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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 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
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2 Answers

up vote 6 down vote accepted

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)"? –  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
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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! –  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
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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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