Lecture 12

Partial and Complete Pivoting¶

Key results¶

Small pivots give huge multipliers and wrong answers.
Partial pivoting: before eliminating column \(k\), swap the row with the largest \(|a_{ik}|\), \(i\ge k\), into the pivot position. Then every \(|m_{ik}|\le1\).
The search costs about \(\frac{n^2}{2}\) comparisons; the row swaps give \(PA=LU\).
Scaled partial pivoting chooses by \(|a_{ik}|/s_i\) with \(s_i=\max_j|a_{ij}|\). Complete pivoting swaps rows and columns, at about \(\frac{n^3}{3}\) comparisons.

A small pivot¶

\[ \begin{aligned} 0.003000\,x_1+59.14\,x_2&=59.17\\ 5.291\,x_1-6.130\,x_2&=46.78 \end{aligned}\qquad (x_1=10,\ x_2=1) \]

4-digit arithmetic	multiplier	\(x_2\)	\(x_1\)
no pivoting	\(1764\)	\(1.001\)	\(-10.00\)
partial pivoting (rows swapped)	\(0.000567\)	\(1.000\)	\(10.00\)

Without pivoting, back substitution computes \(x_1=(59.17-59.14\,x_2)/0.003000\): the error of \(0.001\) in \(x_2\) is multiplied by \(59.14/0.003\approx20\,000\).

Partial pivoting¶

At step \(k\) choose \(p\) with \(|a_{pk}|=\max_{i\ge k}|a_{ik}|\) and swap rows \(p\) and \(k\) (in \(A\) and \(\mathbf b\)), then eliminate as usual. For the system of Lecture 10 the swaps are \(R_1\leftrightarrow R_3\) and then \(R_2\leftrightarrow R_3\), leading to

\[ \underbrace{\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}}_{P}A=\underbrace{\begin{bmatrix}1&0&0\\ \frac23&1&0\\ \frac13&\frac1{11}&1\end{bmatrix}}_{L}\underbrace{\begin{bmatrix}3&4&-2\\0&-\frac{11}{3}&\frac{13}{3}\\0&0&\frac{14}{11}\end{bmatrix}}_{U}, \]

and \(x=1,\ y=2,\ z=3\).

Scaled and complete pivoting¶

Badly scaled rows. Multiplying the first equation by \(10^4\) (\(30.00x_1+591400x_2=591700\)) makes partial pivoting keep row 1, and the 4-digit answer is again \(x_1=-10.00\). Scaled partial pivoting compares \(30.00/591400\approx0.00005\) with \(5.291/6.130\approx0.86\), chooses row 2 and gets \(x_1=10.00\).
Complete pivoting searches the whole remaining submatrix and swaps rows and columns; column swaps reorder the unknowns. For the \(4\times4\) system of the notes the pivots are \(8\), \(\frac72\), \(\frac{15}{14}\), and the solution is \(x=1,\ y=2,\ z=3,\ w=4\).

Stability¶

The entries can grow during elimination. For Wilkinson's matrix, partial pivoting gives growth \(2^{n-1}\) (\(\approx10^{17}\) for \(n=60\)), but such matrices are very rare and partial pivoting is stable in practice. Symmetric positive definite and strictly diagonally dominant matrices need no pivoting at all.

strategy	search cost	safety
none	\(0\)	can fail
partial	\(\frac{n^2}{2}\)	safe in practice
scaled partial	\(\approx n^2\)	also for badly scaled rows
complete	\(\frac{n^3}{3}\)	safest, expensive

Practise this lecture ← Previous: Operation Count Next: LU Factorisation →