September 21, 2023 30 min read

Conjugate gradients is one of the most popular computationally effective methods, used to solve systems of linear equations. Along with other Krylov space methods it can also be used for finding eigenvalues and eigenvectors of a matrix.

## System of linear equations as a quadratic optimization problem

In case $A$ is a Gram matrix (i.e. symmetric and positive definite), so that its eigenvalues are real and positive, it is useful to think about solution of a system of linear equations $Ax = b$ in terms of an equivalent quadratic problem of finding a minimum of a quadratic function, given by a quadratic form:

$x^* : A x - b = 0 \iff x^* = \arg \min \limits_x \phi(x) = x^T A x - b^T x$

Quadratic form $x^T A x$ represents ellipsoid, eigenvectors $\{ e_i \}$ of matrix $A$ are axes, eigenvalues $\{ e_i \}$ are lengths semi-axes, difference between the optimal solution and current point is $x^* - x$, residue vector $r = Ax - b = Ax - Ax^*$ will be discussed later.

## Naive approaches

How to solve a system of linear equations computationally?

We shall briefly consider two naive approches: coordinate descent and steepest descent, before discussing conjugate gradients and their advantages.

### Coordinate descent

First, let us look at the simplest approach: coordinate descent. I.e. we iteratively choose coordinates of the vector $x_i$ and minimize the value of function $\phi(x)$ in that coordinate by solving $\frac{\partial \phi(x)}{\partial x_i} = 0$.

#### Convergence

Coordinate descent clearly does not guarantee convergence in N steps:

Coordinate descent. In this example we have a 2D vector $x$, and coordinate descent does not converge to the solution $x^*$ in 2 steps.

### Steepest descent

A smarter approach is steepest descent.

At each step of coordinate descent we calculate the gradient vector $\nabla \phi$ of function $\phi(x)$ and use it as the direction of descent.

Here and below I will denote $d$ the direction vector for descent. If we are speaking of multiple steps, I will denote it as $d_i$. In case of the steepest descent at each step $d_i = \nabla x_i$, where $x_i$ is the approximation of solution $x^*$, obtained at $i$-th step, starting from an initial approximation $x_0$.

If we are currently at the point $x$, we shall move to some point $x + \alpha d$, where $\alpha$ is the step. We know the direction, but how do we find the optimal step $\alpha$ of descent?

Length of the step is optimal, when the function $\phi(x + \alpha d)$ is minimal in some point $x + \alpha d$. It means that directional derivative $\frac{\partial \phi}{\partial d} = 0$ and the isolevels of the function $\phi(x) = x^T A x - b^T x$ become tangent to the direction $d$ of descent.

Line search. We start from the point $x$ (defined by a vector, starting from origin) and start searching in the direction $d$ for an optimal length $\alpha$ of $d$ vector, so that the directional derivative of quadratic function $\phi(x)$ is $0$ there, i.e. direction vector is tangent to isolevels of function $\phi(x)$ at that point.

Speaking in terms of multivariate calculus $\langle d, \frac{\partial \phi}{\partial x} \rangle = 0$.

Simplifying this expression, we obtain the step length $\alpha$:

$d^T (A (x + \alpha d) - b) = 0$

$d^T A x + \alpha d^T A d - d^T A d - d^T b = 0$

$\alpha = - \frac{d^T (Ax - b)}{d^T A d} = - \frac{d^T r}{ d^T A d }$

Now our steepest descent algorithm is fully defined. By the way, this formula for step length $\alpha$ is going to be useful below.

#### Convergence

Convergence rate of steepest descent varies. For instance, it might converge in one iteration, if $A$ is diagonal and all of its eigenvalues are equal:

Perfect convergence of steepest descent. If matrix $A$ is diagonal with equal eigenvalues, isolevels are spheres and steepest descent converges in 1 iteration.

