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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.

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