Paper 2, Section II, 17B

Numerical Analysis | Part IB, 2021

(a) Define Householder reflections and show that a real Householder reflection is symmetric and orthogonal. Moreover, show that if H,ARn×nH, A \in \mathbb{R}^{n \times n}, where HH is a Householder reflection and AA is a full matrix, then the computational cost (number of arithmetic operations) of computing HAH1H A H^{-1} can be O(n2)\mathcal{O}\left(n^{2}\right) operations, as opposed to O(n3)\mathcal{O}\left(n^{3}\right) for standard matrix products.

(b) Show that for any ARn×nA \in \mathbb{R}^{n \times n} there exists an orthogonal matrix QRn×nQ \in \mathbb{R}^{n \times n} such that

QAQT=T=[t1,1t1,2t1,3t1,nt2,1t2,2t2,3t2,n0t3,2t3,3t3,n00tn,n1tn,n]Q A Q^{T}=T=\left[\begin{array}{ccccc} t_{1,1} & t_{1,2} & t_{1,3} & \cdots & t_{1, n} \\ t_{2,1} & t_{2,2} & t_{2,3} & \cdots & t_{2, n} \\ 0 & t_{3,2} & t_{3,3} & \cdots & t_{3, n} \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ 0 & \cdots & 0 & t_{n, n-1} & t_{n, n} \end{array}\right]

In particular, TT has zero entries below the first subdiagonal. Show that one can compute such a TT and QQ (they may not be unique) using O(n3)\mathcal{O}\left(n^{3}\right) arithmetic operations.

[Hint: Multiply A from the left and right with Householder reflections.]

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