# Conjugate of matrix

Over an arbitrary ring $R$ with unit, is the matrix\begin{pmatrix} a & 0 & b & 0\\ 0 & 0 & 0 &0\\c & 0 & d &0\\ 0 &0&0&0 \end{pmatrix}

conjugate over $GL(R)$ to \begin{pmatrix} a & b & 0&0\\ c & d & 0&0\\0 & 0 &0 &0\\0&0&0&0 \end{pmatrix} ?

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Yes. It should be clear that they represent the same linear transformation but in different bases. –  Qiaochu Yuan Jul 25 '12 at 18:55
yeah, but I am sort of not able to write down the invertible matrix doing the conjugation... –  Cyril Jul 25 '12 at 18:56
Can you write down the correspondence between the bases? It's given by a change of basis matrix. –  Qiaochu Yuan Jul 25 '12 at 19:01
View the matrices as linear maps to see that they are the same (one is the matrix of the other when we switch the second and third vectors). Thus take $$P:=\pmatrix{\mathbf 1_R&0&0&0\\ 0&0&\mathbf 1_R&0\\ 0&\mathbf 1_R&0&0\\ 0&0&0&\mathbf 1_R},$$ where $\mathbf 1_R$ is the unit of $R$.