Let $A$ be an $m\times n$ matrix with $m\ge0$. Prove that if the rank of $A$ equals $n$, $A^TA$ is invertible.
Hint: prove that $A$ and $A^TA$ have the same kernel.
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2What have you tried? Have you showed what the hint says? Can you see how to use the hint? – Tobias Kildetoft Feb 06 '13 at 16:23
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@Tobias I am not sure how to prove what the hint says. Could you give me an idea how I could show this? – dreamer Feb 06 '13 at 16:29
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@MaisamHedyelloo It is asked to prove that AtA is invertible, not just At itself. – dreamer Feb 06 '13 at 16:31
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1Related: http://math.stackexchange.com/q/199981/9464 – Feb 06 '13 at 16:58
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Here's a hint for how to prove the hint ;)
Suppose $v$ is in the kernel of $A^TA$. This means that $A^TAv = 0$, and so $v^TA^TAv = (Av)\cdot(Av)=0$.
Can you take it from there?
EDIT: More explicit help, as requested.
From above we have that if $v\in\ker A^TA$, then $(Av)\cdot (Av) =0$. This means that $\|Av\|^2 = 0$, which is possible only when $Av = 0$. Therefore $\ker A^TA \subset \ker A$.
The other direction is easy: if $w\in \ker A$, obviously $w\in \ker A^TA$.
Therefore $A$ and $A^TA$ have the same kernel, or more importantly, that their kernels have the same dimension. By the rank-nullity theorem, the kernel of $A$ has dimension $n-n=0$, so the kernel of the square matrix $A^TA$ has dimension 0. This means that it is invertible.
user7530
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Thanks for the help. I understand your steps, but unfortunately I still dont see how to proceed from here. How does this show that they share the same kernel? – dreamer Feb 06 '13 at 17:18
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I have one more question. Why does it follow that AtA is invertible from the fact that the kernel had dimension 0? – dreamer Feb 06 '13 at 17:53
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What theorems have you proven about invertible matrices? Usually you show early on that a square matrix is invertible iff it has full rank. – user7530 Feb 06 '13 at 19:06