Two n-by-n matrices A and B are called similar if there exists an invertible n-by-n matrix S such that
Theorem: If λ is an eigenvalue of a square matrix A, then its algebraic multiplicity is at least as large as its geometric multiplicity.    ▣
Let x1, x2, … , xr be all of the linearly independent eigenvectors associated to λ, so that λ has geometric multiplicity r. Let xr+1, xr+2, … , xn complete this list to a basis for ℜn, and let S be the n×n matrix whose columns are all these vectors xs, s = 1, 2, … , n. As usual, consider the product of two matrices AS. Because the first r columns of S are eigenvectors, we have
\[ {\bf A\,S} = \begin{bmatrix} \vdots & \vdots&& \vdots & \vdots&& \vdots \\ \lambda{\bf x}_1 & \lambda{\bf x}_2 & \cdots & \lambda{\bf x}_r & ?& \cdots & ? \\ \vdots & \vdots&& \vdots & \vdots&& \vdots \end{bmatrix} . \]
Now multiply out S-1AS. Matrix S is invertible because its columns are a basis for ℜn. We get that the first r columns of S-1AS are diagonal with &lambda's on the diagonal, but that the rest of the columns are indeterminable. Now S-1AS has the same characteristic polynomial as A. Indeed,
\[ \det \left( {\bf S}^{-1} {\bf AS} - \lambda\,{\bf I} \right) = \det \left( {\bf S}^{-1} {\bf AS} - {\bf S}^{-1} \lambda\,{\bf I}{\bf S} \right) = \det \left( {\bf S}^{-1} \left( {\bf A} - \lambda\,{\bf I} \right) {\bf S} \right) = \det \left( {\bf S}^{-1} \right) \det \left( {\bf A} - \lambda\,{\bf I} \right) \det \det \left( {\bf S} \right) = \det \left( {\bf A} - \lambda\,{\bf I} \right) \]
because the determinants of S and S-1 cancel. So the characteristic polynomials of A and S-1AS are the same. But since the first few columns of S-1AS has a factor of at least (x - λ)r, so the algebraic multiplicity is at least as large as the geometric.    ◂