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Given this matrix A \begin{pmatrix}7+a&2&3&3+a\\2&7&7&11\\3&7&7&2\\3+a&11&2&11\end{pmatrix} where $a \in \mathbb{R}$

Is there a matrix $C \in \mathbb{R^{4x4}}$ with $ AC = CA + A $ ?

Notes:

  • $A$ is symmetric, and Hermitian
  • I've thought of this $AC = CA + A \Rightarrow A = AC - CA$ (can we reach somewhere if we assume that $C = BAB^{-1}$ where $B$ is a regular matrix)
  • $AC = CA + A \Rightarrow A= AC - CA$, if we assume that $C$ is the identity matrix then $CA = AC = I$, so $A = I - I \Rightarrow A = 0$, which is false, so there isn't a matrix $C$ (I am not sure)

Thank you for your time!

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    @J.M.: No, I haven't heard of it :/ And I find it a bit difficult to understand how can I use it. :$2012-05-17

2 Answers 2

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If $A$ and $C$ are two $n\times n$ matrices, then Tr$(AC) =$ Tr$(CA)$, and so Tr$(AC - CA) = 0$. This means that you can't hope to solve $AC - CA = A$ unless $A$ has trace zero. (In your case this happens only for $a = - 32$.)

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    Thank you once again, that's what I did! To sum up, what you provided was a way, to solve this equation right?2012-05-17
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@Chris, I am afraid that my answer does not help you much ; indeed, according to your posts, you are not an eagle in the mathematical field.

Since $AC-CA$ and $A$ commute, $AC-CA$ is nilpotent (this result is due to Jacobson). Then $A$ is necessarily nilpotent. Thus $a=-32$ and $A$ is a real symmetric matrix ; then $A$ is diagonalizable and, consequently, must be the zero matrix. Finally $C$ does not exist.

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    @ Daniel , cf. the introduction of http://jankobracic.files.wordpress.com/2011/02/on-the-jacobsons-lemma.pdf2014-09-14