# Given a matrix $A$, is there a matrix $C$ with $AC = CA + A$?

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 is? –  Ｊ. Ｍ. May 17 '12 at 16:31
@J.M.: No, I haven't heard of it :/ And I find it a bit difficult to understand how can I use it. :$– Chris May 17 '12 at 19:18 ## 2 Answers If$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$.) - Thank you for your reply, but could you explain more why this Tr(AC - CA) = 0 entailes this "This means that you can't hope to solve AC−CA=A" ? How does the equation connect with the trace "Tr(A) = Tr(AC-CA)"? – Chris May 17 '12 at 19:24 @Chris: Dear Chris, If$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$). Regards, – Matt E May 17 '12 at 19:26 So we can apply trace in both parts of the equation! So if a = -32, there is a matrix C, that makes this equation: AC-CA = A true. Should I find the matrix C? Is it possible to do so? Or should I leave the answer as is(until "...true"?) Thank you very much! – Chris May 17 '12 at 19:45 @Chris: Dear Chris, The condition I give (that Tr$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$is a straightforward computation, which I'll leave to you. Regards, – Matt E May 17 '12 at 20:00 Dear Matt, sorry to bug you, but I am not really familiar with this straightforward computation. Could you give me some guidance on how to find C? I've thought considering C = {{a,b,c,d},{e,f,g,h},{i,j,k,l},{m,n,o,p}} and then calculating AC and CA and then subtracting these two. To continue, I would get each row of the result to be equal with the respective row of matrix A. Then I would solve a system of 4 linear equations with 4 variables. Is that right? (It doesn't seem that much, because there a lot of computations) Regards, Chris – Chris May 17 '12 at 20:56 @Chris, I am afraid that my answer does not help you much ; indeed, according to your posts, you are not an eagle in the mathematical field. Since$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? –  user1551 Sep 13 at 10:49
Please, can you give a reference of this result due to Jacobson? Thanks in advance. –  Daniel Sep 14 at 13:33
@ Daniel , cf. the introduction of jankobracic.files.wordpress.com/2011/02/… –  loup blanc Sep 14 at 19:42