Prove that any square matrix is conjugate to its transpose matrix. (You may prove it over \(\mathbb{C}\)).
Solution
We prove the assertion for matrices over any field \(F\).
1. The companion matrix \(C_f\) of any monic polynomial\(f\in F[t]\) of positive degree is similar to its transpose.
Let \(d=\deg f\), and regard \(C_f\) as multiplication by \(t\) on \(E=F[t]/(f)\) in the basis \(1,t,\ldots,t^{d-1}\). Each residue class has a unique representative of degree less than \(d\). Define \(\ell:E\to F\) to be the coefficient of \(t^{d-1}\) in that representative, and set \[ \beta(u,v)=\ell(uv). \] This is bilinear. It is nondegenerate: for a nonzero \(u\) represented by a polynomial of degree \(k<d\), take \(v=t^{d-1-k}\). The product has degree \(d-1\), so needs no reduction modulo \(f\), and \(\beta(u,v)\) is the nonzero leading coefficient of \(u\).
Commutativity and associativity in \(E\) give \(\beta(tu,v)=\ell(tuv)=\beta(u,tv)\). Let \(B_f\) be the matrix of \(\beta\) in the indicated basis. Nondegeneracy makes it invertible, and the last identity reads \[ C_f^{\mathsf T} B_f=B_f C_f. \] Thus \(C_f^{\mathsf T}=B_f C_f B_f^{-1}\).
2. Every square matrix \(A\) over \(F\) is similar to \(A^{\mathsf T}\).
Rational canonical form gives an invertible \(P\) with \(A=PCP^{-1}\), where \(C\) is a direct sum of companion matrices of monic polynomials [@DF04]. Apply step 1 to each block, and let \(B\) be the direct sum of the resulting invertible matrices \(B_f\). Then \(C^{\mathsf T}=BCB^{-1}\).
Put \(Q=(P^{-1})^{\mathsf T}BP^{-1}\). It is invertible, and direct multiplication gives \[ QAQ^{-1} =(P^{-1})^{\mathsf T}BCB^{-1}P^{\mathsf T} =(P^{-1})^{\mathsf T}C^{\mathsf T}P^{\mathsf T} =A^{\mathsf T}. \] This is the required conjugacy, in particular over \(\mathbb C\). The empty matrix, if allowed, is its own transpose and needs no blocks.
Solution
We work over \(\mathbb{C}\), as the problem permits.
1. Every Jordan block is similar to its transpose.
Let \(J=J_k(\lambda)\) be the \(k\times k\) Jordan block, with \(\lambda\) on the diagonal and \(1\) in each entry \((i,i+1)\). Let \(R\) be the permutation matrix with \(1\) in each entry \((i,k+1-i)\); then \(R=R^{-1}\). The entry \((i,j)\) of \(RJR\) is the entry \((k+1-i,k+1-j)\) of \(J\). It equals \(\lambda\) when \(i=j\) and \(1\) when \(j=i-1\), and is zero otherwise. Thus \(RJR^{-1}=J^{\mathsf T}\).
2. Every square complex matrix \(A\) is similar to \(A^{\mathsf T}\).
Jordan canonical form gives an invertible \(P\) with \(A=PJP^{-1}\), where \(J=J_1\oplus\cdots\oplus J_m\) is a direct sum of Jordan blocks. Let \(R_i\) be the matrix of step 1 for \(J_i\) and \(R=R_1\oplus\cdots\oplus R_m\). Then \(RJR^{-1}=J_1^{\mathsf T}\oplus\cdots\oplus J_m^{\mathsf T}=J^{\mathsf T}\). Transposing \(A=PJP^{-1}\) gives \(A^{\mathsf T}=(P^{-1})^{\mathsf T}J^{\mathsf T}P^{\mathsf T}\). With \(Q=(P^{-1})^{\mathsf T}RP^{-1}\), which is invertible, \[ QAQ^{-1}=(P^{-1})^{\mathsf T}RJR^{-1}P^{\mathsf T}=(P^{-1})^{\mathsf T}J^{\mathsf T}P^{\mathsf T}=A^{\mathsf T}. \]