Let's assume that $A$ is a $m \times n$ matrix with linearly independent columns. Why are the columns of $A(A^T)$ also linearly independent? Is this new matrix invertible? What about $(A^T)A$?
Multiplying a Matrix by its Transpose
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linear-algebra
matrices
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4What have you done? Where are you stuck? Are you asking, requesting or demanding? – 2012-10-24
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1Does it help, if you know that $\operatorname{rank}(AA^{T})=\operatorname{rank}(A)=\operatorname{rank}(A^T)$? See this question: [Null space for $AA^{T}$ is the same as Null space for $A^{T}$](http://math.stackexchange.com/questions/66560/null-space-for-aat-is-the-same-as-null-space-for-at) – 2012-10-26