However, in general we cannot guarantee fast convergence for steepest descent. E.g. if the matrix $A$ is ill-conditioned, i.e. the difference in magnitude between its largest and smallest eigenvalues is big, steepest descent will start jumping from one side of the ellipsoid to the other.

Poor convergence of steepest descent. If we have an ill-conditioned matrix, so that even one eigenvalue of matrix $A$ is much smaller than other eigenvalues of matrix $A$, the ellipsoid isolevels look like a ravine, and gradient jumps off the sides of the ravine. Image from J.R.Schewchuk book.

Hence, convergence rate is related to condition numbers $\kappa$ of the matrix. One can actually derive a formula for convergence rate:

$\frac{f(x_{(i)}) - f(x)}{f(x_{(0)}) - f(x)} = \frac{\frac{1}{2} e_{(i)}^T A e_{(i)} }{ \frac{1}{2} e_{(0)}^T A e_{(0)} } \le (\frac{\kappa - 1}{\kappa + 1})^{2i}$

Exact derivation of this fact is somewhat tedious (although, not so hard for understanding) and can be found in J.R. Schewchuk tutorial, part 6.2.

## Conjugate directions and conjugate gradients

As we’ve seen convergence of neither coordinate descent, nor steepest descent in $N$ steps is guaranteed.

Here we discuss a better approach, conjugate directions, which will converge in $N$ steps or less. It has a special case, conjugate gradients, which is even more computationally efficient.

Let us start with an example of orthogonal axes and then transition to a general case.

### Orthogonal axes

Consider the case when all the eigenvectors of our surface are parallel to the coordinate axes. In such a case matrix $A$ is diagonal and performing line search along the coordinate axes guarantees convergence in N iterations.

If matrix $A$ is diagonal, line search along the conjugate directions converges in $N$ steps. Image based on J.R.Schewchuk.

However, in general case we cannot rely on orthogonality of axes. If eigenvectors of matrix $A$ are not orthogonal and parallel to coordinate axes, this approach won’t work and line search along orthogonal directions would not work.

### A-orthogonality

Instead of orthogonality, we introduce the notion of A-orthogonality:

$\langle a, b \rangle_A = 0 \iff \langle A a, b \rangle = 0 \iff \langle a, A b \rangle = 0$

We will apply two transformations of coordinates.

First, apply eigen decomposition of $A$:

$A = E \Lambda E^{-1}$, where $E$ is the matrix of eigenvectors (each eigenvector is a column-vector) and $E$ is orthogonal, because eigenvectors of symmetric matrix are orthogonal; $\Lambda$ is the diagonal matrix of eignevalues, where each eigenvalue $\lambda_i$ is known to be real, positive and related to a corresponding singular value $\sigma_i$ as $\lambda_i = \sigma_i^2$.

$\langle A a, b \rangle = b^T A a = b^T E \Lambda E^{-1} a$

Use the fact that $E$ is orthogonal for symmetric matrix $A$, hence, $E^T = E^{-1}$. Hence, $(E^T b)^T = b^T E = \beta^T$, $E^T a = \alpha$.

What did we do? We changed the coordinates, so that coordinate axes are now the eigenvectors $E$.

In these coordinates vector $a$ has coordinates $\alpha$ and vector $b$ has coordinates $\beta$.

Second, define $\Lambda^{1/2} = \begin{pmatrix} \sqrt{\lambda_1} && 0 && 0 \\ 0 && \sqrt{\lambda_2} && 0 \\ 0 && 0 && \sqrt{\lambda_3} \end{pmatrix}$.

$\langle A a, b \rangle = b^T A a = b^T E \Lambda E^{-1} a = \beta^T \Lambda \alpha = (\Lambda^{1/2} \beta)^T \cdot (\Lambda^{1/2} \alpha)$.

What did we do here? We stretched our coordinate axes by their respective eigenvalues $\lambda_i$, transforming our vectors $\alpha$ into $\Lambda^{1/2} \alpha = \Lambda^{1/2} E^{-1}a$ and $\beta$ into $\Lambda^{1/2} \beta = \Lambda^{1/2} E^{-1}b$.

