Conjugate Gradient Method allows you to solve linear systems whose matrix is symmetric and positive definite.

I will study here the Conjugate Gradient Method. This numerical method allows you to solve linear systems whose matrix is symmetric and positive definite. The search for successive directions makes possible to reach the exact solution of the linear system.


We want to solve the following linear system:


where $A$ is a $n\times n$ symmetric and positive definite matrix ($A^\top =A$ and $x^\top Ax>0$, for all $x\in \mathbf{R}^n$ non-zero). Let $x_\star$ be the exact solution of this linear system.

Conjugate directions

As the matrix $A$ is symmetric and positive definite, we can define the following scalar product on $\mathbf{R}^n$: $\langle u,v \rangle_A = u^\top A v.$ Two elements $u,v\in \mathbf{R}^n$ are $A$-conjuguate if:

\[u^\top A v = 0\]

Conjugate Gradient Method consists in building a vectorial sequence $(p_k)_{k\in\mathbf{N}^*}$ of $n$ $A$-conjugate vectors. Consequently, the sequence $p_1,p_2,\ldots,p_n$ form a basis of $\mathbf{R}^n$. The exact solution $x_\star$ can be expanded like follows:

\[x_\star = \alpha_1 p_1 + \cdots + \alpha_n p_n\]

where $\alpha_k = \dfrac{p_k^\top b}{p_k^\top A p_k},k=1,\ldots,n.$

Construction of Conjugate directions

The exact solution $x_\star$ is also the unique one minimizer of the functionnal

\[J(x) = \frac12 x^\top A x - b^\top x, \quad x\in\mathbf{R}^n\]

We have clearly

\[\nabla J(x) = A x - b, \quad x\in\mathbf{R}^n\]


\[\nabla J(x_\star)=0_{\mathbf{R}^n}\]

We define the residual vector of the linear system

\[r_k = b - Ax_k=-\nabla J(x_k)\]

$r_k$ is the direction of the gradient of the functional $J$ in $x_k$.

The new direction of descent $p_{k+1}$ is the same as its $A$-conjugaison with $p_k$, we have then:

\[p_{k+1} = r_k - \frac{p_k^\top A r_k}{p_k^\top A p_k} p_k\]

It is the choice of the coefficient $\dfrac{p_k^\top A r_k}{p_k^\top A p_k}$ wich allows the $A$-conjugaison of the directions $p_k$. If you want you can calculate $(Ap_{k+1},p_k)$, it is zero !

Conjugate Gradient Algorirthm

We calculate the residual $r_0=b-Ax_0$ for any vector $x_0$. We fix $p_0=r_0$.

For $k=1,2,\ldots$

\[q_k=A p_k\]

If $k=1$



\[\beta_{k-1}= \frac{r_{k-1}^\top r_{k-1}}{r_{k-2}^\top r_{k-2}}\] \[p_k=r_{k-1} + \beta_{k-1}p_{k-1}\]


\[\alpha_k=\frac{r_{k-1}^\top r_{k-1}}{p_k^\top q_k}\] \[x_k=x_{k-1}+\alpha_k p_k\] \[r_k=r_{k-1}-\alpha_k q_k\]