[Math] Algebraic multiplicity = geometric multiplicity

Question

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 ?

Answer 1

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 $dim Ker(A-I)=1$.