I am working on a problem where I see that a bi-diagonal matrix is the solution. I have a sparse real-valued (all either 1 or zeros) square matrix as the input. Can any square real-valued sparse matrix be bi-diagonalizable? Presence of complex numbers are alright.
2026-04-01 23:07:15.1775084835
Can every real valued square matrix be Bi-diagonalizable? (lower or upper, including complex numbers)?
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A real matrix is bidiagonalisable over $\mathbb R$ if and only if it has a real spectrum. If it has a real spectrum, then it is similar over $\mathbb R$ to its Jordan form, which is bidiagonal. Conversely, if it is similar to a bidiagonal matrix over $\mathbb R$, its eigenvalues are identical to the diagonal entries of the bidiagonal matrix, which are real.