Cube root of a matrix 2026-10-25
A cube root of a matrix is a square matrix of the same size over the specified field satisfying . Every complex invertible matrix has one: put it in Jordan normal form, choose a scalar cube root of each nonzero eigenvalue, and apply the truncated binomial cube root of the identity plus a nilpotent matrix on each block. The Nilpotent Jordan block has none, even though its square is nonzero. If , then , making a nilpotent matrix of size three; the Cayley-Hamilton theorem then gives , a contradiction.
We construct a cube root of a matrix on each Jordan block. Since is an invertible matrix over the complex numbers, its Jordan normal form isConsider one block , where is a Nilpotent Jordan block and . Choose a complex cube root of , so .
We need a polynomial with constant term one satisfyingThis can be constructed using only polynomial multiplication. Write . Once have been chosen, the coefficient of in is plus an expression depending only on those earlier coefficients. Choose to make this coefficient one at and zero at . Division by three is possible over the complex numbers, so the recursion always succeeds. Its first coefficients areThis is the truncated binomial cube root of the identity plus a nilpotent matrix; equivalently , where the generalized binomial coefficient is and the empty product is one.
Substitute . The polynomial remainder is a multiple of , hence zero. It follows thatFor , take , so the same construction covers single-entry blocks. Assemble the block diagonal matrix . Taking matrix powers respects these blocks, and conjugating back givesThus every invertible matrix over the complex numbers has at least one cube root of a matrix. The construction uses only finitely many matrix powers.