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March 25th, 2018, 05:44 AM   #1
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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.
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March 25th, 2018, 06:29 AM   #2
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Please use any 2x2 matrices A and B as an example...
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March 27th, 2018, 04:15 PM   #3
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$\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.
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March 27th, 2018, 09:56 PM   #4
SDK
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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
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April 7th, 2018, 01:42 PM   #5
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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.
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April 8th, 2018, 05:19 AM   #6
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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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