Is matrix similarity transitive? What is the difference between a similar matrix and a diagonal matrix? According to my textbook, the definition for both is basically: $B=P^{-1}AP$.
Say if there are three matrices: $A$, $B$ and $C$. If $A$ is similar to $B$, and $B$ is similar to $C$, does this mean that $A$ is similar to $C$?
 A: If $A$ is similar to $B$ and $B$ is similar to $C$, then yes, $A$ is in fact similar to $C$.
Proof By definition, $A$ is similar to $B$ means that there exists an invertible matrix $P$ such that $$B = PAP^{-1}.$$ Similarly (pun not intended), there exists an invertible matrix $Q$ such that $C = QBQ^{-1}$ and therefore $B = Q^{-1}CQ.$ Replace $B$ in the first equation and you get
$$ Q^{-1}CQ = PAP^{-1} $$
$$ QQ^{-1}CQ = QPAP^{-1} $$
$$ CQQ^{-1}= QPAP^{-1}Q^{-1}$$
$$ C = QPAP^{-1}Q^{-1} = QPA(QP)^{-1} $$
Thus, we can see clearly that there exists an invertible matrix $QP$ that makes $A$ similar to $C$ and the proof is done. 
A: Similarity is a special case of equivalence. Equivalence is defined as: 
$A \sim B$ if there are invertible matrices $P$ and $Q$ such that $PAQ=B$. The matrices $A$ and $B$ needn't be square. The reason why it's called equivalence, is that this definition satisfies the requirements of an equivalence relation. 
Now similarity is the special case where $A$ and $B$ are square and the invertible matrices are inverses of each other. So $A$ is similar to $B$ if there is an invertible matrix $P$ so that $P^{-1}AP=B$. And that is really all there is to it...neither $A$ nor $B$ need to be diagonal per se.
Now the significance of similarity becomes apparent when you interpret this in terms of linear algebra, where $A$ and $B$ are matrix representations of linear operators over some vector space, and the matrix $P$ is a change of basis matrix. If you have some matrix $A$ and you want to study how it affects certain vectors, it is a lot easier to do this if the matrix $A$ is diagonal...so if you can find a change of basis matrix so that the representation of $A$ relative to the new basis is diagonal that would be great...that is why we are especially interested in the case where $A$ is similar to $D$, a diagonal matrix. Of course this is not always possible, but then there are other (simple) canonical similarity forms we can try.
