Once again, M is diagonalizable if PM/P = D, where P is nonsingular and D is diagonal.
Let's look at the eigen vectors of D. Take a unit coordinate vector and multiply by D to get a scaled version of that vector. Thus D has n independent eigen vectors, one for each coordinate.
In the last section we showed that eigen vectors in one matrix map onto the eigen vectors of a similar matrix, and this map preserves linear independence. Therefore every diagonalizable matrix (which is similar to a diagonal matrix) has n independent eigen vectors.
Conversely, let M have n independent eigen vectors. Let P map the unit basis onto these n independent vectors. Apply M, which scales these vectors, then apply P inverse, which pulls the vectors back to the coordinate axes. The composite is a matrix that scales the coordinate vectors, in other words, a diagonal matrix. Therefore a matrix is diagonalizable iff it has n independent eigen vectors.
An example of a matrix with too few eigen vectors is [1,1|0,1]. It has one eigen value, 1, and one eigen vector, the y axis. It is not similar to a diagonal matrix, i.e. not diagonalizable.
If D is the identity matrix, or a scale multiple of the identity matrix, it commutes with P. Write PM = DP = PD and cancel P on the left to get M = D. Only the identity matrix is similar to itself.