Eigenvectors, forming the semi-axes of ellipsoids, upon this change of coorindates take identical lengths, so that our ellipsoids become spheres.

So, how does A-orthogonality work: first we go from original coordinates to coordinates, where axes $E$ are eigenvectors of $A$. We apply this transform to both $a$ and $b$ vectors, resulting in their eigendecomposition vectors $E^{-1}a$ and $E^{-1}b$.

Second, we come to a coordinate system, where isolevels are spheres. In these coordinates resulting vectors $\Lambda^{1/2} E^{-1}a$ and $\Lambda^{1/2} E^{-1}b$ must be orthogonal.

A-orthogonality: vectors that are A-orthogonal become orthogonal, if we make two changes of coordinates: first, we make eigenvectors the axes of our coordinate system, obtaining image (a); second, we stretch coordinate axes by eigenvalues, so that our isolevels become concentric spheres, resulting in image (b). Image from J.R.Schewchuk.

### Conjugate directions

Let us prove the guarantees of convergence for conjugate directions algorithm in $N$ steps.

Consider a descent algorithm, which uses vectors $\{ d_i \}$ as descent directions and step sizes $\alpha_i = - \frac{d_i^T r_i }{d_i^T A d_i}$, where $r_i = A x_i - b$ is the current residue.

$x_n = x_0 + \alpha_0 d_0 + \alpha_1 d_1 + ... + \alpha_{n-1} d_{n-1}$

We want to prove that if $\{ d_i \}$ are conjugate directions, then $x_n = x^*$.

Indeed, we know that vectors $\{ d_i \}$ span the whole space. Hence, there exist coefficients $\{ \sigma_i \}$, such that we can come to the desired solution point $x^*$ walking the directions:

$x^* = x_0 + \sigma_0 d_0 + \sigma_1 d_1 + ... + \sigma_{n-1} d_{n-1}$

Pre-multiply this expression by $d_{i} A$:

$d_{i}^T A (x^* - x_0) = \cancel{\sigma_0 d_{i}^T A d_0} + ... + \sigma_1 d_{i}^T A d_{i} + ... + \cancel{\sigma_{n-1} d_{i}^T A d_{n-1}}$

Hence, $\sigma_{i} = \frac{d_{i}^T A (x^* - x_0)}{ d_{i}^T A d_{i} }$.

Recall that $Ax^* = b$ (because $x^*$ is the perfect solution of our system of linear equations in the end).

Thus, $\sigma_{i} = \frac{d_{i}^T A (x^* - x_0)}{ d_{i}^T A d_{i} } = \frac{ d_{i}^T (b - A x_0) }{ d_{i}^T A d_{i} } = \frac{ -d_{i}^T r_0 }{ d_{i}^T A d_{i} }$.

Now, consider $x_i = x_0 + \alpha_0 d_0 + \alpha_1 d_1 + ... + \alpha_{i-1} d_{i-1}$ and pre-multiply it by $d_i^T A$:

$d_i^T A (x_i - x_0) = 0$, hence, $d_i^T A x_i = d_i^T A x_0$.

Then substitute this into $\sigma_{i-1} = \frac{ -d_{i}^T r_0 }{ d_{i}^T A d_{i} }$.

This produces $\sigma_{i} = \frac{ -d_{i}^T r_i }{ d_{i}^T A d_{i} }$, which coincides with $\alpha_i = - \frac{d_i^T r_i }{d_i^T A d_i}$.

Hence, $\sigma_i = \alpha_i$.

#### Conjugate directions as an energy norm minimization

An insightful way of thinking about conjugate gradients is through norm minimization/maximization.

One can show that conjugate directions is similar to PCA in terms of minimization/maximization of a certain norm.

