$AA^T =A$, what can you say about the rank of $A$?
Interview question, and I have no idea. The closest thing I can relate to is the inverse. But that's not the case here.
2026-04-04 08:44:51.1775292291
$AA^T =A$, what can you say about the rank of $A$?
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Robert Israel has already given the answer to the question asked—the rank can be any nonnegative integer up to the dimension of the matrix. However I wanted to point out that one can say a lot more: note that $$ A^T = (AA^T)^T = (A^T)^TA^T = AA^T = A, $$ and so $A$ symmetric. In particular, we now know that $A^2=A$ and so $A^2-A=0$, which means that the minimal polynomial of $A$ divides $t^2-t$. This tells a lot about the matrix $A$ (that it's diagonalizable and that its eiegnvalues can only be $0$ or $1$), and ultimately says that it is a projection matrix.