# Schur factorization from spectral decomposition

Let $$A \in \mathbb C^{d \times d}$$. The Schur factorization of $$A$$ is the decomposition $$A = U T U^\ast$$, where $$T \in \mathbb C^{d \times d}$$ is an upper triangular matrix and $$U \in \mathbb C^{d \times d}$$ is unitary.

If $$A$$ is symmetric, then it follows immediately that $$T$$ is diaogonal. This means, the spectral decomposition of symmetric matrices can be reduced to the Schur factorization of general matrices.

I am wondering about opposite relations. Can I reduce the Schur factorization of general matrices to the spectral decomposition of symmetric matrices?

To illustrate the point, if I can compute the singular value decomposition of any symmetric matrix $$S$$, then I can compute the SVD of any other matrix $$A$$ by computing the SVD of $$S = A^t A$$ and some additional matrix-vector multiplications. I wonder whether anything analogous can be shown for the Schur decomposition.