PCA at each iteration finds a projection that minimizes Frobenius norm of approximation of our matrix by low-rank matrices:

$|| A - \sum_k^i \lambda_k u_k v_k^T ||_F \to \min$

TODO: check PCA analogy

TODO: duality between energy norm and Frobenius norm (maximum of energy norm corresponds to minimum of Frobenius norm)

Similarly, we can introduce a notion of energy norm of vector $x$: $|| x ||_A = x^T A x$.

One may show that at each step conjugate directions algorithm chooses such a vector $d_i$ that minimizes the energy norm of error:

$|| e_i ||_A \to \min$

$|| e_i ||_A = e_i^T A e_i = (\sum \limits_k=i^{n-1} d_k^T ) A (\sum \limits_k=i^{n-1} d_k ) = \sum \limits_{k=i}^{n-1} d_k^T A d_k$.

TODO: check notation and finish this part

#### Conjugate directions and eigenvectors

Let $A = E \Lambda E^T$ be the eigenvector decomposition of $A$.

Consider the definition of conjugate directions: $P^T A P = D$, where $D$ is any diagonal matrix, $P$ is the matrix of conjugate directions row-vector (orthogonal).

$A = P D P^T$

$E \Lambda E^T = P D P^T$

$E \Lambda^{1/2} = P D^{1/2}$

$P = E \Lambda^{1/2} D^{-1/2} = E (\Lambda D^{-1})^{1/2}$

How do we obtain conjugate directions in practice?

Simple and straightforward approach: let us use neg-gradients as directions: $d_{i+1} = - r_{i+1} = - A(x_{i+1} - b)$. Problem is, our “conjugate” directions are not conjugate. $d_i^T A d_{i+1} \ne 0$, directions are not A-orthogonal (although, by the way, we know that $d_{i+1}$ is orthogonal, not A-orthogonal to the previous search direction due to tangency of search direction to isolevels at the point $x_{i+1}$.

Conjugate directions in 2D. Note that $r_{(1)}$ is orthogonal to $d_{(0)}$ and $e_{(1)}$ is A-orthogonal to $d_{(0)}$. Image from J.R. Schewchuk.

So, we correct the gradient direction by subtracting $d_{i}$ with some coefficient to A-orthogonalize $d_{i+1}$ and $d_{i}$:

We know that $d_{i+1}$ is orthogonal to each of the previous directions $\{ d_1, ..., d_{i-1} \}$.

$x^* = x_0 + \alpha_0 d_0 + \alpha_1 d_1 + ... + \alpha_{n-1} d_{n-1}$

$x_{i+1} = x_0 + \alpha_0 d_0 + ... + \alpha_i d_i$

$r_{i+1} = Ax_{i+1} - b = Ax_{i+1} - Ax^* = A (x_0 + \alpha_0 d_0 + ... + \alpha_i d_i) - A (x_0 + \alpha_0 d_0 + \alpha_1 d_1 + ... + \alpha_{n-1} d_{n-1}) = - A (\alpha_{i+1}d_{i+1} + ... + \alpha_{n-1} d_{n-1})$

$d_k^T r_{i+1} = 0$

Hence, $r_{i+1}$ is already orthogonal to $\{ d_0, ..., d_{i} \}$.

But more importantly for conjugate gradients, it is not only orthogonal, but also A-orthogonal to $\{ d_0, ..., d_{i-1} \}$!

$x_{i+1} = x_{i} + \alpha_i d_{i}$

$r_{i+1} = r_{i} + \alpha_i A d_{i}$

$\alpha_i A d_i = r_{i+1} - r_{i}$

Pre-multiply this with $r^T_k$: $\alpha_i r^T_k A d_i = r^T_k r_{i+1} - r^T_k r_{i}$.

