How can I understand that ATA is invertible if A has independent columns? I found a similar question, phrased the other way around, so I tried to use the theorem

rank(ATA)≤min(rank(AT),rank(A))

Given rank(A)=rank(AT)=n and ATA produces an n×n matrix, I can’t seem to prove that rank(ATA) is actually n.

I also tried to look at the question another way with the matrices

ATA=[aT1aT2…aTn][a1a2…an]=[ATa1ATa2…ATan]

But I still can’t seem to show that ATA is invertible. So, how should I get a better understanding of why ATA is invertible if A has independent columns?

**Answer**

Consider the following:

ATAx=0

Here, Ax, an element in the range of A, is in the null space of AT. However, the null space of AT and the range of A are orthogonal complements, so Ax=0.

If A has linearly independent columns, then Ax=0⟹x=0, so the null space of ATA={0}. Since ATA is a square matrix, this means ATA is invertible.

**Attribution***Source : Link , Question Author : Chewers Jingoist , Answer Author : Noble Mushtak*