Eigenvectors corresponding to distinct eigenvalues are linearly independent.
Suppose $M$ has $n$ linearly independent eigenvectors,${ v_1, v_2, … v_n}$, with corresponding eigenvalues ${ \lambda_1, \lambda_2, … \lambda_n}$
Suppose \(w = \alpha_1 v_1 + \alpha_2 v_2 + ... + \alpha_n v_n\) \(\implies Mw = \alpha_1 \lambda_1 v_1 + \alpha_2 \lambda_2 v_2 + ... + \alpha_n \lambda_n v_n\)
In this representation $(\alpha_1, … \alpha_n)$ are co-ordinates of $W$ in ${v_1, … v_n}$. If we change basis from ${ e_1, e_2, … e_n}$ to ${ v_1, v_2, … v_n }$, then the linear transformation corresponding to $M$ acts like a diagonal matrix.
So in order to change our basis, we must premultiply by a change of basis matrix, say $P$, where $P$ converts from the new basis to our old standard basis. Thus, if in the old basis, we had $w \mapsto Mw$, then in our new basis, we have $\bar w = P^{-1}w$, and the new mapping $\bar w \mapsto P^{-1}MP \bar w$
So if we wished to think of this linear transform in our new basis only, as some transform $\bar M$, such that $\bar w \mapsto \bar M \bar w$, then clearly we have $\bar M = P^{-1}MP$.
Since application happens from right to left, we can think of what happens to $\bar w$ sequentially, for a more intuitive understanding.
Note that the eigenvalues of $\bar M$ will be given directly by the diagonal elements of the matrix, which are simply all the $\lambda_i$’s, i.e the same as the eigenvalues of of $\bar M$
Consider a relation for $n \times n$ matrix defined as such
$M \equiv \bar M$ iff there exists invertible $P$ such that $P^{-1}MP = \bar M$.
$M$ is said to be diagonalizable iff $M \equiv$ to a diagonal matrix.
Diagonalizable matrices have some useful properties which stem from the diagonal matrices they are equivalent to.
\(M^tv\) \(= (PDP^{-1})^tv\) \(= (PD^tP^{-1})v\)
And computing $D^t$ itself is trivial given $D$ is a diagonal matrix; we simply take the elements along the diagonal to the power $t$ for each one.