Let $L$ be a lower-triangular matrix ($n \times n$) and $U$ be an upper-triangular matrix (also $n \times n$).

We've discussed how to solve $Ux = y$ with backward substitution:

$$x_i = \frac{\displaystyle y_i - \sum_{j = i+1}^n u_{ij}x_j}{u_{ii}},\quad i = n, n-1, \ldots, 2,1.$$

We also discussed how to solve $Ly = b$ with forward substitution:

$$y_i = \frac{\displaystyle b_i - \sum_{j=1}^{i-1} l_{ij} y_j }{l_{ii}}, \quad i = 1,2,\ldots,n-1,n.$$

Using an LU factorization¶

Assume that we can express $A$ as $LU$ and we want to solve $Ax=b$:

$$A x = LU x = b = L(Ux) = b.$$

1. Solve $Ly = b$ for $y = Ux$ by forward substitution.
2. Solve $Ux = y$ for $x$ by backward substitution.

Operation count¶

Recall from Lecture 9 that backward substitution requires $O(n^2)$ operations. The same is true of forward substitution.

But what if the $LU$ factorization is not known in advance? How much effort is required to compute it? The answer is $O(n^3)$ because Gaussian elimination is the means by which we $LU$-factorize, as we will see next.

The LU factorization algorithm¶

Assume $A$ is an $n\times n$ matrix that can be put in upper-triangular form without row swaps (we'll deal with those next lecture).

Consider the $3\times 3$ case first

$$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$ $$R_2 - \frac{a_{21}}{a_{11}}R_1 \to R_2$$

$$R_2 - \frac{a_{21}}{a_{11}}R_1 \to R_1$$ $$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

Here $a^{(1)}_{22} = a_{22} - a_{12} \frac{a_{21}}{a_{11}}$ and $a^{(1)}_{23}= a_{23} - a_{13} \frac{a_{21}}{a_{11}}$. The question that will lead us to an $LU$ factorization is: What type of matrix does this row operation correspond to?

$$\begin{bmatrix} 1 & 0 & 0 \\ -a_{21}/a_{11} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

Recall that the goal is to express $A = LU$, so we find the inverse of the lower-triangular matrix:

$$\begin{bmatrix} 1 & 0 & 0 \\ a_{21}/a_{11} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ -a_{21}/a_{11} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

So, after a single step, we have found

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ a_{21}/a_{11} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$$

This is closer to a lower/upper factorization.

Next we perform the row operation $R_{3} - \frac{a_{31}}{a_{11}} R_1 \to R_3$ which corresponds to

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - a_{31}/a_{11} & 0 & 1 \end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & a^{(1)}_{32} & a^{(1)}_{33} \end{bmatrix}$$

where $a^{(1)}_{32} = a_{32} - a_{12} \frac{a_{31}}{a_{11}}$ and $a^{(1)}_{33} = a_{33} - a_{13} \frac{a_{31}}{a_{11}}$

The inverse of the lower-triangular factor can be confirmed

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a_{31}/a_{11} & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ - a_{31}/a_{11} & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0& 0 & 1 \end{bmatrix}$$

Thus:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a_{31}/a_{11} & 0 & 1 \end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & a^{(1)}_{32} & a^{(1)}_{33} \end{bmatrix}$$

Combine the two row operations to yield:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ a_{21}/a_{11} & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a_{31}/a_{11} & 0 & 1 \end{bmatrix}\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & a^{(1)}_{32} & a^{(1)}_{33} \end{bmatrix}$$

For the last step, we need to perform $R_3 - \frac{a^{(1)}_{32}}{a^{(1)}_{22}} R_2 \to R_3$

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -a^{(1)}_{32}/a^{(1)}_{22} & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & a^{(1)}_{32} & a^{(1)}_{33} \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & 0 & a^{(2)}_{33} \end{bmatrix}$$

where $a^{(2)}_{33} = a^{(1)}_{33} - a^{(1)}_{23} \frac{a^{(1)}_{32}}{a^{(1)}_{22}}$.

Again, you can check that the inverse of the matrix on the left is simple

$$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & a^{(1)}_{32}/a^{(1)}_{22} & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -a^{(1)}_{32}/a^{(1)}_{22} & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

so that we get:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & a^{(1)}_{32} & a^{(1)}_{33} \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & a^{(1)}_{32}/a^{(1)}_{22} & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & 0 & a^{(2)}_{33} \end{bmatrix}$$

Inserting this into the existing factorization:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ a_{21}/a_{11} & 1 & 0 \\ a_{31}/a_{11} & 0 & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & a^{(1)}_{32} & a^{(1)}_{33} \end{bmatrix}$$

yields:

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ a_{21}/a_{11} & 1 & 0 \\ a_{31}/a_{11} & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & a^{(1)}_{32}/a^{(1)}_{22} & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & 0 & a^{(2)}_{33} \end{bmatrix}$$

The inner factors simplify and we have an LU factorization

$$\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0 \\ a_{21}/a_{11} & 1 & 0 \\ a_{31}/a_{11} & a^{(1)}_{32}/a^{(1)}_{22} & 1 \end{bmatrix} \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ 0 & a^{(1)}_{22} & a^{(1)}_{23} \\ 0 & 0 & a^{(2)}_{33} \end{bmatrix}$$