Multiplying a Matrix by its Transpose

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 $(A^T)A$?

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What have you done? Where are you stuck? Are you asking, requesting or demanding? –  Pragabhava Oct 24 '12 at 23:40
Does 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 $A^{T}$ –  Martin Sleziak Oct 26 '12 at 4:22
The columns of $AA^T$ cannot be linearly independent unless $m=n$. If the columns of $A$ are linearly independent, then necessarily $m\ge n$.
If $m>n$, then $AA^T$ has $m$ columns, each of which is a linear combination of the columns of $A$, and there are only $n$ of those, so you have more than $n$ vectors in a space of dimension $n$, so they're not linearly independent.