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 March 25th, 2018, 05:44 AM #1 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Finding transformation matrix for similar matrices I scouted through online but I am unable to find any comprehensive explanation of finding the transformation matrix T A and B are n x n matrices. A and B are 'similar' matrices. Here is the definition of similar matrices $\displaystyle B = T^{-1} A T$ Find T. Last edited by zollen; March 25th, 2018 at 05:46 AM. March 25th, 2018, 06:29 AM #2 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 Please use any 2x2 matrices A and B as an example... March 27th, 2018, 04:15 PM #3 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 $\displaystyle \mathbf{B} = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & -1 & 3 \end{bmatrix}$ $\displaystyle \mathbf{D} = \begin{bmatrix} 3 & 2 & 2 \\ 0 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$ $\displaystyle D = P^{-1} B P$ Find P. I need help on this one. How to use change of bases to find out the matrix P?? Last edited by zollen; March 27th, 2018 at 04:23 PM. March 27th, 2018, 09:56 PM #4 Senior Member   Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics The Jordan normal form of a linear operator is unique and thus similar matrices have the same Jordan normal form (up to a possible reordering of the Jordan blocks). This means if $A,B$ are similar matrices, and $J$ is their shared Jordan normal form with a fixed choice of ordering of the Jordan blocks, then you can find two invertible matrices satisfying $P_1^{-1} A P_1 = J \qquad \text{and} P_2^{-1}BP_2 = J$ This is explicitly computable, see for example: https://en.wikipedia.org/wiki/Jordan_normal_form With $P_{1,2}$ computed, the equality above implies that setting $T = P_1P_2^{-1}$ will satisfy $T^{-1} A T = B$. Thanks from topsquark and zollen April 7th, 2018, 01:42 PM #5 Senior Member   Joined: Jan 2017 From: Toronto Posts: 209 Thanks: 3 This took me a while but I finally got it!!!!! Both matrix B and D have the same eigenvalues and both matrices are diagonalizable. $\displaystyle D = P^{-1} ~B~ P \\ B = P_B~ \lambda ~P_B^{-1} \\ D = P_D~ \lambda ~P_D^{-1} \\ P_B^{-1} ~B~ P_B = P_D^{-1} ~D~ P_D \\ (P_D P_B^{-1}) ~B~ (P_B P_D^{-1}) = D \\ (P_D P_B^{-1}) ~B~ (P_D P_B^{-1})^{-1} = D \\ \\ P_{final} = P_D P_B^{-1}$ Last edited by zollen; April 7th, 2018 at 01:45 PM. April 8th, 2018, 05:19 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 In this particular problem, neither "B" nor "D" is in "Jordan normal form". I would do this by "brute strength". (It titillates me to think of myself as having "brute strength"!) If $\displaystyle D= P^{-1}BP$ then $\displaystyle PD= BP$. Writing P as $\displaystyle \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ Then $\displaystyle \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}\begin{bmatrix}3 & 2 & 2 \\ 0 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix}= \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 1 & -1 & 3 \end{bmatrix}\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i\end{bmatrix}$ Doing the multiplication on each side, $\displaystyle \begin{bmatrix}3a & 2a- b+ c & 2a+ c \\ 3d & 2d- e+ f & 2d+ f \\ 3g & 2g- h - i & 2g+ i \end{bmatrix}= \begin{bmatrix} b & e & f \\ a & b & c \\ a- d+ 3g & b- e+ 3h& d- f+ 3i \end{bmatrix}$. So we have the 9 equations 3a= b, 2a- b= e, 2a+ c= f, 3d= a, 2d- e+ f= b, 2d+ f= c, 3g= a- d+ 3g, 2g- h- i= b- e+ 3h, and 2g+ i= d- f+ 3i to solve for the 9 unknowns, a, b, c, d, e, f, g, h, and i. Thanks from zollen Last edited by Country Boy; April 8th, 2018 at 05:56 AM. Tags finding, matrices, matrix, similar, transformation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jezb5 Linear Algebra 1 February 9th, 2013 03:40 PM Wil Algebra 1 April 11th, 2011 08:03 AM hobochu Linear Algebra 1 July 26th, 2010 03:10 PM remeday86 Linear Algebra 3 June 30th, 2010 10:19 AM snarkle34 Linear Algebra 2 May 7th, 2007 09:35 PM

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