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Two matrices $X$ and $Y$ are similar if and only if there is an invertible matrix $T$ such that $X=TYT^{-1}$.

Given that $X$ and $Y$ are similar, how would you find a value of $T$?

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  • $\begingroup$ What do you mean by "the value of T"? Do you mean an explicit real or complex valued matrix that corresponds to the linear transformation T with respect to a chosen basis? It's not clear what you're asking since the question is stated on abstract vector spaces-you need to be a little more exact in stating the question. $\endgroup$ Commented Nov 11, 2014 at 23:48
  • $\begingroup$ @Mathemagician1234 Thanks! I do mean an explicit value. $\endgroup$ Commented Nov 11, 2014 at 23:49
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    $\begingroup$ In general $T$ is hardly unique; e.g. if $X=Y=0$ then $T$ can be any invertible matrix. $\endgroup$
    – vadim123
    Commented Nov 11, 2014 at 23:56

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You obtain it from the Jordan normal form of both matrices. The $J$ matrix in the JNF is the same for both matrices (this is a consequence/definition of similarity).

Lets write the Jordan normal forms as

$J = Q_X^{-1} X Q_X$

and

$J = Q_Y^{-1} Y Q_Y$

Then you have

$T = Q_X Q_Y^{-1}$.

In case you have diagonalizable matrices $X$ and $Y$ the $J$ is a diagonal matrix containing the eigenvalues of both matrices. (Note that similar matrices need to have the same eigenvalues). In the general case, the $J$ matrix is block diagonal.

But note that the Jordan normal forms need to have the same ordering of eigenvalues (but this can always be achieved with a proper permutation).

EDIT: vadim123 made an important point in the comments. Since the $Q$ matrix in the JNF is not unique (there is some flexibility choosing the rows corresponding to one Jordan block) also the resulting $T$ matrix is not unique.

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  • $\begingroup$ Damn.beat me to it............lol $\endgroup$ Commented Nov 11, 2014 at 23:53
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    $\begingroup$ The important thing to remember is that similar matrices are matrices that actually represent the same linear transformation that differ by a choice of basis for the vector space. So the invertible matrix is actually a change of basis transformation. That's the main point that needs to be understood here. That's why the Jordan form is so useful because the characteristic equation for Y can be read immediately and the resulting eigenvectors form a basis that can be chosen to yield X! It's important to understand why we go through all this trouble. $\endgroup$ Commented Nov 11, 2014 at 23:58
  • $\begingroup$ @Mathemagician1234: Yes, I agree $\endgroup$
    – Andreas H.
    Commented Nov 12, 2014 at 0:05
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    $\begingroup$ Basically yes. More precisely it is the generalized eigenvectors. Eigenvectors it is only when the matrices are diagonalizable (i.e. the geometric rank is equal to the algebraic rank of the eigenvalues). $\endgroup$
    – Andreas H.
    Commented Nov 12, 2014 at 0:13
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    $\begingroup$ @DiaMcThrees: Well I suggest that you should look into construction of Jordan normal form - this will explain it. The generalized eigenvalues are, to my knowledge, only used for the JNF. Generalized eigenvalues are needed if the geometric multiplicity (the actual number of linearly independent eigenvectors) is smaller than the algebratic multiplicty (the needed number of independent eigenvectors). Then eigenvectors are "missing" to construct a diagonal normal form. As a workaround one uses generalized eigenvectors, which allows for an almost diagonal normal form (i.e. the JNF). $\endgroup$
    – Andreas H.
    Commented Nov 12, 2014 at 0:23
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What you need to do is find a pair of bases e=(e_1 \dots e_n) and f=(f_1 \dots f_n) such that the matrix of X with respect to e is identical to the matrix of Y with respect to f. The first thing you can do is look at the eigenvalues of X and Y: they have the same eigenvalues, so a correspondence between their eigenvectors is the first place you can look to build your bases. If the matrices are diagonalizable, then in fact this is all you have to do, as their eigenvectors will form a basis for each, and obviously you are finished.

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  • $\begingroup$ I selected the other one because it was more clear, but this was just as helpful. Thanks for the in-depth answer! $\endgroup$ Commented Nov 11, 2014 at 23:54
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By the definition of similarity, both $X$ and $Y$ have an identical Jordan Forms $J$. Jordan forms are similar to the original matrix, so $$J=Q^{-1}_XXQ_X \textrm{ and }J=Q^{-1}_YYQ_Y$$ so $$Q^{-1}_XXQ_X=Q^{-1}_YYQ_Y$$ $$X=Q_XQ^{-1}_YYQ_YQ^{-1}_X$$ $$X=Q_XQ^{-1}_YY(Q_XQ^{-1}_Y)^{-1}$$ $$T=Q_XQ^{-1}_Y$$

The columns of $Q_X$ are made of of the eigenvectors for each eigenvalue of $X$. The same holds for $Q_Y$. (Generalized eigenvectors if algebraic multiplicity for an eigenvalue is greater than 1)

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