Conditioning of linear systems

2.8. Conditioning of linear systems

We are ready to consider the conditioning of solving the square linear system \(\mathbf{A}\mathbf{x}=\mathbf{b}\). In this problem, the data are \(\mathbf{A}\) and \(\mathbf{b}\), and the solution is \(\mathbf{x}\). Both data and result are multidimensional, so we will use norms to measure their magnitudes.

The motivation for the definition of relative condition number in Chapter 1 was to quantify the response of the result to perturbations of the data. For simplicity, we start by allowing perturbations to \(\mathbf{b}\) only while \(\mathbf{A}\) remains fixed.

Let \(\mathbf{A}\mathbf{x}=\mathbf{b}\) be perturbed to

\[ \mathbf{A}(\mathbf{x}+\mathbf{h}) = \mathbf{b}+\mathbf{d}.\]

The condition number should be the relative change in the solution divided by relative change in the data,

\[ \frac{\quad\dfrac{\| \mathbf{h} \| }{\| \mathbf{x} \| }\quad}{\dfrac{\| \mathbf{d} \| }{\| \mathbf{b} \|}} = \frac{\| \mathbf{h} \|\;\| \mathbf{b} \| }{\| \mathbf{d} \|\; \| \mathbf{x} \| }.\]

We can bound \(\| \mathbf{h} \|\) in terms of \(\| \mathbf{d} \|\):

\[\begin{split}\begin{split} \mathbf{A}\mathbf{x} + \mathbf{A} \mathbf{h} &= \mathbf{b} + \mathbf{d}, \\ \mathbf{A} \mathbf{h} &= \mathbf{d},\\ \mathbf{h} &= \mathbf{A}^{-1} \mathbf{d},\\ \| \mathbf{h} \| &\le \| \mathbf{A}^{-1}\| \,\| \mathbf{d} \|, \end{split}\end{split}\]

where we have applied \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and (2.7.6). Since also \(\mathbf{b}=\mathbf{A}\mathbf{x}\) implies \(\| \mathbf{b} \|\le \| \mathbf{A} \|\, \| \mathbf{x} \|\), we derive

\[ \frac{\| \mathbf{h} \|\; \| \mathbf{b} \|}{\| \mathbf{d} \|\; \| \mathbf{x} \|} \le \frac{\bigl(\| \mathbf{A}^{-1} \|\, \| \mathbf{d} \|\bigr) \bigl(\| \mathbf{A} \|\,\| \mathbf{x} \|\bigr)}{\| \mathbf{d} \|\,\| \mathbf{x} \|} = \| \mathbf{A}^{-1}\| \, \| \mathbf{A} \|.\]

It is possible to show that this bound is tight, in the sense that the inequalities are in fact equalities for some choices of \(\mathbf{b}\) and \(\mathbf{d}\). This result motivates a new definition.

Definition 2.8.1 :  Matrix condition number

The matrix condition number of an invertible square matrix \(\mathbf{A}\) is

(2.8.1)\[\kappa(\mathbf{A}) = \| \mathbf{A}^{-1}\| \, \| \mathbf{A} \|.\]

This value depends on the choice of norm; a subscript on \(\kappa\) such as \(1\), \(2\), or \(\infty\) is used if clarification is needed. If \(\mathbf{A}\) is singular, we define \(\kappa(\mathbf{A}) = \infty\).

Main result

The matrix condition number (2.8.1) is equal to the condition number of solving a linear system of equations. Although we derived this fact above only for perturbations of \(\mathbf{b}\), a similar statement holds when \(\mathbf{A}\) is perturbed.

Using a traditional \(\Delta\) notation for the perturbation in a quantity, we can write the following.

Theorem 2.8.2 :  Conditioning of linear systems

If \(\mathbf{A}(\mathbf{x} + \Delta \mathbf{x}) = \mathbf{b} + \Delta \mathbf{b}\), then

(2.8.2)\[\frac{\| \Delta \mathbf{x} \|}{\| \mathbf{x} \|} \le \kappa(\mathbf{A}) \frac{\| \Delta \mathbf{b} \|}{\| \mathbf{b} \|}.\]

If \((\mathbf{A}+\Delta \mathbf{A}) (\mathbf{x} + \Delta \mathbf{x}) = \mathbf{b}\), then

(2.8.3)\[\frac{\| \Delta \mathbf{x} \|}{\| \mathbf{x} \|} \le \kappa(\mathbf{A}) \frac{\| \Delta \mathbf{A} \|}{\| \mathbf{A} \|},\]

in the limit \(\| \Delta \mathbf{A} \| \to 0\).

