Definition 1: The matrices A and B are called similar if they represent the same linear transformation but in (possibly) different bases.
Definition 2: The matrices A and B are called similar if there is an invertible matrix P such that A = PBP^−1.
I'm supposed to show whether or not these two definitions are equivalent (saying the same thing). I know that they are but I am having trouble putting it into words. I'm also supposed to describe what it is the matrices P and P inverse do. At one point this was a homework question but I'm now just trying to understand the concepts. Any help is appreciated!
I'll give an analogy that will tell you what the role of $P$.
Imagine that Harry Potter has a spell that takes (only) black coloured broken chairs and fixes it , keeping it black after fixing . However, you have pink coloured chairs at home (I do so), and would like them to come out pink as well. Then, you do the following:
Paint your pink coloured chair black.
Ask Harry Potter to fix the broken black chair.
Take the fixed black chair and paint it pink.
Now, that is precisely what is happening here, with $P$ being the "paint black chairs pink" operation on chairs, and $B$ being Harry Potter. The first operation is the inverse of $P$. $A$, is then the procedure of fixing pink chairs.
Done with the analogy?
Now, remember the following very crucial statement : the matrix representation of a linear transformation depends on the basis with respect to which the matrix is written.
That is, when you try to figure out what $B$ does to a vector, you need to present the vector in the right form to $B$. So, for example, if $A$ is written with respect to the basis $\mathcal{A}$ and $B$ with respect to $\mathcal{B}$, then giving to $B$, a vector expressed in the basis $\mathcal{A}$, and expecting it to operate correctly, is an incorrect expectation. At least Harry Potter would refuse you if you gave him a pick chair : $B$ will complain very little, and give you the wrong answer!
So, what is the role of $P$? Note that $A$ is represented with respect to $\mathcal{A}$, so it expects only vectors written in the basis $\mathcal{A}$ to be fed to it. But if $A$ wants the help of $B$, then it cannot directly feed the vector that is coming to it, to $B$. Rather, it needs to repaint(or convert) the vector in the basis $\mathcal{B}$. Then, $B$ will see the repainted vector, do the right job on it, and return it, but in basis $\mathcal{B}$. So now, you repaint : get back the vector in basis $\mathcal{A}$.
Hence, $P$ is referred to as a change of basis matrix, for the obvious reasons I have mentioned earlier.
Apart from the fact that Harry takes black chairs and the procedure $A$ takes pink chairs, the sort of crucial thing between both of them is the fact that they fix the chair. That is, up to the job of painting/repainting the chair, which is, a reversible operation, they don't really differ in what they do. This is why the procedures are *similar" : they are doing essentially the same thing, but to different coloured chairs, and it is easy to go from one colour to the other.
Now, to your question : if $A,B$ are similar, then they are the same linear transformation in different bases, if and only if $A = PBP^{-1}$ for some invertible $P$.
What does "the same linear transformation" mean, without invoking change of basis? Note that it is not equality of matrices. What it means is the following : if you fix coefficients $c_i$, and $A$ changes $\sum c_ia_i$ to $\sum d_ia_i$, then $B$ changes $\sum c_ib_i$ to $\sum d_ib_i$.
That is, in some "coordinate free" sense, $A$ and $B$ are the same. I can explain it better, but wish not to.
So, we proceed with this notion.
(Allow me, from this point, to be a little more precise).
One direction : Let $A,B$ be the same linear transformation, but in different bases, say $\mathcal{A}$ and $\mathcal{B}$. Let $P$ be the matrix formed as follows : since $\mathcal{B}$ is a basis, we can express each element $a_i$ of $\mathcal{A}$ as $a_i = \sum_{j=1}^n c_{ij}b_j$, where $b_j$ are the elements of $\mathcal{B}$. Now, the matrix $P = c_{ij}$ does the job. I leave you to see this.
The other way is even simpler, and I leave you to do it. But please proceed only once you have understood what I have said before this section of the answer.