from numpy import *
from scipy import linalg
from scipy.linalg import norm
from matplotlib.pyplot import *
from prettytable import PrettyTable
from timeit import default_timer as timer
import sys
sys.path.append('fncbook/')
import fncbook as FNC

# This (optional) block is for improving the display of plots.
# from IPython.display import set_matplotlib_formats
# set_matplotlib_formats("svg","pdf")
# %config InlineBackend.figure_format = 'svg'
rcParams["figure.figsize"] = [7, 4]
rcParams["lines.linewidth"] = 2
rcParams["lines.markersize"] = 4
rcParams['animation.html'] = "jshtml"  # or try "html5"

2.3.Linear systems¶

We now attend to the central problem of this chapter: Given a square, $n\times n$ matrix $\mathbf{A}$ and an $n$ -vector $\mathbf{b}$ , find an $n$ -vector $\mathbf{x}$ such that $\mathbf{A}\mathbf{x}=\mathbf{b}$ . Writing out these equations, we obtain

\begin{split} A_{11}x_1 + A_{12}x_2 + \cdots + A_{1n}x_n &= b_1, \\ A_{21}x_1 + A_{22}x_2 + \cdots + A_{2n}x_n &= b_2, \\ \vdots \\ A_{n1}x_1 + A_{n2}x_2 + \cdots + A_{nn}x_n &= b_n. \end{split}

(2.3.1)

If $\mathbf{A}$ is invertible, then the mathematical expression of the solution is $\mathbf{x}=\mathbf{A}^{-1}\mathbf{b}$ because

\begin{split} \mathbf{A}^{-1}\mathbf{b} = \mathbf{A}^{-1} (\mathbf{A} \mathbf{x}) = (\mathbf{A}^{-1}\mathbf{A}) \mathbf{x} = \mathbf{I} \mathbf{x} = \mathbf{x}. \end{split}

(2.3.2)

When $\mathbf{A}$ is singular, then $\mathbf{A}\mathbf{x}=\mathbf{b}$ may have no solution or infinitely many solutions.

2.3.1Don’t use the inverse¶

Matrix inverses are indispensable for mathematical discussion and derivations. However, as you may remember from a linear algebra course, they are not trivial to compute from the entries of the original matrix. You might be surprised to learn that matrix inverses play almost no role in scientific computing.

In fact, when we encounter an expression such as $\mathbf{x} = \mathbf{A}^{-1} \mathbf{b}$ in computing, we interpret it as “solve the linear system $\mathbf{A} \mathbf{x} = \mathbf{b}$ ” and apply whatever algorithm is most expedient based on what we know about $\mathbf{A}$ .

In Python, the numpy.linalg.solve function is used to solve linear systems.

Example 2.3.2 (Solving linear systems)

For a square matrix $A$ , the command solve(A, b) from scipy.linalg is mathematically equivalent to $\mathbf{A}^{-1} \mathbf{b}$ .

A = array([[1, 0, -1], [2, 2, 1], [-1, -3, 0]])
b = array([1, 2, 3])

from scipy import linalg
x = linalg.solve(A, b)
print(x)

[ 2.14285714 -1.71428571  1.14285714]

One way to check the answer is to compute a quantity known as the residual. It is (ideally) close to machine precision(relative to the elements in the data).

residual = b - A @ x
print(residual)

[-4.4408921e-16 -4.4408921e-16  0.0000000e+00]

If the matrix $\mathbf{A}$ is singular, you may get an error.

A = array([[0, 1], [0, 0]])
b = array([1, -1])
linalg.solve(A, b)    # error, singular matrix

---------------------------------------------------------------------------
LinAlgError                               Traceback (most recent call last)
Cell In[5], line 3
      1 A = array([[0, 1], [0, 0]])
      2 b = array([1, -1])
----> 3 linalg.solve(A, b)    # error, singular matrix

File ~/miniforge3/envs/myst/lib/python3.13/site-packages/scipy/_lib/_util.py:1233, in _apply_over_batch.<locals>.decorator.<locals>.wrapper(*args, **kwargs)
   1231 # Early exit if call is not batched
   1232 if not any(batch_shapes):
-> 1233     return f(*arrays, *other_args, **kwargs)
   1235 # Determine broadcasted batch shape
   1236 batch_shape = np.broadcast_shapes(*batch_shapes)  # Gives OK error message

File ~/miniforge3/envs/myst/lib/python3.13/site-packages/scipy/linalg/_basic.py:329, in solve(a, b, lower, overwrite_a, overwrite_b, check_finite, assume_a, transposed)
    327 elif assume_a in {'lower triangular', 'upper triangular'}:
    328     lower = assume_a == 'lower triangular'
--> 329     x, info = _solve_triangular(a1, b1, lower=lower, overwrite_b=overwrite_b,
    330                                 trans=transposed)
    331     _solve_check(n, info)
    332     _trcon = get_lapack_funcs(('trcon'), (a1, b1))

File ~/miniforge3/envs/myst/lib/python3.13/site-packages/scipy/linalg/_basic.py:528, in _solve_triangular(a1, b1, trans, lower, unit_diagonal, overwrite_b)
    526     return x, info
    527 if info > 0:
--> 528     raise LinAlgError(f"singular matrix: resolution failed at diagonal {info-1}")
    529 raise ValueError(f'illegal value in {-info}-th argument of internal trtrs')

LinAlgError: singular matrix: resolution failed at diagonal 0

A linear system with a singular matrix might have no solution or infinitely many solutions, but in either case, a numerical solution becomes trickier. Detecting singularity is a lot like checking whether two floating-point numbers are exactly equal: because of roundoff, it could be missed. We’re headed toward a more robust way to fully describe this situation.

