I need to show that $\mathrm{rank}\mathsf{T} = \mathrm{rank}\mathsf{T}^\ast$ for a linear operator $\mathsf{T}$ on a finite-dimensional inner product space $\mathsf{V}$.
Let $\beta = \{v_1,\dots,v_n\}$ be a basis for $\mathsf{V}$ and $[\mathsf{T}]_\beta$ be the matrix representation of $\mathsf{T}$. I know that $[\mathsf{T}^\ast]_\beta = [\mathsf{T}]_\beta^\ast = \overline{[\mathsf{T}]_\beta^t}$ and that $\mathrm{rank}([\mathsf{T}]_\beta) = \mathrm{rank}([\mathsf{T}]_\beta^t)$. I am not sure how to begin going about showing that $\mathrm{rank}([\mathsf{T}]_\beta^t) = \mathrm{rank}(\overline{[\mathsf{T}]_\beta})$ though. Can anyone hint at how to start?
In proving this with the matrix representation, if you know how to prove $\mathrm{rank}([\mathsf{T}]_\beta) = \mathrm{rank}([\mathsf{T}]_\beta^t)$, then it is basically the same method using the conjugate transpose. So let's say you have the matrix equivalence $PAQ=$ Dg$[I_r,0]$ with $r$ the unique rank of $A$...then
$(PAQ)^*=Q^*A^*P^*=$ Dg$[I_r,0]^*=$ Dg$[I_r,0]$
And since $Q$ and $P$ are invertible, so are $Q^*$ and $P^*$, so that the above equation is also an equivalence of matrices between $A^*$ and Dg$[I_r,0]$...so $A^*$ has rank $r$, which is the same as the rank of $A$.