TDSM 8.9

LU factorization of a matrix is not necessarily unique. Example: proof for [math]2 \times 2[/math] square matrix:

Let [math] L = \begin{bmatrix} 1 & 0\\ l & 1 \end{bmatrix} [/math], [math] U = \begin{bmatrix} u_1 & u_3\\ 0 & u_2 \end{bmatrix} [/math]

[math] \Rightarrow LU = \begin{bmatrix} u_1 & u_3\\ lu_1 & lu_3+u_2 \end{bmatrix} [/math]

Let [math] M = \begin{bmatrix} m_1 & m_3\\ m_2 & m_4 \end{bmatrix} = LU [/math]

[math]\Rightarrow \left\{ \begin{array}{lr} u_1 = m_1 & \\ lu_1 = m_2 & \\ u_3 = m_3& \\ lu_3 + u_2 = m_4 \end{array} \right. [/math]

Let [math]m_1 = m_2 = 0 \Rightarrow[/math] There are 3 equations for 4 variables [math]\Rightarrow[/math] There are many value for [math]l[/math] satisfies the equations.

[math]\Rightarrow[/math] LU factorization of [math]M[/math] not unique.