# linear algebra – How to show that if \$A^{T}Amathbf{x}=mathbf{0}\$ then \$Amathbf{x}=mathbf{0}\$

Assuming that you are dealing with real matrices, note that if $$A^T A mathbf{x} = mathbf{0}$$, then $$A mathbf{x}$$ must be in the null space of $$A^T$$, which in particular is orthogonal to the row space of $$A^T$$ (and hence orthogonal to the column space of $$A$$). But $$A mathbf{x}$$ is by construction in the column space of $$A$$!

All this is to show that $$A mathbf{x}$$ is both in the column space of $$A$$ and the space orthogonal to the column space of $$A$$. These two facts together imply that $$A mathbf{x} = mathbf{0}$$.