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. $%(I don't know why the conjugate transpose expression is not working with mathjax in my computer.. sorry)$

  • $\begingroup$ 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. $\endgroup$
    – CrSb0001
    May 31 at 13:48
  • 2
    $\begingroup$ A is m×n complex matrix, A*A is invertible iff A has full column rank, is it true or false? $\endgroup$ May 31 at 13:50
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    $\begingroup$ Note that $A^*A$ is always square, even if $A$ is not. $\endgroup$ May 31 at 13:52

1 Answer 1


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.


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