As was already mentioned, the term “diagonalize” comes from a matrix-based perspective. Let $M$ be a matrix representation of $T$ relative to some basis $B$. Let

be a matrix whose column vectors are eigenvectors expressed relative to $B$. Thus,

where $\lambda_{i}$ is the eigenvalue associated to $v_{i}$. The above $n$ equations are more succinctly as the matrix equation

where $D$ is the diagonal matrix with $\lambda_{i}$ in the $i$-th position. Now the eigenvectors in question form a basis, if and only if $P$ is invertible. In that case, we may write

Thus in the matrix-based approach, to “diagonalize” a matrix $M$ is to find an invertible matrix $P$ and a diagonal matrix $D$ such that equation (1) is satisfied.

There are two fundamental reasons why a transformation $T$ can fail to be diagonalizable.

1. 1.

   The characteristic polynomial of $T$ does not factor into linear factors over $K$.

2. 2.

   There exists an eigenvalue $\lambda$, such that the kernel of $(T-\lambda I)^{2}$ is strictly greater than the kernel of $(T-\lambda I)$. Equivalently, there exists an invariant subspace where $T$ acts as a nilpotent transformation plus some multiple of the identity. Such subspaces manifest as non-trivial Jordan blocks in the Jordan canonical form of the transformation.