A, B and C are matrices. Prove that $A^{T}AB = A^{T}AC$ iff AB = AC
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Let $A in mathbb{R}^{m times n}$ , $B in mathbb{R}^{n times p}$ , and $C in mathbb{R}^{n times p}$ . Show that $A^{T}AB = A^{T}AC$ iff $AB=AC$ . What I have done so far : The direction where I assume $AB=AC$ is trivial. So far, I want to claim that if $A^{T}AB = A^{T}AC$ holds for all $B$ and $C$ , then A^{T}A must be a full rank matrix. The reason I suspect this is if $AB neq AC$ and $A^{T}AB = A^{T}AC$ , I would have at least one column of $B$ and $C$ , say $b_i$ and $c_i$ , such that $Ab_i neq Ac_i$ , but $A^{T}Ab_i = A^{T}Ac_i$ . This isn't really too formal, but it gives me some intuition as to why I expect this result to be true.
linear-algebra matrices
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