# multiplicity of eigenvalues and dimension of vector space

Given a diagonalizable linear operator $T$ on vector space $V$,

The characteristic polynomial of $T$ splits and the multiplicities of eigenvalues for $T$ sums up to the $\dim(V)$.

I'm having difficulty in understanding why the multiplicities of eigenvalues for $T$ sums up to the $\dim(V)$.

## 1 Answer

a) As $T$ is diagonalizable, tehre exists a basis according to which $T$ is represented by a diagonal matrix. The mukltiplicitites of eigenvalues are the multiplicities of the various values occuring on the diagonal. The sum of these multiplicities is tehrefore tha number of diagonal entries, i.e., $\dim V$.

b) The degree of the characteristic polynomial is $\dim V$. The multiplicities of eigenvalues are the multiplcities of the roots of the characteristic polynomial. The total number of roots (counting multiplicities) is the degree (as by assumption all roots exist in the field we work in)

• hmm.. im clueless still – Little Rookie Jan 9 '17 at 14:42