intuition for similar matrix If matrices $A$ and $B$ satisfy the definition of similar matrix, i.e. $B=PAP^{-1}$, then we can interpret $A$ and $B$ are the same linear transformation under different basis. But my question is how to "grok" this interpretation by just look at the definition of similar matrix?
 A: Sure. $P$ is a linear transformation which takes things to a new basis. after applying $B$, you bring your (transformed) basis back to where it used to be (transformed). If this is the same thing as your original transformation ($A$), then $A$ and $B$ must fundamentally be the same transform, in the different basis. I visualize it with a rotation example: rotate to a new basis ($P$), do your rotation $B$, and rotate back $(P^{-1})$. If this is the same as one rotation $A$, then they were the same.
$Start\rightarrow^P Rotated$
$\downarrow A\hspace{40 pt}\downarrow B$
$End \leftarrow^{P^{-1}}  Twice$
If you end at the same place then $A$ and $B$ must be the same.
A: I always understood that the matrix $P$ is a change of basis matrix from the domain to itself. So as the codomain and the domain are the same vector space, the definition is saying $A$ and $B$ are similar if transforming by $A$ is the same as changing basis, then transforming via $B$, then changing back again.
Hope it helps.
A: My current understanding, not sure this helps:
$B=PAP^{-1}$ means $BP=PA$, so if $A$ is a diagonal, $BP=PA$ is just the definition of eigenvetors, which is easy to understand. Because we know a matrix and its diagonal form are the same transformation under different basis, so to extrapolate to non-diagonal cases, we can define any square matrix of the form $B'P=PA'$ or $B'=PA'P^{-1}$ representing the same transformation under different basis, called similar matrix.
Edit: $B=PAP^{-1}$ means $A=P^{-1}BP$, and think of $P$ as the change of basis matrix, then $P^{-1}BP$ means change of basis first, then do the linear transformation, then change the basis back, so $A$ and $B$ represent the same linear transformation under different bases. See here: https://youtu.be/P2LTAUO1TdA?t=9m12s
A: I'm not sure if this helps, but...
The way I think of it is that if $\pi \in \mathbb{R}^n$ is the representation of the vector $v$ in the basis formed from the columns of $P$ (ie, $v = \sum \pi_i p_i$), then $P\pi$ is the representation of $v$ in the basis $e_1,...e_n$ (since $v = \sum \pi_i p_i = \sum \pi_i P e_i = P \sum \pi_i e_i = \sum [P \pi]_i e_i$).
Specifically, this means that $P$ is the identity mapping with different bases for the domain and range ($v$ in basis $p_1,...,p_n$ maps to $v$ in basis $e_1,...,e_n$).
Similarly, $P^{-1}$ is the identity mapping with the domain and range bases swapped.
So, I think of the $P$ and $P^{-1}$ in the expression $P^{-1}AP$  as identity mappings, and both $A$ and $B$ represent the same linear operator in some 'coordinate-free' space.
A: Consider the extreme case, i.e $A = Q\Lambda Q^{-1}$. Since Q is orthonormal, $Q' = Q^{-1}$. When apply A to vector x, first x is projected to another Cartesian coordinate represented by Q, $\{q_1, q_2, ...\}$, then stretched, and then rotated back. It's a change of basis operation. The stretch is controlled by $\Lambda$, geometrically the shape of the hyperellipse in $\{e_1, e_2, ...\}$ (after projection)coordinate. A is a hyperellipse in original coordinate. So similar matrix means two matrices have the same shape of hyperellipse.
