Can we define the eigenvectors of $M$ as the columns of a matrix $P$, such that $P \cdot \text{Diagonal}\cdot P^{-1} = M$ I understand the standard definition of eigenvectors, and how we calculate them. But a question in a book I'm reading has left me a bit confused:
He gives a matrix $M$, and asks:
"Find a basis with respect to which this matrix has that diagonal representation"
He finds the solution simply by calculating the eigenvectors.
I'm very well aware that diagonalizability and eigenvectors are closely connected, but, can we define the eigenvectors of $M,$ as vectors whose concatenation is the change of basis matrix $P,$ taking the diagonal matrix to $M?$
If true, this seems to make reasonable sense to me, and I quite like this definition. But I cannot prove that vectors with this property are only scaled after being transformed by $M.$ I've tried from both directions but failed.
(the book in question can be found here: http://joshua.smcvt.edu/linearalgebra/book.pdf. The question can be seen as an imgur picture here: https://imgur.com/a/BBqpc , and can be found in Chapter five, II.3, question 3.32 b. Solution manual is here: http://joshua.smcvt.edu/linearalgebra/jhanswer.pdf )
Any help is appreciated. Have a nice day!
 A: The best you can do in general is defined by the Jordan Canonical form.
In general you can only write: $$AM=MJ$$
https://en.wikipedia.org/wiki/Jordan_normal_form
A: You wrote "i quite like this definition", but it does not appear that "definition" is the right word for what you're trying to say.
Let $e_i$ by the column vector whose $i$th component is $1$ and whose other components are $0.$
Let $v_i$ be the $i$th eigenvector, with eigenvalue $\lambda_i.$
The $Mv_i = \lambda_i v_i.$
Further suppose there is a basis consisting of independent eigenvectors. (For some matrices, there are not enough eigenvectors for that.)
Let $P$ be a matrix whose columns are independent eigenvectors. Then, since there are "enough" eigenvectors, the columns span the image space, so $P$ is invertible. Let the diagonal matrix $D$ have $\lambda_i$ as its $(i,i)$ entry.  Then we have
$$
P^{-1} v_i = e_i, \quad DP^{-1} v_i = D e_i = \lambda_i e_i, \quad PDP^{-1} v_i = P(\lambda_i e_i) = \lambda_i P e_i = \lambda_i v_i.
$$
Since this works for every eigenvector and those form a basis, we must have $M = PDP^{-1}.$
A: If you write the identity as $MP=PD$ it follows at once that the columns of $P$ are eigenvectors of $M$. (Just look at the matrix products column by column.)
A: In general we can not do this. Because many matrices which are not diagonalizable (there exists no such diagonal $\bf D$ matrix) do still have eigenvectors (vectors which fulfill ${\bf Mv=}\lambda {\bf v}$).
A famous example is 
$${\bf M} = \left[\begin{array}{cc}1&1\\0&1\end{array}\right]$$
${\bf v} = [1,0]^T$ is an eigenvector, but you will not be able to diagonalize $\bf M$