The following cases arise: $\begin{cases} k < i: r^T_k A d_i = 0 \\ k = i: r^T_i A d_i = - r^T_i r_{i} \\ k = i + 1: r^T_{i+1} A d_i = r^T_{i+1} r_{i+1} \end{cases}$

Hence, $r_{i+1}$ is already A-orthogonal to $d_k$ for $k < i$. To A-orthogonalize $d_{i+1}$ derived from $r_{i+1}$, all we need to do is to orthogonalize it with respect to $d_i$. This greatly simplifies the calculation process and saves us a lot of memory:

$d_{i+1} = - r_{i+1} + \beta d_{i}$

Calculation of $\beta$ is evident from A-orthogonality of $d_{i+1}$ and $d_{i}$: pre-multiply this expression with $d^T_{i} A$:

$\cancel{d^T_{i} A d_{i+1}} = - d^T_{i} A r_{i+1} + \beta d^T_{i} A d_{i}$ (all other $d^T_{k} A d_{i} = 0$)

$d^T_{i} A r_{i+1} = \beta d^T_{i} A d_{i}$

$\beta = \frac{ d^T_{i} A r_{i+1} }{ d^T_{i} A d_{i} }$

This is just a special case of Schmidt orthogonalization.

#### Krylov space

One more important aspect: assume $x_0 = 0$ and consider $-r_0 = b - Ax = b$. Hence, $x_1 = b$.

$x_2 - x_1 = - r_{2} + \beta (x_1 - x_0) = (b - A x_2) + \beta x_1$

$x_2 (I + A) = b + (1 + \beta) x_1 = b + (1 + \beta) b = 2b + \beta b$

$x_2 = b (2 + \beta)$

Similarly, if we consider $x_3$, we get a polynomial of $A^2 b$, $A b$ and $b$. And so on.

Hence, in case of $x_i$ we are working with a space of vectors $[ A^{i-1} b, A^{i-2} b, ..., A b, b ]$

This space is called Krylov space and is useful in related problems, such as numeric eigenvector/eigenvalue estimation. Such problems are often solved with a related family of algorithms, such as Arnoldi iteration and Lanczos algorithm.

TODO: fix bugs

#### Convergence rate: inexact arithmetic

If we allow for a certain error, conjugate gradients might not even require $N$ iterations. Consider the following clustering of eigenvalues of our matrix $A$:

Two clusters of eigenvalues. From Nocedal-Wright’s Numerical optimization.

Recall the convergence rate formula: $|| x_{n-k} - x^* ||_A \le (\frac{\lambda_{n-k} - \lambda_1}{ \lambda_{n-k} + \lambda_1 })^2 || x_0 - x^* ||_A$.

Suppose that we’ve already made $m+1$ iterations. Then $\lambda_{n-(m+1)} - \lambda_1$ is very small and $\lambda_{n-(m+1)} + \lambda_1 \approx 2 \lambda_1$.

Hence, $|| x_{n-k} - x^* ||_A \le \epsilon$, where $\epsilon > 0$ is small, and we’ve basically converged to the solution.

### Preconditioning and clusters of eigenvalues

If the matrix $A$ is ill-conditioned, iterative multiplication of a vector $b$ by a matrix $A$ would accumulate the floating point rounding errors, resuling in a failure to converge to the correct solution. Hence, we need to “normalize” the matrix $A$ similarly to how we apply batch norm or layer norm to normalize the layers of neural networks.

This process is called pre-conditioning. We are multiplying both sides of our equation by a pre-conditioner matrix $P$: $P A x = P b$.

An ideal pre-conditioner would convert our matrix to a perfect sphere. Then pre-conditioning axes should be the eigenvectors and coefficients be eigenvalues. So, the perfect pre-conditioner is the matrix inverse. However, this is computationally impractical, as it is more-or-less equivalent to the solution of the initial system $Ax = b$ and would take $O(N^3)$ operations.

The simplest pre-conditioner matrix is a diagonal matrix of the values at the diagonal of the matrix $A$. This is called Jacobi pre-conditioner. It is equivalent to scaling the matrix along the coordinate axes by values at the diagonal of matrix $A$. Obviously, this is an imperfect solution.

