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Give this matrix A: \begin{pmatrix}-25&2&3&-29\\2&7&7&11\\3&7&7&2\\-29&11&2&11\end{pmatrix}

How can we calculate C matrix when A = AC - CA without extensive computations? Thought of doing these steps:

  1. Let C be {{a,b,c,d},{e,f,g,h},{i,j,k,l},{m,n,o,p}}
  2. Calculate (via wolfram) AC, CA
  3. Subtract AC, CA (cannot do it with wolfram, there's a limit on the characters of input
  4. Then assign the 1st row of the result with the respective row of matrix A
  5. Solve the system of 4 linear equations with 4 variables

Is that correct? How can I find a workaround for step 3?

Thank you for your time!

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Why only the look at the first row of the equality A=AC-CA? Wouldn't you need equality in all four rows? Also: wouldn't C be unique only up to addition by matrices B such that AB-BA=0 (ie, the kernel of the adjoint, $\operatorname{ker}\,\operatorname{ad}_A$)? – anon May 17 '12 at 21:53
@anon: "Solve the system of 4 linear equations with 4 variables", that is what I mean, all the 4 rows (I only mentioned the 1st row). – Chris May 17 '12 at 22:12

I'm assuming this is homework, so you need to do some work.

First, a good question to ask is how this can be true with scalars (this is can be particularly effective with symmetric matrices)? The answer here is: only if $A$ is zero. This suggests a direction...

Suppose $v$ is a unit eigenvector of $A$ corresponding to the eigenvalue $\lambda$. Consider the value of $v^T A v$: $$ v^T A v = v^T A C v - v^T C A v $$ The left hand side has value $\lambda$, and the right hand side is straightforward to compute.

What does this tell you about $\lambda$?

Since $\lambda$ is an arbitrary eigenvalue, this means all eigenvalues satisfy this condition. What does that tell you about the symmetric matrix $A$?

Given the above, what can you conclude about the solutions of the equation?

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+1 A very nice hint! For some reason I momentarily thought that you need $C$ to be symmetric as well for this to work, but then I reconsidered... – Jyrki Lahtonen May 17 '12 at 22:25
This is a nice hint; however, you seem to assume that $A$ is symmetric. The question doesn't say that explicitly (although the example given is symmetric). – Fixee May 17 '12 at 22:45
I believe it says 'Give [sic.] this matrix'? (Yes, I do assume $A$ is symmetric.) – copper.hat May 17 '12 at 22:56
Why is that "The left hand side has value λ" ? I am new to these concepts and I am a bit confused. But I've thought as a solution to my problem, proving that C exists or not, by using traces only. – Chris May 18 '12 at 12:53
Because $v$ is an eigenvector of $A$ corresponding to $\lambda$ we have $Av = \lambda v$, Since $v$ is a unit vector, $v^T \lambda v = \lambda v^T v = \lambda$. Also, its not clear to me how you would use the trace in this instance. – copper.hat May 18 '12 at 15:02

You can also use vectorization trick: $A=ACI-ICA$ $vec(A)=(A\otimes I)vec(C)-(I\otimes A)vec(C)$ Then C is on one side of the equation so you can solve it.

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