Note that for any induced matrix norm,

\[ 1 = \| \mathbf{I} \| = \| \mathbf{A} \mathbf{A}^{-1} \| \le \| \mathbf{A} \|\, \| \mathbf{A}^{-1} \| = \kappa(\mathbf{A}).\]

A condition number of 1 is the best we can hope for—in that case, the relative perturbation of the solution has the same size as that of the data. A condition number of size \(10^t\) indicates that in floating-point arithmetic, roughly \(t\) digits are lost (i.e., become incorrect) in computing the solution \(\mathbf{x}\). And if \(\kappa(\mathbf{A}) > \epsilon_\text{mach}^{-1}\), then for computational purposes the matrix is effectively singular.

Demo 2.8.3

Julia has a function cond to compute matrix condition numbers. By default, the 2-norm is used. As an example, the family of Hilbert matrices is famously badly conditioned. Here is the \(6\times 6\) case.

Type \kappa followed by Tab to get the Greek letter \(\kappa\).

A = [ 1/(i+j) for i in 1:6, j in 1:6 ]
κ = cond(A)

Because \(\kappa\approx 10^8\), it’s possible to lose nearly 8 digits of accuracy in the process of passing from \(\mathbf{A}\) and \(\mathbf{b}\) to \(\mathbf{x}\). That fact is independent of the algorithm; it’s inevitable once the data are expressed in finite precision.

Let’s engineer a linear system problem to observe the effect of a perturbation. We will make sure we know the exact answer.

x = 1:6
b = A*x
6-element Vector{Float64}:

Now we perturb the system matrix and vector randomly by \(10^{-10}\) in norm.

ΔA = randn(size(A));  ΔA = 1e-10*(ΔA/opnorm(ΔA));
Δb = randn(size(b));  Δb = 1e-10*normalize(Δb);

We solve the perturbed problem using pivoted LU and see how the solution was changed.

new_x = ((A+ΔA) \ (b+Δb))
Δx = new_x - x
6-element Vector{Float64}:

Here is the relative error in the solution.

@show relative_error = norm(Δx) / norm(x);
relative_error = norm(Δx) / norm(x) = 0.0017093353064577405

And here are upper bounds predicted using the condition number of the original matrix.

println("Upper bound due to b: $(κ*norm(Δb)/norm(b))")
println("Upper bound due to A: $(κ*opnorm(ΔA)/opnorm(A))")
Upper bound due to b: 0.0006723667712613617
Upper bound due to A: 0.004566989872745153

Even if we didn’t make any manual perturbations to the data, machine roundoff does so at the relative level of \(\macheps\).

Δx = A\b - x
@show relative_error = norm(Δx) / norm(x);
@show rounding_bound = κ*eps();
relative_error = norm(Δx) / norm(x) = 1.2278255878021855e-9
rounding_bound = κ * eps() = 1.134607140820324e-8

Larger Hilbert matrices are even more poorly conditioned:

A = [ 1/(i+j) for i=1:14, j=1:14 ];
κ = cond(A)

Note that \(\kappa\) exceeds \(1/\macheps\). In principle we therefore may end up with an answer that has relative error greater than 100%.

rounding_bound = κ*eps()

Let’s put that prediction to the test.

x = 1:14
b = A*x  
Δx = A\b - x
@show relative_error = norm(Δx) / norm(x);
relative_error = norm(Δx) / norm(x) = 2.3271868502658917

As anticipated, the solution has zero accurate digits in the 2-norm.

Residual and backward error

Suppose that \(\mathbf{A}\mathbf{x}=\mathbf{b}\) and \(\tilde{\mathbf{x}}\) is a computed estimate of the solution \(\mathbf{x}\). The most natural quantity to study is the error, \(\mathbf{x}-\tilde{\mathbf{x}}\). Normally we can’t compute it because we don’t know the exact solution. However, we can compute something related.

Definition 2.8.4 :  Residual of a linear system

For the problem \(\mathbf{A}\mathbf{x}=\mathbf{b}\), the residual at a solution estimate \(\tilde{\mathbf{x}}\) is

(2.8.4)\[ \mathbf{r} = \mathbf{b} - \mathbf{A}\tilde{\mathbf{x}}.\]

Obviously, a zero residual means that \(\tilde{\mathbf{x}}=\mathbf{x}\), and we have the exact solution. What happens more generally? Note that \(\mathbf{A}\tilde{\mathbf{x}}=\mathbf{b}-\mathbf{r}\). That is, \(\tilde{\mathbf{x}}\) solves the linear system problem for a right-hand side that is changed by \(-\mathbf{r}\). This is precisely what is meant by backward error.

Hence residual and backward error are the same thing for a linear system. What is the connection to the (forward) error? We can reconnect with (2.8.2) by the definition \(\mathbf{h} = \tilde{\mathbf{x}}-\mathbf{x}\), in which case

\[\mathbf{d} = \mathbf{A}(\mathbf{x}+\mathbf{h})-\mathbf{b}=\mathbf{A}\mathbf{h} = -\mathbf{r}.\]

Thus (2.8.2) is equivalent to

(2.8.5)\[ \frac{\| \mathbf{x}-\tilde{\mathbf{x}} \|}{\| \mathbf{x} \|} \le \kappa(\mathbf{A}) \frac{\| \mathbf{r} \|}{\| \mathbf{b} \|}.\]

Equation (2.8.5) says that the gap between relative error and the relative residual is a multiplication by the matrix condition number.

