Every square matrix is similar to its transpose

Posted: March 15, 2022 in Basic Algebra, Matrices
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Throughout this post, M_n(\mathbb{C}) denotes the ring of n \times n matrices with entries from \mathbb{C}, the field of complex numbers, and I_n is the n \times n identity matrix.

One of the most fundamental theorems in linear algebra is the Jordan canonical form theorem. We are now going to use the theorem to show that every A \in M_n(\mathbb{C}) is similar to A^T, the transpose of A. But first, let’s use some notations.

Notations. We define the matrices B_n, H_n \in M_n(\mathbb{C}) as follows

\displaystyle B_n=\begin{pmatrix} 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 1 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 1 & 0 & 0 & \cdots & 0 & 0\end{pmatrix}, \ \ \ \ H_n=\begin{pmatrix} 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 0 & 1 \\ 0 & 0 & 0 & \cdots & 0 & 0\end{pmatrix}

Remark. Clearly B_n is invertible because swapping columns of B_n will eventually give the identity matrix, and it is easy to see that B_nH_n=H_n^TB_n. Thus H_n=B_n^{-1}H_n^TB_n.

Definition. A Jordan block matrix J_n is any n \times n matrix in the form J_n(\lambda)=\lambda I_n+ H_n, \ \lambda \in \mathbb{C}.

Problem. Let A \in M_n(\mathbb{C}). Show that A and A^T are similar.

Solution. By the Jordan canonical form theorem, there exist an invertible matrix P \in M_n(\mathbb{C}), positive integers n_1, \cdots , n_k, and \lambda_1, \cdots , \lambda_k \in \mathbb{C} such that

PAP^{-1}= \begin{pmatrix}J_{n_1}(\lambda_1) & & & \\ & J_{n_2}(\lambda_2) & & \\ & & \ddots \\ & & & J_{n_k}(\lambda_k)\end{pmatrix},

where all the entries in the empty space are zero. So, by the above Definition and Remark,

PAP^{-1}=\begin{pmatrix}\lambda_1 I_{n_1}+ H_{n_1} & & & \\ & \lambda_2 I_{n_2}+ H_{n_2} & & \\ & & \ddots \\ & & & \lambda_k I_{n_k}+H_{n_k} \end{pmatrix}

=\begin{pmatrix}\lambda_1 I_{n_1}+B_{n_1}^{-1}H_{n_1}^TB_{n_1} & & & \\ & \lambda_2 I_{n_2}+ B_{n_2}^{-1}H_{n_2}^TB_{n_2}& & \\ & & \ddots \\ & & & \lambda_k I_{n_k}+ B_{n_k}^{-1}H_{n_k}^TB_{n_k}\end{pmatrix}.

Thus if we put

Q:=\begin{pmatrix}B_{n_1}& & & \\ & B_{n_2} & & \\ & & \ddots \\ & & & B_{n_k}\end{pmatrix},

then, since

Q^{-1}=\begin{pmatrix}B_{n_1}^{-1}& & & \\ & B_{n_2}^{-1} & & \\ & & \ddots \\ & & & B_{n_k}^{-1}\end{pmatrix},

we get that

PAP^{-1}=Q^{-1}\begin{pmatrix}\lambda_1 I_{n_1}+H_{n_1}^T & & & \\ & \lambda_2 I_{n_2}+ H_{n_2}^T& & \\ & & \ddots \\ & & & \lambda_k I_{n_k}+ H_{n_k}^T\end{pmatrix}Q

=Q^{-1}(PAP^{-1})^TQ=Q^{-1}(P^T)^{-1}A^TP^TQ

and so A=P^{-1}Q^{-1}(P^T)^{-1}A^TP^TQP=(P^TQP)^{-1}A^TP^TQP. \ \Box

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