2.3.2Triangular systems¶

The solution process is especially easy to demonstrate for a system with a triangular matrix. For example, consider the lower triangular system

\begin{bmatrix} 4 & 0 & 0 & 0 \\ 3 & -1 & 0 & 0 \\ -1 & 0 & 3 & 0 \\ 1 & -1 & -1 & 2 \end{bmatrix} \mathbf{x} = \begin{bmatrix} 8 \\ 5 \\ 0 \\ 1 \end{bmatrix}.

(2.3.5)

The first row of this system states simply that $4x_1=8$ , which is easily solved as $x_1=8/4=2$ . Now, the second row states that $3x_1-x_2=5$ . As $x_1$ is already known, it can be replaced to find that $x_2 = -(5-3\cdot 2)=1$ . Similarly, the third row gives $x_3=(0+1\cdot 2)/3 = 2/3$ , and the last row yields $x_4=(1-1\cdot 2 + 1\cdot 1 + 1\cdot 2/3)/2 = 1/3$ . Hence, the solution is

\mathbf{x} = \begin{bmatrix} 2 \\ 1 \\ 2/3 \\ 1/3 \end{bmatrix}.

(2.3.6)

The process just described is called forward substitution. In the $4\times 4$ lower triangular case of $\mathbf{L}\mathbf{x}=\mathbf{b}$ it leads to the formulas

\begin{split} x_1 &= \frac{b_1}{L_{11}}, \\ x_2 &= \frac{b_2 - L_{21}x_1}{L_{22}}, \\ x_3 &= \frac{b_3 - L_{31}x_1 - L_{32}x_2}{L_{33}}, \\ x_4 &= \frac{b_4 - L_{41}x_1 - L_{42}x_2 - L_{43}x_3}{L_{44}}. \end{split}

(2.3.7)

For upper triangular systems $\mathbf{U}\mathbf{x}=\mathbf{b}$ an analogous process of backward substitution begins by solving for the last component $x_n=b_n/U_{nn}$ and working backward. For the $4\times 4$ case we have

\begin{bmatrix} U_{11} & U_{12} & U_{13} & U_{14} \\ 0 & U_{22} & U_{23} & U_{24} \\ 0 & 0 & U_{33} & U_{34} \\ 0 & 0 & 0 & U_{44} \end{bmatrix} \mathbf{x} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \\ b_4 \end{bmatrix}.

(2.3.8)

Solving the system backward, starting with $x_4$ first and then proceeding in descending order, gives

\begin{split} x_4 &= \frac{b_4}{U_{44}}, \\ x_3 &= \frac{b_3 - U_{34}x_4}{U_{33}}, \\ x_2 &= \frac{b_2 - U_{23}x_3 - U_{24}x_4}{U_{22}}, \\ x_1 &= \frac{b_1 - U_{12}x_2 - U_{13}x_3 - U_{14}x_4}{U_{11}}. \end{split}

(2.3.9)

It should be clear that forward or backward substitution fails if and only if one of the diagonal entries of the system matrix is zero. We have essentially proved the following theorem.

2.3.3Implementation¶

Consider how to implement the sequential process implied by Equation (2.3.7). It seems clear that we want to loop through the elements of $\mathbf{x}$ in order. Within each iteration of that loop, we have an expression whose length depends on the iteration number. This leads to a nested loop structure.

Algorithm 2.3.1 (forwardsub)

forwardsub.py

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def forwardsub(L,b):
    """
     forwardsub(L,b)

    Solve the lower-triangular linear system with matrix L and right-hand side
    vector b.
    """
    n = len(b)
    x = np.zeros(n)
    for i in range(n):
        s = L[i,:i] @ x[:i]
        x[i] = ( b[i] - s ) / L[i, i]
    return x

The implementation of backward substitution is much like forward substitution and is given in Function 2.3.2.

Algorithm 2.3.2 (backsub)

backsub.py

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def backsub(U,b):
    """
    backsub(U,b)

    Solve the upper-triangular linear system with matrix U and right-hand side
    vector b.
    """
    n = len(b)
    x = np.zeros(n)
    for i in range(n-1, -1, -1):
        s = U[i, i+1:] @ x[i+1:]
        x[i] = ( b[i] - s ) / U[i, i]
    return x

Example 2.3.3 (Triangular systems of equations)

It’s easy to get just the lower triangular part of any matrix using the tril function.

A = 1 + floor(9 * random.rand(5, 5))
L = tril(A)
print(L)

[[3. 0. 0. 0. 0.]
 [2. 7. 0. 0. 0.]
 [4. 1. 1. 0. 0.]
 [9. 7. 3. 2. 0.]
 [3. 9. 3. 1. 4.]]