Observation 2.8.5

When solving a linear system, all that can be expected is that the backward error, not the error, is small.


  1. ⌨ Refer to Demo 2.8.3 for the definition of a Hilbert matrix. Make a table of the values of \(\kappa(\mathbf{H}_n)\) in the 2-norm for \(n=2,3,\ldots,16\). Speculate as to why the growth of \(\kappa\) appears to slow down at \(n=13\).

  2. ⌨ The purpose of this problem is to verify, like in Demo 2.8.3, the error bound

    \[\frac{\| \mathbf{x}-\tilde{\mathbf{x} \|}}{\| \mathbf{x} \|} \le \kappa(\mathbf{A}) \frac{\| \mathbf{h} \|}{\| \mathbf{b} \|}.\]

    Here \(\tilde{\mathbf{x}}\) is a numerical approximation to the exact solution \(\mathbf{x}\), and \(\mathbf{h}\) is an unknown perturbation caused by machine roundoff. We will assume that \(\| \mathbf{d} \|/\| \mathbf{b} \|\) is roughly eps().

    For each \(n=10,20,\ldots,70\) let A = matrixdepot("prolate",n,0.4) and let \(\mathbf{x}\) have components \(x_k=k/n\) for \(k=1,\ldots,n\). Define b=A*x and let \(\tilde{\mathbf{x}}\) be the solution produced numerically by backslash.

    Make a table including columns for \(n\), the condition number of \(\mathbf{A}\), the observed relative error in \(\tilde{\mathbf{x}}\), and the right-hand side of the inequality above. You should find that the inequality holds in every case.

  3. Exercise 2.3.7 suggests that the solutions of linear systems

    \[\begin{split}\mathbf{A} = \begin{bmatrix} 1 & -1 & 0 & \alpha-\beta & \beta \\ 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} \alpha \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}\end{split}\]

    become less accurate as \(\beta\) increases. Using \(\alpha=0.1\) and \(\beta=10,100,\ldots,10^{12}\), make a table with columns for \(\beta\), \(|x_1-1|\), and the condition number of the matrix.

  4. ⌨ Let \(\mathbf{A}_n\) denote the \((n+1)\times(n+1)\) version of the Vandermonde matrix in Equation (2.1.2) based on the equally spaced interpolation nodes \(t_i=i/n\) for \(i=0,\ldots,n\). Using the 1-norm, graph \(\kappa(\mathbf{A}_n)\) as a function of \(n\) for \(n=4,5,6,\ldots,20\), using a log scale on the \(y\)-axis. (The graph is nearly a straight line.)

  5. ⌨ The matrix \(\mathbf{A}\) in (2.6.1) has unpivoted LU factors given in (2.6.2) as a function of parameter \(\epsilon\). For \(\epsilon = 10^{-2},10^{-4},\ldots,10^{-10}\), make a table with columns for \(\epsilon\), \(\kappa(\mathbf{A})\), \(\kappa(\mathbf{L})\), and \(\kappa(\mathbf{U})\). (This shows that solution via unpivoted LU factorization is arbitrarily unstable.)

  6. ✍ Define \(\mathbf{A}_n\) as the \(n\times n\) matrix \(\displaystyle\begin{bmatrix} 1 & -2 & & &\\ & 1 & -2 & & \\ & & \ddots & \ddots & \\ & & & 1 & -2 \\ & & & & 1 \end{bmatrix}.\)

    (a) Write out \(\mathbf{A}_2^{-1}\) and \(\mathbf{A}_3^{-1}\).

    (b) Write out \(\mathbf{A}_n^{-1}\) in the general case \(n>1\). (If necessary, look at a few more cases in Julia until you are certain of the pattern.) Make a clear argument why it is correct.

    (c) Using the \(\infty\)-norm, find \(\kappa(\mathbf{A}_n)\).

  7. (a) Prove that for \(n\times n\) nonsingular matrices \(\mathbf{A}\) and \(\mathbf{B}\), \(\kappa(\mathbf{A}\mathbf{B})\le \kappa(\mathbf{A})\kappa(\mathbf{B})\).

    (b) Show by means of an example that the result of part (a) cannot be an equality in general.

  8. ✍ Let \(\mathbf{D}\) be a diagonal \(n\times n\) matrix, not necessarily invertible. Prove that in the 1-norm,

    \[\kappa(\mathbf{D}) = \frac{\max_i |D_{ii}|}{\min_i |D_{ii}|}.\]

    (Hint: See Exercise 2.7.10.)