Let $A \in \Bbb M_{n \times n} (\Bbb R), b \in \Bbb R^n $. If $A^T$ represents the transpose of the matrix $A$, does the system of linear equation $A^TAz=A^Tb$ always have a solution?
Attempt: I worked out a few examples and I suspect the above statement is true. In order to prove this statement, we need to prove that $\operatorname{rank} (A^TA:A^Tb) =\operatorname{rank}(A^TA)$ where $(A^TA:A^Tb)$ is the augmented matrix.
Since, $\operatorname{rank}(A^TA)=\operatorname{rank}(A)=\operatorname{rank}(A^T),$ it would suffice to prove that $\operatorname{rank} (A^TA:A^Tb) =\operatorname{rank}(A)$
Could someone please give me hints to move forward. Thanks a lot for the help!
Best Answer
Hint: The statement is true. It suffices to note/show that the column space of $A^TA$ is a subspace of the column space of $A^T$ and that $\operatorname{rank}(A^TA) = \operatorname{rank}(A^T)$.
Regarding the second statement: since we know that $\operatorname{rank}(A^T) = \operatorname{rank}(A)$, the quickest approach is to show that $A^TAx = 0 \implies Ax = 0$. From there, the rank-nullity theorem gives us the desired result.