We’ll set up and solve a linear system with this matrix.

b = ones(5)
x = FNC.forwardsub(L, b)
print(x)

[ 0.33333333  0.04761905 -0.38095238 -0.5952381   0.32738095]

It’s not clear how accurate this answer is. However, the residual should be zero or comparable to $\macheps$ .

b - L @ x

array([0., 0., 0., 0., 0.])

Next we’ll engineer a problem to which we know the exact answer.

alpha = 0.3;
beta = 2.2;
U = diag(ones(5)) + diag([-1, -1, -1, -1], k=1)
U[0, 3:5] = [ alpha - beta, beta ]
print(U)

[[ 1.  -1.   0.  -1.9  2.2]
 [ 0.   1.  -1.   0.   0. ]
 [ 0.   0.   1.  -1.   0. ]
 [ 0.   0.   0.   1.  -1. ]
 [ 0.   0.   0.   0.   1. ]]

x_exact = ones(5)
b = array([alpha, 0, 0, 0, 1])
x = FNC.backsub(U, b)
print("error:", x - x_exact)

error: [2.22044605e-16 0.00000000e+00 0.00000000e+00 0.00000000e+00
 0.00000000e+00]

Everything seems OK here. But another example, with a different value for $\beta$ , is more troubling.

alpha = 0.3;
beta = 1e12;
U = diag(ones(5)) + diag([-1, -1, -1, -1], k=1)
U[0, 3:5] = [ alpha - beta, beta ]
b = array([alpha, 0, 0, 0, 1])

x = FNC.backsub(U, b)
print("error:", x - x_exact)

error: [-4.8828125e-05  0.0000000e+00  0.0000000e+00  0.0000000e+00
  0.0000000e+00]

It’s not so good to get 4 digits of accuracy after starting with sixteen! But the source of the error is not hard to track down. Solving for $x_1$ performs $(\alpha-\beta)+\beta$ in the first row. Since $|\alpha|$ is so much smaller than $|\beta|$ , this a recipe for losing digits to subtractive cancellation.

The example in Example 2.3.3 is our first clue that linear system problems may have large condition numbers, making inaccurate solutions inevitable in floating-point arithmetic. We will learn how to spot such problems in Conditioning of linear systems. Before reaching that point, however, we need to discuss how to solve general linear systems, not just triangular ones.

2.3.4Exercises¶

Exercise 2.3.5

Suppose a string is stretched with tension $\tau$ horizontally between two anchors at $x=0$ and $x=1$ . At each of the $n-1$ equally spaced positions $x_k=k/n$ , $k=1,\ldots,n-1$ , we attach a little mass $m_i$ and allow the string to come to equilibrium. This causes vertical displacement of the string. Let $q_k$ be the amount of displacement at $x_k$ . If the displacements are not too large, then an approximate force balance equation is

n \tau (q_k - q_{k-1}) + n\tau (q_k - q_{k+1}) = m_k g, \qquad k=1,\ldots,n-1,

where $g=-9.8$ m/s $^2$ is the acceleration due to gravity, and we define $q_0=0$ and $q_n=0$ due to the anchors. This defines a linear system for $q_1,\ldots,q_{n-1}$ .

(a) ✍ Show that the force balance equations can be written as a linear system $\mathbf{A}\mathbf{q}=\mathbf{f}$ , where $\mathbf{q}$ is a vector of the unknown displacements and $\mathbf{A}$ is a tridiagonal matrix (i.e., $A_{ij}=0$ if $|i-j|>1$ ) of size $(n-1)\times(n-1)$ .

(b) ⌨ Let $\tau=10$ N, and $m_k=(1/10n)$ kg for every $k$ . Using backslash, find the displacements when $n=8$ and $n=40$ , and superimpose plots of $\mathbf{q}$ over $0\le x \le 1$ for the two cases. (Be sure to include the zero values at $x=0$ and $x=1$ in your plots.)

(c) ⌨ Repeat (b) for the case $m_k = (k/5n^2)$ kg.

Exercise 2.3.6

⌨ If $\mathbf{B}\in\real^{n \times p}$ has columns $\mathbf{b}_1,\ldots,\mathbf{b}_p$ , then we can pose $p$ linear systems at once by writing $\mathbf{A} \mathbf{X} = \mathbf{B}$ , where $\mathbf{X}$ is $n\times p$ . Specifically, this equation implies $\mathbf{A} \mathbf{x}_j = \mathbf{b}_j$ for $j=1,\ldots,p$ .

(a) Modify Function 2.3.1 and Function 2.3.2 so that they solve the case where the second input is $n\times p$ for $p\ge 1$ .

(b) If $\mathbf{A} \mathbf{X}=\mathbf{I}$ , then $\mathbf{X}=\mathbf{A}^{-1}$ . Use this fact to write a function ltinverse that uses your modified forwardsub to compute the inverse of a lower triangular matrix. Test your function on at least two nontrivial matrices. (We remind you here that this is just an exercise; matrix inverses are rarely a good idea in numerical practice!)