# Algebraic multiplicity = geometric multiplicity?

I was wondering if algebraic multiplicity was equal to the geometric multiplicity. If the matrix (of size $n\times n$) is diagonalisable, i.e. the characteristic polynomial is of the form $$p(x)=(x-\lambda_1)^{m_1}\cdot ...\cdot (x-\lambda_k)^{m_k}$$ with $m_1+...+m_k=n$, I think that indeed algebraic multiplicity of $\lambda_i$ (i.e. $m_i$) and geometric multiplicity are the same. But is there cases where it doesn't hold ?

• If the matrix is not diagonalizable, the two are unequal. – Gerry Myerson Jun 24 '16 at 7:22
• A matrix is diagonalisable if its minimal polynomial has distinct roots, not the characteristic polynomial splits. And if you mean the usual definition of diagonalizability, then its algebraic and geometric multiplicity coincide. – cjackal Jun 24 '16 at 7:29

## 2 Answers

Sure, I can give a simple example:

$$A=\begin{pmatrix} 1 & 1\\ 0 & 1 \end{pmatrix}.$$

The characteristic polynomial is $(\lambda - 1)^2$, so the algebraic multiplicity os $2$, however, geometric multiplicity is $1$, indeed $dim Ker(A-I)=1$.

If $M$ is diagonalizable, then, $$V=E_1\oplus...\oplus E_k.$$ If it's not diagonalisable, then $$V\supset E_1\oplus...\oplus E_k$$ but $$V\neq E_1\oplus...\oplus E_k.$$ Therefore, $$m_1+...+m_k=n:=\dim(V)>\dim(E_1)+...+\dim(E_k),$$ and thus $$(m_1-\dim(E_1))+...+(m_k-\dim (E_k))>0.$$

You can easily show by induction that $m_k\geq \dim(E_k)$ for all $k$. But you can't say more.