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Schur decomposition theorem states that for any linear operator A on a complex vector space V, there exists a unitary operator U such that, when we apply U on A, it transforms A into an upper triangular matrix T, i.e., U^(-1)AU = T. Since A and T are similar matrices, they will have the same eigenvalues and determinant. Hence, we can find a cube root for T, say R, and use it to compute the cube root for A.

Let A and T be as described in step 1. That is, A is the given invertible operator and T is the upper triangular matrix similar to A obtained using Schur decomposition. To find a cube root for T (denote it as R), we first look at its diagonal entries. The diagonal entries of T are its eigenvalues, and we know that every complex number has a unique cube root. So, we can define R to have diagonal entries as the cube roots of the diagonal entries of T. Now, we need to show that the upper triangular nature is preserved under matrix multiplication (in particular, under matrix exponentiation). This can be done by induction, proving that if P and Q are upper triangular matrices, then so are PQ and P^k for k = 1, 2, 3. For any upper triangular matrix, multiplying it by another upper triangular matrix will always result in an upper triangular matrix. Thus, if we take R to be an upper triangular matrix whose diagonal entries are the cube roots of the diagonal entries of T, then R^3 will also be an upper triangular matrix with the same eigenvalues as T.

Now that we've found a cube root R for the upper triangular matrix T, we can use the unitary operator U from the Schur decomposition to find a cube root for A. Since U is unitary, its inverse is simply its adjoint, i.e., U^(-1) = U*. We know that R^3 = T, so U^(-1)AU = T ^<=> AU = U*R^3. To prove that A has a cube root, we need to find an operator S such that S^3 = A. Let's define S = UR. Then S^3 = URRRU* = U*T*U* = A. Thus, we have found a cube root S for the invertible operator A. In conclusion, we have shown that for any invertible operator A on a complex vector space V, there exists an operator S such that S^3 = A. This proves that every invertible operator on a complex vector space has a cube root.