A more practical pre-conditioner would be an incomplete Cholesky decomposition. I will first discuss a regular Cholesky decomposition in the context of conjugate directions, and then its sparse variation.

#### Cholesky decomposition as a special case of conjugate directions

Cholesky decomposition is a representation of our matrix $A$ as a product of lower-diagonal matrix $L$ and its transpose upper-diagonal matrix $L^T$: $A = L L^T$.

One can see that the vectors of inverse of Cholesky decomposition are actually conjugate directions of matrix $A$, such that the first direction is parallel to the first coordinate axis, second direction is in the plan of the first two axes etc. Consider an upper-triangular matrix $V$:

$V^T A V = I$

$V^T A = V^{-1}$

$A = {V^{-1}}^T V^{-1}$

Now we can denote $L = {(V^{-1})}^T$ and $L^T = V^{-1}$, and we get $A = L L^T$.

To see that inverse of an upper-triangular matrix is upper-triangular, just apply the textbook method of matrix inversion, i.e. write down a matrix and a unit matrix next to each other, Gauss-eliminate the left, and you will obtain its inverse from the right:

$\begin{pmatrix} a^1_1 & a^1_2 & \cdots & a^1_{n-1} & a^1_n & 1 & 0 & \cdots & 0 & 0 \\\ & a^2_2 & \cdots & a^2_{n-1} & a^2_n & & 1 & \cdots & 0 & 0 \\\ & & \ddots & \vdots & \vdots & & & \ddots & \vdots & \vdots \\\ & & & a^{n-1}_{n-1} & a^{n-1}_n & & & & 1 & 0 \\\ & & & & a^n_n & & & & & 1 \end{pmatrix}$

#### Cholesky decomposition algorithm

The algorithm for Cholesky decomposition is actually pretty straightforward.

$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} l_{1,1} & 0 & 0 \\ l_{2,1} & l_{2,2} & 0 \\ l_{3,1} & l_{3,2} & l_{3,3} \\ \end{pmatrix} \cdot \begin{pmatrix} l_{1,1} & l_{2,1} & l_{3,1} \\ 0 & l_{2,2} & l_{3,2} \\ 0 & 0 & l_{3,3} \\ \end{pmatrix}$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} l_{1,1} & 0 & 0 \\ ? & ? & 0 \\ ? & ? & ? \\ \end{pmatrix} \cdot \begin{pmatrix} l_{1,1} & ? & ? \\ 0 & ? & ? \\ 0 & 0 & ? \\ \end{pmatrix} \implies l^2_{1,1} = a_{1,1}$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} l_{1,1} & 0 & 0 \\ ? & ? & 0 \\ ? & ? & ? \\ \end{pmatrix} \cdot \begin{pmatrix} ? & l_{2,1} & ? \\ 0 & ? & ? \\ 0 & 0 & ? \\ \end{pmatrix} \implies l_{2,1} = a_{1,2} / l_{1,1}$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} l_{1,1} & 0 & 0 \\ ? & ? & 0 \\ ? & ? & ? \\ \end{pmatrix} \cdot \begin{pmatrix} ? & ? & l_{3,1} \\ 0 & ? & ? \\ 0 & 0 & ? \\ \end{pmatrix} \implies l_{3,1} = a_{1,3} / l_{1,1}$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} ? & 0 & 0 \\ l_{2,1} & l_{2,2} & 0 \\ ? & ? & ? \\ \end{pmatrix} \cdot \begin{pmatrix} ? & l_{2,1} & ? \\ 0 & l_{2,2} & ? \\ 0 & 0 & ? \\ \end{pmatrix} \implies l_{2,2}^2 = a_{2,2} - l_{1,1}^2$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} ? & 0 & 0 \\ ? & ? & 0 \\ l_{3,1} & ? & ? \\ \end{pmatrix} \cdot \begin{pmatrix} l_{1,1} & ? & ? \\ 0 & ? & ? \\ 0 & 0 & ? \\ \end{pmatrix} \implies l_{3,1} = a_{3,1} / l_{1,1}$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} ? & 0 & 0 \\ l_{1,2} & l_{2,2} & 0 \\ ? & ? & ? \\ \end{pmatrix} \cdot \begin{pmatrix} ? & ? & l_{3,1} \\ 0 & ? & l_{3,2} \\ 0 & 0 & l_{3,3} \\ \end{pmatrix} \implies l_{1,2} l_{3,1} + l_{2,2} l_{3,2} = a_{2,3} \implies l_{3,2} = \frac{ a_{2,3} - l_{1,2} l_{3,1} } { l_{2,2} }$
$\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ a_{2,1} & a_{2,2} & a_{2,3} \\ a_{3,1} & a_{3,2} & a_{3,3} \\ \end{pmatrix} = \begin{pmatrix} ? & 0 & 0 \\ ? & ? & 0 \\ l_{3,1} & l_{3,2} & l_{3,3} \\ \end{pmatrix} \cdot \begin{pmatrix} ? & ? & l_{3,1} \\ 0 & ? & l_{3,2} \\ 0 & 0 & l_{3,3} \\ \end{pmatrix} \implies l_{3,3}^2 = a_{3,3} - l_{3,1}^2 - l_{3,2}^2$

