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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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    Do you know what a [Sylvester equation](http://en.wikipedia.org/wiki/Sylvester_equation) is?2012-05-17
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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 for your reply, but could you explain more why this Tr(AC - CA) = 0 entailes this "This means that you can't hope to solve AC−CA=A" ? How does the equation connect with the trace "Tr(A) = Tr(AC-CA)"?2012-05-17
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    @Chris: Dear Chris, If $A = AC - CA$, then since the right hand side has trace zero, so does the left hand side. So if Tr$(A) \neq 0$ then we can't write $A$ in the form $AC - CA$ (or indeed in the form $BC - CB$ for any matrices $B$ and $C$). Regards,2012-05-17
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    So we can apply trace in both parts of the equation! So if a = -32, there is a matrix C, that makes this equation: AC-CA = A true. Should I find the matrix C? Is it possible to do so? Or should I leave the answer as is(until "...true"?) Thank you very much!2012-05-17
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    @Chris: Dear Chris, The condition I give (that Tr$A = 0$) is necessary, but not obviously sufficient, and I didn't think about whether it is or isn't sufficient in your particular case. But plugging in $a = -32$ and then attempting to solve for $C$ is a straightforward computation, which I'll leave to you. Regards,2012-05-17
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    Dear Matt, sorry to bug you, but I am not really familiar with this straightforward computation. Could you give me some guidance on how to find C? I've thought considering C = {{a,b,c,d},{e,f,g,h},{i,j,k,l},{m,n,o,p}} and then calculating AC and CA and then subtracting these two. To continue, I would get each row of the result to be equal with the respective row of matrix A. Then I would solve a system of 4 linear equations with 4 variables. Is that right? (It doesn't seem that much, because there a lot of computations) Regards, Chris2012-05-17
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    @Chris: Dear Chris, Yes, this is what you should do. There are tricks to speed it up, and it would be easier on a machine rather than by hand. If you need more help, why don't you ask another question specifically about how to solve this kind of equation efficiently? Someone here probably knows some tricks. Cheers,2012-05-17
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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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    Very succinct! Why didn't anyone up vote this answer?2014-09-13
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    Please, can you give a reference of this result due to Jacobson? Thanks in advance.2014-09-14
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    @ Daniel , cf. the introduction of http://jankobracic.files.wordpress.com/2011/02/on-the-jacobsons-lemma.pdf2014-09-14