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This question already has an answer here:

Some square matrices, like $ \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)$, don't have a complete set of eigenvectors. By complete I mean that the eigenvectors span the entire space. The only eigenvector of the previous matrix is $\left( \begin{array}{c} 1 \\ 0 \end{array} \right)$. These are called defective matrices. How can one prove whether a matrix is or is not defective?

The reason that I ask is that I know stochastic matrices are supposed to have an eigen-decomposition but I don't know how to show it?

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marked as duplicate by Git Gud, Carl Mummert, drhab, apnorton, Hamou Oct 13 '14 at 0:14

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A matrix is not defective if and only if is diagonalizable, so if you calculate the characteristic polynomial of a matrix $A$, that is $det(A-tI)$, you can find the algebraic multiplicity of any eigenvalue. Now you must calculate the geometric multiplicity of any eigenvalue $\lambda$ that is the size of $Ker(A-\lambda I)$ and if it is equal to the geometric multiplicity for any eigen value $\lambda$ you have that the matrix is diagonalizable and is not defective.

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