#### Incomplete Cholesky decomposition

Full Cholesky decomposition is expensive and takes ~$O(N^3)$ operations.

In case of sparse matrices (or introducing sparsification ourselves), we can achieve the same result with a much smaller compute.

The simplest form of incomplete Cholesky decomposition repeats the sparsity patter of the original matrix. E.g. if the original matrix is:

$A = \begin{pmatrix} 1 && 0 && 0 && 2 \\ 0 && 3 && 0 && 4 \\ 0 && 0 && 5 && 0 \\ 2 && 4 && 0 && 6 \end{pmatrix}$

its incomplete Cholesky decomposition would have a sparsity pattern of:

$L = \begin{pmatrix} ? && 0 && 0 && 0 \\ 0 && ? && 0 && 0 \\ 0 && 0 && ? && 0 \\ ? && ? && 0 && ? \end{pmatrix}$

Zeros above main diagonal stem from the fact that $L$ is lower diagonal, zeros below the main diagonal reproduce the sparsity pattern of matrix $A$:

$A = L L^T = \begin{pmatrix} ? && 0 && 0 && 0 \\ 0 && ? && 0 && 0 \\ 0 && 0 && ? && 0 \\ ? && ? && 0 && ? \end{pmatrix} \cdot \begin{pmatrix} ? && 0 && 0 && ? \\ 0 && ? && 0 && ? \\ 0 && 0 && ? && 0 \\ 0 && 0 && 0 && ? \end{pmatrix}$

This approach is known as $IC(0)$. If you want more precision and less sparsity, you can take the sparsity pattern of $A^2$, which has less 0 elements, and calculate Cholesky decomposition for it. This would be called $IC(1)$. And so on, then it would be called $IC(k)$. Greater $k$ would result in a better pre-conditioning at the cost of more compute.

If the matrix $A$ is not sparse at all, we can employ $IC(\tau)$ approach: choose some threshold of absolute value of an element of matrix $A$, called $\tau$ and if the element is smaller than that, consider it 0. Thus, we achieve a sparse matrix with a Frobenius norm more or less close to the original matrix $A$. Hence, we get a reasonable pre-conditioner at an $O(N^2)$ compute and $O(N)$ memory cost.

## References:

Written by Boris Burkov who lives in Moscow, Russia, loves to take part in development of cutting-edge technologies, reflects on how the world works and admires the giants of the past. You can follow me in Telegram