Linear Algebra - LU Decomposition Example

Let’s perform an LU decomposition on a $4\times 4$ matrix $A$. Here’s an example:

$$A=\left(\begin{array}{cccc}1& 2& 4& 3\\ 3& 8& 14& 11\\ 2& 6& 13& 9\\ 4& 12& 29& 18\end{array}\right)$$

1. Transform $A$ into $U$
   Eliminate $a_{21},a_{31}$, and $a_{41}$ :

• To eliminate 3 under 1 (first column), multiply the first row by 3 and subtract it from the second row.
• To eliminate 2 under 1 , multiply the first row by 2 and subtract it from the third row.
• To eliminate 4 under 1 , multiply the first row by 4 and subtract it from the fourth row.

Proceed to the second column:

• Now focus on making the matrix below the main diagonal of the second column zeros, considering the matrix already modified in step 1.

And so on for the third column:

• You only need to eliminate $a_{43}$ in this step.

2. Construct $U$
   The matrix $U$ is the result of these elimination steps. Since the actual arithmetic is not shown here, let’s directly jump to constructing $L$ from the multipliers used.

3. Construct $L$
   The matrix $L$ captures the multipliers used in each step:
   For $a_{21}$, if the multiplier was 3 (you subtracted 3 times the first row from the second row), then $l_{21}=3$.
   For $a_{31}$, if the multiplier was 2 , then $l_{31}=2$.
   For $a_{41}$, if the multiplier was 4 , then $l_{41}=4$.
   The process continues similarly for other subdiagonal elements based on the multipliers used.

$$L=\left(\begin{array}{cccc}1& 0& 0& 0\\ l_{21}& 1& 0& 0\\ l_{31}& l_{32}& 1& 0\\ l_{41}& l_{42}& l_{43}& 1\end{array}\right),U=\left(\begin{array}{cccc}{u}_{11}& u_{12}& u_{13}& u_{14}\\ 0& u_{22}& u_{23}& u_{24}\\ 0& 0& u_{33}& u_{34}\\ 0& 0& 0& u_{44}\end{array}\right)$$

Gradient
$\nabla F(x,y,z)=(\frac{dF}{dx},\frac{dF}{dy},\frac{dF}{dz})$
Example:

$$\begin{array}{c}F(x,y,z)=x+y^2+z^3\\ \nabla F(x,y,z)=\left(\frac{dF}{dx},\frac{dF}{dy},\frac{dF}{dz}\right)=\left(1,2y,3z^2\right)\end{array}$$

If we want to find the direction to move to increase our function the fastest, we plug in our current coordinates (such as 3,4,5) into the gradient and get:

$$\text{ direction }=\left(1,2(4),3(5{)}^2\right)=(1,8,75)$$

The gradient points to the direction of greatest increase;