what condition of A makes transpose(A)*A nonsingular?

Question

What contidion of A makes $$A^TA$$ nonsingular?

If so, that is $$A^TA$$ is non-singular than a unique solution exists.

Answer 1

Let $A$ be a $m \times n$ matrix. The rank of $A^T A$ turns out to be the same as the rank of $A$, and $A^T A$ is a $n \times n$ matrix. So $A^T A$ is nonsingular if and only if the rank of $A$ is $n$, i.e. if $A$ has linearly independent columns.

Answer 2

Here is a conceptual approach. Note that

$$\langle Au, Av \rangle = \langle A^T A u, v \rangle$$

where, say, $u, v \in \mathbb{R}^n$ and $Au, Av \in \mathbb{R}^m$. Then $A^T A$ is nonsingular iff the bilinear form on the RHS, and hence the bilinear form on the LHS, is nondegenerate, which is true iff $A$ has full rank, or equivalently is injective as a linear map $\mathbb{R}^n \to \mathbb{R}^m$.