Algebric and geometric multiplicity and the way it affects the matrix Given a matrix $A$. Suppose $A$ has $\lambda_1,\dots,\lambda_n$ eigenvalues each with $g_i$ geometric multiplicity and $r_1,\dots,r_n$ algebric multiplicity, $g_i\leq r_i$.
Given this information alone, can I understand the way the matrix might look like?
And more in general, how does the way algebric and geomtric multiplicties affect the matrix?
 A: All square matrices $A$ satisfy that the algebraic multiplicity is larger or equal than the geometric multiplicity. Hence $g_i\le r_i$ does not give any new information on how the matrices may look like. If a matrix of size $n$ has the sum of geometric multiplicities equal to $n$, then it is diagonalizable. Then also the sum of the algebraic multiplicites is equal to $n$.
A: The difference between algebraic and geometric comes from the number of linearly independent eigenvectors.
Geometric multiplicity is strictly less than algebraic multiplicity if and only if the number of linearly independent eigenvectors is less than $n$ and some eigenvectors have to be repeated in an eigendecomposition of $A$.
The eigenvectors then do not span the space and do not give a basis.
This means that $A$ can not be diagonalized.
Eigenvalue multiplicities are basis independent so they do not say much about the appearance of $A$ itself, since any transformation will leave them unchanged. They do constrain the Jordan normal form though, but the full specification of the Jordan normal form requires the actual eigenvector multiplicities. For each distinct eigenvector in a given eigendecomposition there is a block with the size of that eigenvectors multiplicity filled with the corresponding eigenvalue.
