Find a matrix such that the systems are equivalentTake the system{w=v′v″−μv′+v=0&nbsp;for&nbsp;μ∈RFind a constant matrix&nbsp;Aμ&nbsp;such that the above system is equivalent tov'w'=Aμv'w'My first thought was to find a solution to&nbsp;v&nbsp;″−μv′+v=0. Taking the characteristic polynomial as&nbsp;r2−μr+1=0, I get&nbsp;r=12(μ±μ2−4). Not really sure what to do at this point, however. Another way to solve this might be first to replace v' in the second equation with w to get&nbsp;v&nbsp;″−μw+v=0&nbsp;but I'm not sure where to go from there either.

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Find a matrix such that the systems are equivalentTake the system{w=v′v″−μv′+v=0&amp;nbsp;for&amp;nbsp;μ∈RFind a constant matrix&amp;nbsp;Aμ&amp;nbsp;such that the above system is equivalent tov&#039;w&#039;=Aμv&#039;w&#039;My first thought was to find a solution to&amp;nbsp;v&amp;nbsp;″−μv′+v=0. Taking the characteristic polynomial as&amp;nbsp;r2−μr+1=0, I get&amp;nbsp;r=12(μ±μ2−4). Not really sure what to do at this point, however. Another way to solve this might be first to replace v&#039; in the second equation with w to get&amp;nbsp;v&amp;nbsp;″−μw+v=0&amp;nbsp;but I&#039;m not sure where to go from there either.

anghoelv1lw · Accepted Answer

Step 1(I&#039;m assuming here that you meant&amp;nbsp;v&#039;w&#039;=Aμvw&amp;nbsp;instead of&amp;nbsp;v&#039;w&#039;=Aμv&#039;w&#039;; in the latter case,&amp;nbsp;Aμ&amp;nbsp;would just be the identity matrix.)If you replace v&#039; with w in the second line of your equation and isolate each equation for v&#039; and w&#039;, you get:{v′=ww′=−v+μw&amp;nbsp;Step 2Now all you have to do is find a matrix of coefficients a through d such that:v&#039;w&#039;=abcdvwFor this matrix to correspond to your set of equations, it must be the case that&amp;nbsp;a=0,b=1,c=−1, and&amp;nbsp;d=μ.

Find a matrix such that the systems are

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