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I've read somewhere that Hessenberg decomposition is not unique unless the first column of $Q$ in $Q^{T}AQ =H$ is specified. But then, if I am given a matrix $A \in R^{n \times n}$, I can apply the Householder reduction algorithm to reduce $A$ to Hessenberg form $H$ which is the unique output of the algorithm. Then how is that not unique?

Amzoti
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Elnaz
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    But you may also apply the Hessenberg reduction to $PAP$ where $P=I-2vv^T/v^Tv$ is a Householder reflector. This premultiplication can not be undone by the reflectors of the Hessenberg reduction algorithm. See "bulge chasing methods" in implicitely shifted QR algorithms. – Lutz Lehmann Apr 22 '14 at 22:45
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    For those who wonder where the statement for Hessenberg decomposition is not unique presents: Golub, G.H. & Van Loan, C.F. (2013), Matrix Computations, 4-Ed., p. 381. – Herpes Free Engineer Apr 28 '19 at 10:28

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Every $2\times 2$ matrix $A$ is already in Hessenberg form. If you choose a transformation matrix $Q$ (it can be also orthogonal) and compute

$$B = QAQ^{-1}\text{,}$$

then probably, $A\neq B$ for some $Q$s.

Antoine
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