# Is my proof correct? m×n complex matrix A*A is invertible iff A has full column rank.

I am a high school student in South Korea. I am currently learning Linear Algebra in school. In this question, $$A^*$$ means conjugate transpose. I am wondering if this sentence is true or false. [m×n complex matrix $$A^*A$$ is invertible iff $$A$$ has full column rank.] If A is real, it is true and the proof is simple. Since $$null(A^{T}A)=null(A)$$, $$rank(A^{T}A)=rank(A)$$ so $$A^{T}A$$ is invertible iff A has full column rank. Is it okay if I use this result directly in complex field?(Assuming standard inner product in a complex field, $$A^*Ax=0 \implies x ^* A ^* Ax=0,||Ax||^{2}=0\,\,Ax=0$$ so $$A ^* A$$ and $$A$$ has the same nullspace, and so on. ) Or if there is a counterexample, please tell me. $$%(I don't know why the conjugate transpose expression is not working with mathjax in my computer.. sorry)$$

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. May 31 at 13:48
• A is m×n complex matrix, A*A is invertible iff A has full column rank, is it true or false? May 31 at 13:50
• Note that $A^*A$ is always square, even if $A$ is not. May 31 at 13:52

## 1 Answer

Yes, the statement holds and the idea behind your proof is correct. Here is one complete version of the proof.

If $$A^*A$$ is invertible, then $$Ax = 0 \implies A^*Ax = 0 \implies (A^*A)^{-1}(A^*A)x = 0 \implies x = 0,$$ which means that the columns of $$A$$ are linearly independent, which means that $$A$$ has full column rank.

Conversely, if $$A$$ has full column rank, then we find that $$A^*Ax = 0 \implies x^*A^*Ax = 0 \implies (Ax)^*(Ax) = 0 \\\implies \|Ax\|^2 = 0 \implies Ax = 0 \implies x = 0.$$ Thus, $$A^*A$$ is a square matrix with linearly independent columns, which means that $$A^*A$$ is invertible.