Proof that $AA^T$ is invertible if $A$ has independent rows

Question

While looking for the answer for my question I came across this post. It may be a silly idea, but if $A^{t}$ has independent rows can I just transpose it and get $A$ with independent columns and proceed as is shown in the linked solution? Or it is not working this way?

Yes, your approach is fine. – Ben Grossmann Feb 01 '21 at 17:55 — Ben Grossmann, Feb 01 '21 at 17:55
Thank you for such a fast answer wow! – Bearnardd Feb 01 '21 at 17:58 — Bearnardd, Feb 01 '21 at 17:58

score 1 · Accepted Answer · answered Feb 02 '21 at 09:51

Suppose $A$ is $m\times n$ and that $AA^T$ is not invertible. Then there exists $v\ne0$ (an $m\times1$ column vector) such that $$ AA^Tv=0 $$ This implies $v^TAA^Tv=0$, hence $(A^Tv)(A^Tv)=0$, so $A^Tv=0$. Thus $v^TA=0$ and therefore $A$ doesn't have linearly independent rows.

score 0 · Answer 2 · answered Feb 02 '21 at 07:58

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For every matrix $A$ with independent rows, we have $det A \neq 0$ also $det A= det A^t$so both $A$ and $A^t$ are invertible. Then $AA^t$ is invertible if and only if $A$ is.

answered Feb 02 '21 at 07:58

Ali

2,277

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The matrix $A$ is not supposed to be square. – egreg Feb 02 '21 at 09:48
You are right $A$ is not supposed to be square. – Ali Feb 03 '21 at 08:43

Proof that $AA^T$ is invertible if $A$ has independent rows

2 Answers2