Let $F$ be an edomorphism of a vector space. If there exists an eigenvalue $\lambda$ that has algebraic multiplicty of $>1$ then F is not diagonalizable.

I need to show whether or not the statement above is true but the part about the algebraic multiplicity just throws me off. In the lecture note it says:

$F \in End(V)$ is diagonalizable, if there exists a basis $B=(b_1,\dots,b_n)$ of $V$ so that the matrix of $F$ has a diagonal shape:

\begin{equation} MB (F) = diag(\lambda_1,\dots,\lambda_n):=\left( \begin{array}{ccc} \lambda_1 & \dots &0 \\ 0& \ddots&0\\ 0 &\dots&\lambda_n \\ \end{array} \right) \end{equation} with $\lambda_1\dots\lambda_n\in\Bbb K\,$.

I don't see how the algebraic multiplicty of an eigenvalue comes into play.

In the script there's an example: (It was pointed out in the comments that the example I used indeed showed an eigenvalue with algebraic multiplicty 2. Since it's not helpful I've editted it out.)

Could someone help me understand the relation between the multiplicity of eigenvalues and the diagonlizability of an endomorphism?

  • 1
    $\begingroup$ The statement is not true, the 2x2 identity matrix has eigenvalue 1 with algebraic multiplicity of 2 and its diagonal. A well known theorem states that a matrix is diagonalizable iff the algebraic multiplicity coincides with the geometric multiplicity for each eigenvalue. $\endgroup$ May 6, 2021 at 23:10
  • $\begingroup$ @Ramita Oh okay could you please help me understand that theorem? $\endgroup$
    – ggraann
    May 6, 2021 at 23:55

1 Answer 1


Here's a full proof for the case of $\mathbb{R}$ https://www.math.tamu.edu/~mike.stecher/323/diagonalizationTheorems.pdf

The idea behind it is the following and it's similar in other spaces:

If both multiplicities coincide, then for each eigenspace (whose dimension is exactly it's eigenvalues multiplicity) you have a base of eigenvectors. As eigenspaces are in direct sum you have that the union of those basis is a base of the whole vector space i.e. you have a basis of eigenectors i.e. $F$ is diagonalizable.

Conversely, if $F$ is diagonalizable, you can prove that as $F=CDC^{-1}$ with $D$ diagonal then the characteristic polinomials of $F$ and $D$ are the same. Now for each eigenvalue $\lambda$, it must appear the exactly the same number of times in $D$'s diagonal as it's multiplicity (algebraic) in $F$'s ($=D$'s) characteristic polinomial, this is by definition of the polinomial.

As $D$ is diagonal if $\lambda$ appears $k$ times in it's diagonal then $D$ has a $k$-dimensional eigenspace associated to $\lambda$ thus $\lambda$'s geometric multiplicity (in $D$) is exaclty $k$. As $F$ and $D$ are simmilar matrices they are "essentialy the same matix" in the sense that their eigenspaces are "the same" via a change of basis (of course this can be formalized) so the $\lambda$-eigenspace of $F$ also has dimension $k$ thus $\lambda$'s geometric multiplicity (in $F$) is $k$ which is it's algebraic multiplicity.


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