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March 25th, 2018, 05:44 AM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 193 Thanks: 2  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: 193 Thanks: 2 
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: 193 Thanks: 2 
$\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: 476 Thanks: 263 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 $T^{1} A T = B$. 
April 7th, 2018, 01:42 PM  #5 
Senior Member Joined: Jan 2017 From: Toronto Posts: 193 Thanks: 2 
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: 3,261 Thanks: 894 
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. Last edited by Country Boy; April 8th, 2018 at 05:56 AM. 

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