It seems the most usual way to define matrix similarity is as follows:

"Let $A$ and $A'$ be two n-by-n matrices, we say they are similar if there exist some invertible n-by-n matrix $P$ such that $A=PA'P^{-1}$"

However in my first course of linear algebra we defined it this way:

"Let $A$ and $A'$ be two n-by-n matrices, we say they are similar if there exist two invertible n-by-n matrices $C$ and $D$ such that $A=CA'D^{-1}$"

And I guess the reason we defined it this way was because we were thinking of matrices as representing linear transformations, so from that definition (the latter one) we deduced that if $A$ and $A'$ were two matrices associated to a linear operator in different bases, then $C$ and $D$ were just the change of basis matrices.

Now, what I want to understand is how can I find a matrix $P$, like in the first definition, given the matrices $C$ and $D$ of the second definition, is $P$ a function of $C$ and $D$? I want to prove the equivalence between both definitions, the implication from the first one to the second one is obvious, but I don't realize how to prove the other implication.

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    $\begingroup$ They aren't equivalent. The latter condition is used if you are viewing a matrix as a linear transformation $V \to W$ between two unrelated vector spaces. The former is used if you are viewing a matrix as a linear transformation $V \to V$ with the same domain and codomain. $\endgroup$ – Dustan Levenstein Feb 26 '16 at 16:37
  • $\begingroup$ A simple test: the alternative definition would allow the determinant to change $\endgroup$ – user251257 Feb 26 '16 at 16:48
  • $\begingroup$ @DustanLevenstein but then if we are in the case of matrices being seen as endomorphisms, are those definitions equivalent? And in the affirmative case how can I prove the second one implies the first one? $\endgroup$ – la flaca Feb 26 '16 at 16:48
  • $\begingroup$ Your question makes no sense. Your first definition of similarity is the one we obtain if we view a matrix as an endomorphism $V \to V$, allowing for a change of basis. Your second definition is the one obtained by viewing it as a linear transformation $V \to W$, and allowing for independent changes of bases on both vector spaces. You can't separate the definitions of "similarity" from how we view the matrices. $\endgroup$ – Dustan Levenstein Feb 26 '16 at 17:08

The first definition you cite is the usual definition for matrix similiarity, under this definition, two matrices are similar, if they can represent the same linear transformation $V \to V$ under a change of basis, where in domain and codomain the same basis is used.

The second definition defines another thing, two matrices fulfilling it are usually called equivalent, not similar. Two matrices are equivalent if they can represent the same linear transformation $V \to W$ under a change of basis in domain and codomain, where both bases are unrelated.

For example $$ A = \begin{pmatrix} 0 & 1\\ 0 & 0\end{pmatrix}, \quad B = \begin{pmatrix} 1 & 0\\ 0 & 0\end{pmatrix} $$ and not similar, but equivalent.

  • $\begingroup$ That makes sense, actually I've realized that when we defined "similarity" (the second definition) in our course, we defined if for n-by-m matrices. In your example, how do you realize those matrices are not similar, how could we prove that we can't find such a $P$ that makes them similar? $\endgroup$ – la flaca Feb 26 '16 at 17:13
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    $\begingroup$ @Eliana They are not similar because they have different sets of eigenvalues ($B$ has $1$ as an eigenvalue while $A$ does not). The are equivalent because you can swap two columns to convert one to the other; this corresponds to right multiplication by the invertible matrix $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$. $\endgroup$ – Dustan Levenstein Feb 26 '16 at 17:15

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