Triangularisation

Statement

Let $F$ be a field and let $A$ be an $n \times n$ matrix over $F$. Then $A$ is similar to an upper triangular matrix over $F$ if and only if the characteristic polynomial of $A$ splits over $F$. Here, to split means to be a product of linear factors. In particular, over an algebraically closed field, such as $\C$, each square matrix is similar to an upper triangular matrix.

Remarks

The diagonal entries of the triangular matrix are the eigenvalues of $A$, with their algebraic multiplicities. The proof is by induction: choose an eigenvector, extend it to a basis, and work on the quotient space. Over $\C$, the matrix $P$ can even be chosen unitary; this is the Schur decomposition.

Explanations

L1 human

Not every matrix can be diagonalised, but over $\C$ every matrix can be made triangular by a change of basis. A triangular form is almost as useful: its eigenvalues are on the diagonal. Over $\R$, a rotation matrix cannot be triangularised, because it has no real eigenvalue.