An $n \times n$ matrix is diagonalizable if there exists a basis of ${\bf C}^n$ consisting of eigenvectors of $A$. (We use ${\bf C}^n$ rather than ${\bf R}^n$ in order to allow complex eigenvectors.) Not every matrix is diagonalizable, but almost every matrix is. This has a precise mathematical meaning. The space of $n \times n$ real matrices is $n^2$-dimensional, and only an $(n^2-1)$-dimensional subset fails to be diagonalizable.

The standard example of a non-diagonalizable matrix is $A= \begin{pmatrix} 1&1 \cr 0&1 \end{pmatrix}$. The characteristic polynomial is $p_A(\lambda) = (\lambda-1)^2$, whose only root is $\lambda=1$, but $E_1$ is only 1-dimensional, being all multiples of $\begin{pmatrix} 1 \cr 0 \end{pmatrix}$.

• The algebraic multiplicity $m_a(\lambda)$ of an eigenvalue $\lambda$ is its multiplicity as a root of the characteristic polynomial. In this case 1 is a double root, so $m_a(1)=2$.
• The geometric multiplicity $m_g(\lambda)$ is the dimension of $E_\lambda$. In this case $m_g(1)=1$.

There are two important results about algebraic multiplicity, with the first being a stepping-stone towards the second:

Theorem 1: For any eigenvalue $\lambda$, $1 \le m_g(\lambda) \le m_a(\lambda)$.


Main Theorem: The following three conditions are equivalent.

1. $A$ is diagonalizable.
2. The sum of the geometric multipicities is $n$
3. For each eigenvalue $\lambda$, $m_g(\lambda)=m_a(\lambda)$.

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