LU Decomposition - Example

Example

We factorize the following 2-by-2 matrix:

\begin{bmatrix} 4 & 3 \\ 6 & 3 \\ \end{bmatrix} = \begin{bmatrix} l_{11} & 0 \\ l_{21} & l_{22} \\ \end{bmatrix} \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \\ \end{bmatrix}.

One way to find the LU decomposition of this simple matrix would be to simply solve the linear equations by inspection. Expanding the matrix multiplication gives

This system of equations is underdetermined. In this case any two non-zero elements of L and U matrices are parameters of the solution and can be set arbitrarily to any non-zero value. Therefore to find the unique LU decomposition, it is necessary to put some restriction on L and U matrices. For example, we can conveniently require the lower triangular matrix L to be a unit one (i.e. set all the entries of its main diagonal to ones). Then the system of equations has the following solution:

Substituting these values into the LU decomposition above yields

\begin{bmatrix} 4 & 3 \\ 6 & 3 \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1.5 & 1 \\ \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 0 & -1.5 \\ \end{bmatrix}.

Read more about this topic:  LU Decomposition

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