 & 0  \\ 0 & 16 & 0 \\ 0 & 0 & 16\end{bmatrix}.$$
A: If we denote $a_{ij}$ the term of $A$ at the intersection of the $i^{th}$ row and $j^{th}$ column we have
$$(A^TA)_{ij}=\sum_{k=1}^na_{ki}a_{kj}$$
So for $i\neq j$ we have the scalar product of two distinct columns of $A$ i.e $0$ because the rows are orthogonal and on the diagonal we have the square of the 2-norm of the corresponding column i.e $16$ and so we have
$$A^TA=16I_n$$
where $I_n$ is the identity matrix of size $n\times n$
A: You can denote $A=(v_1, v_2,v_3)$ where the $v_i$ are the column vectors of $A$. Now $A^T$ is a matrix having 3 lines and each line is the transpose vector of a $v_i$.
So the coefficient $(A^TA)_{ij}$ of the product $A^TA$ is $v_i^T.v_j$ which is equal to $0$ if $i \neq j$ as by hypothesis the column vectors are orthogonal and to $16$ when $i =j$ as each column vector is supposed to have a $2$-norm equal to $4$.
Finally $$A^TA= 16 I_3$$ where $I_3$ is the identity matrix of dimension $3$.
A: You just need to exploit the fact that $(A B)_{ij} = A^{row}_i \cdot B^{col}_j$ and recall that $v^T v = \|v\|_2^2$ for every vector $v$. Now you have 
$$(A^T A)_{ij} = (A^T)^{row}_i \cdot A^{col}_j = (A^{col}_i)^T \cdot A^{col}_j = 16 \delta_{ij}.$$
