We define matrices $A$ and $B$ to be equivalent iff there exist invertible matrices $P$ and $Q$ such that $B = Q^{-1}AP$.

If $A$ and $B$ are both square matrices, then we consider a different equivalence relation: similarity ($A$ and $B$ similar iff $A = CBC^{-1}$ for some invertible $C$).

Equivalent matrices represent the same mapping under a different choice of basis for the domain and range. When we deal with similar matrices, clearly the domain and the range are the same. But why is it necessary to introduce a new relation? Why doesn't equivalence suffice?

  • $\begingroup$ So similarity is a special type of "equivalence" where you use the change change of basis for the domain and codomain. Similarity is important for a lot of reasons, one of them being that it preserves eigenvalues. $\endgroup$ – Ian Feb 12 '17 at 19:51

A way to reframe the definitions: matrices $A,B: \Bbb R^n \to \Bbb R^m$ are equivalent if there exists bases $\mathcal B_1, \mathcal B_2$ of $\Bbb R^n,\Bbb R^m$ such that $B$ is the matrix of $A$ with respect to these new bases.

However, in the case of square matrices, we might say that choosing different bases for the first and second space loses too much structure from the original map $A$. For instance, take $$ A = \pmatrix{0&1\\1&0} $$ (the reflection through $y=x$). This map is equivalent to the identity map. However, $A$ is fundamentally different from $I$ in that $A$ changes the space, and $I$ doesn't. The problem here is that if we can change our perspective in the output space, we can make it seem as though $A$ didn't do anything at all; we can take the image of our input basis to be our output basis.

In order to get an idea of how $A$ changes a space, we restrict ourselves to similarity. That is, we impose the constraint that the basis we choose for the output space should be the same as what we choose for the input space; this basis provides us with a fixed frame of reference. A useful choice of basis, in this instance, is $\mathcal B = \{(1,1),(1,-1)\}$. Relative to this basis, we find that $$ [A]_{\mathcal B} = B = \pmatrix{1&0\\0&-1} $$ we see that the similar matrix $B$ is clearly a reflection since it flips the sign of the $y$-coordinate.

One notable property is that if $A$ is similar to $B$, then $A^n$ is similar to $B^n$ for any $n$.

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    $\begingroup$ Thanks for the help! What do you mean when you say that "$B$ is the matrix of $A$ with respect to these new matrices"? $\endgroup$ – user17902 Feb 12 '17 at 20:15
  • $\begingroup$ With respect to the new bases! Typo there. $\endgroup$ – Ben Grossmann Feb 12 '17 at 20:22
  • $\begingroup$ Still a little confused as to what you mean with the phrase "$B$ is the matrix of $A$." $\endgroup$ – user17902 Feb 12 '17 at 20:28
  • $\begingroup$ @user17902 do you know what is mean by "the matrix of a linear transformation" with respect to a basis? $\endgroup$ – Ben Grossmann Feb 12 '17 at 20:36
  • $\begingroup$ I think I see...just the matrix defined by the image of the basis vectors in $\mathbb{R}^n$ using the basis vectors of $\mathbb{R}^m$? $\endgroup$ – user17902 Feb 12 '17 at 20:39

Two rectangular matrices (with the same format) are equivalent iff they have the same rank.

Similarity is also an equivalence relation but is finer than matrix equivalence (similarity implies matrix equivalence but the converse is false) : for example, similar matrices have the same eigenvalues with the same multiplicity (since they have the same characteristic polynomial).

  • $\begingroup$ I would say "restrictive" or "strict" rather than "accurate" $\endgroup$ – Ben Grossmann Feb 12 '17 at 20:06
  • $\begingroup$ @Omnomnomnom: Thank you. I was trying to translate from the french "relation plus fine que ...". $\endgroup$ – Adren Feb 12 '17 at 20:09
  • $\begingroup$ "a finer relation than..." works here $\endgroup$ – Ben Grossmann Feb 12 '17 at 20:11
  • $\begingroup$ @Omnomnomnom: Ok :) Editing ... $\endgroup$ – Adren Feb 12 '17 at 20:27

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