If $A^TA$ is a diagonal matrix, are the columns of $A$ linearly independent? If I have two matrices, $A$ and $A^T$, and $A^TA$ creates a diagonal matrix, how can I tell and argue that the columns of $A$ are linearly independent? My professor mentioned that the dot product and the fact that the columns are orthogonal can be used to solve the problem, if that helps. Any comments would be appreciated. 
 A: Are you sure you don't have another condition?
$\begin{bmatrix}1&0\\0&0\end{bmatrix}\begin{bmatrix}1&0\\0&0\end{bmatrix}=\begin{bmatrix}1&0\\0&0\end{bmatrix}$
My guess is that the diagonal matrix is meant to have non-zero entries on the main diagonal. If so, the lengths of each of the columns is non-zero, and they are orthogonal. Can you prove that if $v_1, v_2, \dots, v_n$ are non-zero, mutually orthogonal vectors then they are linearly independent?
A: Let $A_i$ be the $i$-th column of $A$ and suppose $A_i \neq 0$ for every $i$. Note that $AA^T$ diagonal implies ($A^TA$ diagonal and thus) $\langle A_i,A_j\rangle =0$ for every $i\neq j$.
Now, let $c_1,\ldots,c_n$ such that $\sum_{i=1}^n c_iA_i=0$.then we have 
$$0=\left\|\sum_{i=1}^n c_iA_i\right\|_2^2= \left\langle\sum_{i=1}^n c_iA_i,\sum_{j=1}^n c_jA_j\right\rangle = \sum_{i,j=1}^nc_ic_j\langle A_i,A_j\rangle= \sum_{i=1}^n c_i^2 \langle A_i,A_i\rangle = \sum_{i=1}^n c_i^2 \|A_i\|_2^2$$
Since $A_i \neq 0$ we have $\|A_i\|_2^2>0$ and thus $c_i =0$ for every $i$.
