# Homework Help: Make the representative (diagonal)matrix of L

1. Sep 19, 2012

### damabo

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

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)

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

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. ${ v_{1,1},v_{1,2},...,v_{1,m_1},v_{2,1},...,v_{2,m_2},...,v_{k,1},...,v_{k,m_k} }$ 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.

2. Sep 19, 2012

### HallsofIvy

I don't understand why you would need to "make the diagonal matrix of L" to prove this. What you have is correct- there exist a basis of eigenvectors of L so there are a total of n independent eigenvectors. Since the number of eigenvectors corresponding to any one eigenvector cannot be larger than its algebraic multiplicity, in order to get all n eigenvectors you must have each eigenvalue must have number of eigenvectors (its geometric multiplicity) equal to its algebraic eigenvalue.

3. Sep 19, 2012

### damabo

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 $d_i ≤ m_i$, 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.
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 ?