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This question already has an answer here:

Given $A\in\Bbb C^{n \times n}$,I want to prove $\text{rank} A=\text{rank} A^H$ where $A^H$ is the conjugate transpose matrix of $A$. We only need to prove:

$\text{rank} A=\text{rank} \bar A$

where $\bar A$ is the conjugate matrix of $A$. I have tried but was not able to solve the problem.


I found a similar question here rank of complex conjugate transpose matrix property proof. I know rank-nullity theorem, but I have not learn any theorem about maps that preserve the dimension of nullity. How do we prove $A \to \bar A$ map preserves dimension of nullity?

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marked as duplicate by Jack, user223391, Claude Leibovici, user91500, John B Dec 13 '16 at 11:23

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    $\begingroup$ $Ax=0$ iff $\bar A\bar x=\bar 0=0$. $\endgroup$ – A.Γ. Dec 12 '16 at 22:59
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    $\begingroup$ if $A$ is a matrix representing a linear transformation, then the physical meaning of conjugate transpose is the matrix for the adjoint of the transformation. Hope this helps! $\endgroup$ – Li Chun Min Dec 12 '16 at 23:00
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    $\begingroup$ done? to prove what A.G. says pick any vector in the kernel of A and compute $\langle Av, v \rangle $. $\endgroup$ – Li Chun Min Dec 12 '16 at 23:13
  • $\begingroup$ Thanks for your guys' hints! I got it. $\endgroup$ – River Dec 12 '16 at 23:20
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If $U$ is the reduced row echelon form for $A$, then we know that $U=FA$ for some invertible matrix $A$. Since matrix multiplication cannot increase the rank, $$\DeclareMathOperator{\rk}{rk} \rk(A)=\rk(F^{-1}FA)\le\rk(FA)\le\rk(A) $$ so $\rk(A)=\rk(U)$. Similarly, $\rk(A^H)=\rk(U^H)$, because $F^H$ is invertible.

Thus you only need to show that $U$ and $U^H$ have equal rank. It's easy to see that the nonzero columns of $U^H$ are linearly independent; their number is the same as the number of pivot columns of $U$, which is the rank of $U$.

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