# Make the representative (diagonal)matrix of L

damabo

## Homework Statement

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

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

## Homework 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 $\phi = det(X I_n - A)$ . Because phi equals a product of first degree factors, I guess that $\phi=(X-λ_i)^{m_i}$

## 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. ${ v_{1,1},v_{1,2},...,v_{1,m_1},v_{2,1},...,v_{2,m_2},...,v_{k,1},...,v_{k,m_k} }$ is the set of eigenvectors. Since β is a basis of V, it should have n eigenvectors (thus $m_1+m_2+...+m_k = n$).
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.

I understand there is a theorem which says that $d_i ≤ m_i$, 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 $Ʃd_i = n$ and $m_i ≤ d_i$ for all i.