Contexts and definitions of the $LU$ method:

Consider a matrix $A \in \mathbb{R}^{n \times n}$:

$$ A = \begin{bmatrix}a_{11} & \dots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \dots & a_{nn} \end{bmatrix} $$

We define: $$ \ell_{i,j} := \frac{a_{ij}}{a_{jj}}, \,\,\, m_j = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ \ell_{j+1, j} \\ \ell_{j+2,j} \\ \vdots \\ \ell_{n,j} \end{bmatrix}, e_j = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 (j_{th} \text{ line}) \\ 0 \\ \vdots \\ 0 \end{bmatrix}, M_j = I - m_je^T_j$$

$M = M_{n-1}P_{n-1} \dots M_1P_1 $, where each $ P_k$ is a permutation matrix and $P = P_{n-1} \dots P_1$.

The $LU$ method with pivoting says: $$ PA = LU \\ U = MA \text{ is upper triangular}\\ L = PM^{-1} \text{ is unit lower triangular}$$

I want to proof the following: $$ L = PM^{-1} \text{ is unit lower triangular}$$

Reference: this is an exercice left to the reader of the book "Numerical Mathematics" by Alfio Quarteroni, second edition (page 89).


Given that we are talking about LU factorisation, I assume that the permutation matrix $P_i$ exchanges line $j$ and $k$ with $j, k \ge i$ (Typically $j=i$ and $k>i$).

With this assumption, we can rewrite the matrix $M$ in the following form:

$$ M = M_{n-1}P_{n-1} M_{n-2}P_{n-2} \dots M_2P_2M_1P_1 = L_{n-1}L_{n-2}\dots L_2L_1 \cdot P_{n-1}P_{n-2}\dots P_2P_1$$

with $$L_{n-1} = M_{n-1}, \quad L_{n-2} = P_{n-1}M_{n-2}P_{n-1}^{-1},\quad L_{n-3} = P_{n-1}P_{n-2}M_{n-3}P_{n-2}^{-1}P_{n-1}^{-1}, \quad \text{etc}$$

thus having $M_i$ equal to $L_i$, but with the sub-diagonal nonzero entries permuted by $\dots P_{i+2}P_{i+1}M_i$. This is a result of the assumption that $P_i$ does not change the order of the rows before row $i$. If this hypothesis is not respected, then $L$ is no longer unit triangular. It follows that the matrices $L_i$ are lower triangular and unit diagonal, and thus so is their product $L^{-1}$.

$$ M = \underbrace{L_{n-1}L_{n-2}\dots L_2L_1}_ {L^{-1}} \cdot \underbrace{P_{n-1}P_{n-2}\dots P_2P_1}_{P}$$

The inverse of a unit diagonal triangular matrix is also a unit triangular matrix which implies that $L$ is unit triangular.

Complementary note N°1: Here is an example (with n = 4) showing that the $M$ is actually equal to $L_{n-1}L_{n-2}\dots L_2L_1 \cdot P_{n-1}P_{n-2}\dots P_2P_1$, as it might be useful to convince yourself:

Using the definition for $L_i$:

$$L_3 L_2 L_1 \cdot P_3 P_2 P_1 = M_3 (P_3 M_2 P_3^{-1})(P_3 P_2 M_1 P_2^{-1} P_3^{-1}) \cdot P_3 P_2 P_1$$

cancelling the permutation matrices gives the expected matrix $M$:

$$L_3 L_2 L_1 \cdot P_3 P_2 P_1 = M_3 P_3 M_2 P_2 M_1 P_1 = M$$

Complementary note N°2: Example showing why $L_i$ and $M_i$ have the same structure. Let's have a look at $L_2$ in the same example: $$ L_2 = P_3 M_2 P_3^{-1}$$ with $M_2$ being of the form \begin{pmatrix} 1 & & & \newline & 1 & & \newline & x & 1& \newline & x && 1 \end{pmatrix}

($x$ means some nonzero entry). $P_3$ will shuffle rows 3 and 4, thus $P_3 L_2$ will be of the form: \begin{pmatrix} 1 & & & \newline & 1 & & \newline & x & 1 \text{ or } 0 & 1 \text{ or } 0\newline & x & 1 \text{ or } 0 & 1 \text{ or } 0 \end{pmatrix}

$P_3^{-1}$ will then shuffle columns 3 and 4 in an opposite way, resetting the 2x2 lower right bloc back to an identity sub-matrix.

Edit: if you want more details, I would suggest reading lectures 20 and 21 of the book "Numerical linear algebra" by Lloyd N. Trefethen & David Bau III, the whole algorithm for LU is explained and it's surprisingly pleasant to read


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.