Let be a finite-dimensional inner product space, and let be a linear transformation. Then in the table below, the statements within the same row are equivalent.

Operator kind | Description in terms of eigenvectors | Description in terms of diagonalizability | Notes | Examples |
---|---|---|---|---|

is diagonalizable | There exists a basis of consisting of eigenvectors of | is diagonalizable (there exists a basis of with respect to which is a diagonal matrix) | This basis is not unique because we can reorder the vectors and also scale eigenvectors by a non-zero number to obtain an eigenvector. But there are at most distinct eigenvalues so the diagonal matrix should be unique up to order? | If is the identity map, then every non-zero vector is an eigenvector of with eigenvalue because . Thus every basis diagonalizes . The matrix of with respect to is the identity matrix. |

is normal | There exists an orthonormal basis of consisting of eigenvectors of | is diagonalizable using an orthonormal basis | ||

self-adjoint ( is Hermitian) | There exists an orthonormal basis of consisting of eigenvectors of with real eigenvalues | is diagonalizable using an orthonormal basis and the diagonal entries are all real | ||

is an isometry | There exists an orthonormal basis of consisting of eigenvectors of whose eigenvalues all have absolute value 1 | is diagonalizable using an orthonormal basis and the diagonal entries all have absolute values 1 | This only works when the field of scalars is the complex numbers | |

is positive | There exists an orthonormal basis of consisting of eigenvectors of with positive real eigenvalues | is diagonalizable using an orthonormal basis and the diagonal entries are all positive real numbers |