Remark

Tag 05X1, permanent. Cite it as [Bourbaki, Tag 05X1].

(1) Two square matrices which differ only in the order of their rows (or the order of their columns) are equivalent, but in general not similar. A matrix similar to a square matrix $X = (x_{ij})$ can be obtained by performing the same permutation $\sigma^{-1}$ on the rows and columns, that is by considering the matrix $X' = (x'_{ij})$ , where $x'_{ij} = x_{\sigma(i), \sigma(j)}$ for every ordered pair of indices; for if X is the matrix of an endomorphism u of $A^n_d$ with respect to a basis $(e_i)_{1 \leq i \leq n}$ , $X'$ is the matrix of u with respect to the basis $(e_{\sigma(i)})_{1 \leq i \leq n}$ .

(2) Let X and $X'$ be two square matrices of order n which can be written in the form of diagonal matrices of square matrices (no. 7, Example IV):

X = \begin{pmatrix} X_1 & 0 & \ldots & 0 \\ 0 & X_2 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & X_p \end{pmatrix} \qquad X' = \begin{pmatrix} X'_1 & 0 & \ldots & 0 \\ 0 & X'_2 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & X'_p \end{pmatrix}

corresponding to the same partition of the indexing set $[1, n]$ for X and $X'$ . If, for $1 \leq i \leq p$ , $X_i$ and $X'_i$ are equivalent (resp. similar), then X and $X'$ are equivalent (resp. similar): for, if $X'_i = P_i X_i Q_i$ for $1 \leq i \leq p$ , then $X' = PXQ$ where

P = \begin{pmatrix} P_1 & 0 & \ldots & 0 \\ 0 & P_2 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & P_p \end{pmatrix} \qquad Q = \begin{pmatrix} Q_1 & 0 & \ldots & 0 \\ 0 & Q_2 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & Q_p \end{pmatrix}

as follows from calculating the "block" product (no. 6). Moreover, if $Q_i = P_i^{-1}$ for all i, then $Q = P^{-1}$ .

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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