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Suppose we are given a symmetric tridiagonal $4 \times 4$ matrix $A$ and perform QR factorization

$$A=QR$$

Then we define $A' := RQ$. Matrix $A'$ still possesses the tridiagonal structure where we can again repeat the steps to find the eigenvalues of $A$.

I know $A$ has to be symmetric tridiagonal (not just only tridiagonal), but is $A$ necessarily invertible? If so, please illustrate the proof or idea that invertibility is required. If not, please provide an example. Thank you very much.

Rodrigo de Azevedo
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    The QR algorithm is guaranteed to converge, if $A$ is symmetric tridiagonal and unreduced, that is, the entries on the lower and upper diagonal are nonzero. Such a matrix can still be singular, see also http://math.stackexchange.com/questions/521188/symmetric-tridiagonal-matrix?rq=1 and your other question :) – Algebraic Pavel Oct 13 '13 at 19:18
  • Actually $A$ does not need to be necessarily tridiagonal, but the assumption of distinct eigenvalues (or unreduced tridiagonal form) is quite important. – Algebraic Pavel Oct 13 '13 at 19:22

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