 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2017/digital-systems-linear-algebra-introduction-to-formal-proof/6/b/iv/solution

Yes. Part (a) diagonalizes

G = B

3/8 < 1

. Thus

G^{n} = S diag (0^{n}, (3/8)^{n}) S^{- 1} ⟶ 0 (n \geq 1),

and every initial error converges to zero. For the specified initial vector, solving

A x = b

gives

x = (1, - 1)^{T}

and

e_{0} = (- 1, 1)^{T}

. In fact,

e_{n} = (3/8)^{n} (- 2 1), x_{n} = (1 - 2 (3/8)^{n} - 1 + (3/8)^{n}) (n \geq 1) .

The special formula is not asserted for

n = 0

, because the zero-eigenvalue component disappears only after the first step.

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