# User:IssaRice/Linear algebra/Classification of operators

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Let $V$ be a finite-dimensional inner product space, and let $T : V \to V$ be a linear transformation. Then in the table below, the statements within the same row are equivalent. $T$ is diagonalizable There exists a basis of $V$ consisting of eigenvectors of $T$ $T$ is diagonalizable (there exists a basis $\beta$ of $V$ with respect to which $[T]_\beta^\beta$ 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 $\dim V$ distinct eigenvalues so the diagonal matrix should be unique up to order? If $T$ is the identity map, then every non-zero vector $v \in V$ is an eigenvector of $T$ with eigenvalue $1$ because $Tv = 1v$. Thus every basis $\beta = (v_1,\ldots,v_n)$ diagonalizes $T$. The matrix of $T$ with respect to $\beta$ is the identity matrix. $T$ is normal There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ $T$ is diagonalizable using an orthonormal basis $T$ self-adjoint ( $T$ is Hermitian) There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ with real eigenvalues $T$ is diagonalizable using an orthonormal basis and the diagonal entries are all real $T$ is an isometry There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ whose eigenvalues all have absolute value 1 $T$ 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 $T$ is positive There exists an orthonormal basis of $V$ consisting of eigenvectors of $T$ with positive real eigenvalues $T$ is diagonalizable using an orthonormal basis and the diagonal entries are all positive real numbers