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If we let $A$ be an $m × n$ matrix. Suppose that the rows of $A$ are linearly independent.
(a) Is $A^TA$ invertible? Explain.
(b) Is $AA^T$ invertible? Explain.

So if the rows of $A$ are linearly independent then $Ax=0$. But how do I go about proving a and b?

mhelp
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1 Answers1

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I assume that you are working over the reals.

Assume the rows of $A$ are linearly independent. That means that the only solution to $xA = 0$ is $x = 0$ where $x$ is a row vector of length $m$.

Now assume that $AA^T$ is not invertible. Then there exists a row vector $v$ such that $vAA^T = 0$ (easy to see if we are over the reals or any other field, a little harder over arbitrary rings). But then also $vAA^Tv^T = 0$. Setting $x := vA$ we thus have $xx^T = 0$, which might ring some bells and then lead to a contradiction.

Dirk
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  • @ Bemte what contradiction is this? – mhelp Oct 24 '16 at 12:47
  • There's technically no need to appeal to contradiction here. Just let $v$ be such that $vAA^T=0$, deduce that $vA=0$. Then conclude that $v=0$ by linear independence of rows in $A$. I like a proof by contradiction as much as the next guy, but some times starting a proof with an assumption and concluding with a contradiction is just unnecessary filler, and the proof itself works just as well and looks better without it. – Arthur Oct 24 '16 at 12:52