Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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} ?

share|cite|improve this question
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
up vote 4 down vote accepted

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 $R$.

share|cite|improve this answer

You can get from one to the other by swapping the second and third rows and then the second and third columns. This corresponds to multiplying on the left by a certain row-switching matrix and then on the right by a certain column-switching matrix, as described in Wikipedia. What are they? Note that such matrices are their own inverses.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.