关于m×n复矩阵$A^*A$可逆性与列满秩等价性的证明正确性问询

Hey there! Let's break this down clearly for you—your reasoning is totally on point, and you can absolutely extend the real matrix result to the complex case without any issues.

First, let's recap your correct logic step by step, since it's solid:

• For the standard inner product on complex vector spaces, if $A^Ax = 0$, left-multiplying both sides by $x^$ gives $x^*A*Ax = 0$. This simplifies to $\langle Ax, Ax \rangle = |Ax|^2 = 0$.
• In complex spaces, a vector has norm 0 if and only if it's the zero vector, so this immediately implies $Ax = 0$. This tells us $\text{null}(A^*A) \subseteq \text{null}(A)$.
• Conversely, if $Ax = 0$, then $A^Ax = A^(0) = 0$, so $\text{null}(A) \subseteq \text{null}(A^*A)$.
• Putting these together, we get $\text{null}(A^*A) = \text{null}(A)$—the null spaces are identical.

Now using the rank-nullity theorem:

• For your $m \times n$ matrix $A$, $\text{rank}(A) + \text{nullity}(A) = n$.
• For the $n \times n$ matrix $A^*A$, $\text{rank}(A^*A) + \text{nullity}(A^*A) = n$.

Since their nullities are equal (because their null spaces are the same), their ranks must also be equal: $\text{rank}(A^*A) = \text{rank}(A)$.

From here, the equivalence follows directly:

• $A^*A$ is invertible if and only if $\text{rank}(A^*A) = n$ (since it's an $n \times n$ matrix).
• This is exactly the condition that $\text{rank}(A) = n$, meaning $A$ has full column rank.

There are no counterexamples here—your extension of the real matrix proof to the complex case is perfectly valid. The key difference (and what makes it work) is using the conjugate transpose paired with the complex inner product's norm property, which you already correctly applied.

备注：内容来源于stack exchange，提问作者starlight