# How come a matrix not diagonalizable if the geometric multiplicity is less than the algebraic multiplicity?

I understand that the algebraic multiplicity is the number of times an eigenvalue appears as a root of the characteristic polynomial. I also understand that the geometric multiplicity is the dimension of the eigenspace corresponding to some eigenvalue. I do not understand why having an eigenvalue appear m+1 times in the characteristic polynomial while having a dimension of m means that it is not diagonalizable. Is these something I am missing in my understanding of one of these concepts? It is not clear to me how they are connected.

• Contrapositive. Suppose $A$ is diagonalisable, so that say $X^{-1}AX = \Lambda$ (diagonal matrix of eigenvalues). Then columnwise, $AX = X\Lambda$ is of the form $Ax = \lambda x$, and you get as many independent eigenvectors for each $\lambda$ as the multiplicity of $\lambda$ (the number of times it appears along the diagonal of $\Lambda$) – the vectors being independent because they're columns of the invertible matrix $X$. – M. Vinay Mar 19 at 2:42
• – John Douma Mar 19 at 2:47
• Following what M. Vinay said, now that means that there are $n$ independent eigenvectors if the ambient space has dimension $n$. But the sum of the algebraic multiplicities is $n$, so this cannot happen unless all the geometric multiplicities equal the corresponding algebraic multiplicities. – Ian Mar 19 at 2:49
• diagonalizable if and only if the minimal polynomial is squarefree ( product of distinct linear terms) – Will Jagy Mar 19 at 3:00
• Simply because the sum of the dimension of the eigen spaces don't add up to the dimension of the vector space. Hence there's no basis consisting of eigenvectors ... – Michael Hoppe Mar 19 at 11:15