Make the representative (diagonal)matrix of L

1. The problem statement, all variables and given/known data

V is a linear vector space of dimension n
[itex]\phi = det(X I_n -A) [/itex] equals a product of first degree factors.
[itex]Spec(L)={λ_1,...,λ_k}[/itex] is the set of eigenvalues

show that: if L is diagonizable than d(λ_i)=m(λ_i)

2. Relevant equations

d_i=d(λ_i)= geometric multiplicity = dim E_i = dim {v in V| λ_i v = L(v) = A.v}
m_i=m(λ_i)= algebraic multiplicity = multiplicity at which λ_i is a zero-point of [itex] \phi = det(X I_n - A) [/itex] . Because phi equals a product of first degree factors, I guess that [itex]\phi=(X-λ_i)^{m_i}[/itex]

3. The attempt at a solution

If L is diagonizable, then V has a basis β of eigenvectors of L. write the eigenvectors v so that they correspond to the eigenvalues they belong to. the first m_1 eigenvectors (note that m_1= m(λ_1) will correspond to the eigenvector λ_1. [itex]{ v_{1,1},v_{1,2},...,v_{1,m_1},v_{2,1},...,v_{2,m_2},...,v_{k,1},...,v_{k,m_k} } [/itex] is the set of eigenvectors. Since β is a basis of V, it should have n eigenvectors (thus [itex] m_1+m_2+...+m_k = n[/itex]).
Now I am supposed to make the diagonal matrix of L with respect to β to show that d_i=m_i for all λ_i.

 

So, first, we need n independent eigenvectors to form a basis for V. this means that the sum of d_i's will be n.
I understand there is a theorem which says that [itex] d_i ≤ m_i[/itex], because phi will be of the form ( X - λ )^d * p(X). This means that there could be more zero-points of phi, than d.
So we know [itex]Ʃd_i = n[/itex] and [itex] m_i ≤ d_i [/itex] for all i.
since the Ʃm_i≤ n - the number of zeropoints will be smaller or equal to the total number of dimensions - we know that m_i=d_i for all i ?