I am solving some exercises, one of which I need to prove the following:
Suppose $A\epsilon M_n(\Re)$
if $A^T.A = 0$ then $A=0$
The solution I have come up with is the following:
Let $B = A = 0$ (1), show that $A^T.B = 0$
$A^T.B$ = $\sum_{ k \mathop =1}^na^T_{ij}b_{i,j}$
By (1) $b_{i,j} = 0$ so summation = 0, thus $A^T.B = 0$.
Is this the correct way to prove it?
Hint: The $(i,i)$ entry of $A^T A$ is
$$ (A^T A)_{i,i} = \sum_{k=1}^n (A^T)_{i,k} A_{k, i} = \sum_{k=1}^n A_{k,i} A_{k,i} = \sum_{k=1}^n (A_{k,i})^2. $$