Given A any matrix, prove that $A^tA$ is positive definite if $\ker(A)=0$, and always positive semi-definite.
2026-03-25 15:54:28.1774454068
Matrix $A^tA$ properties
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$ \langle A^{t}Ax, x \rangle =\|Ax\|^{2} \geq 0$ so $A^{t}A$ is always positive semidefinite. It is positive definite iff $\|Ax\|=0$ only for $x=0$ which means $ker (A)=\{0\}$.