TDSM 8.11

Proof for [math] 2\times 2[/math] square matrix (higher degree matrixes proof is similar):

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]

[math] M = \begin{bmatrix} m_1 & m_3\\ m_2 & m_4 \end{bmatrix} [/math] is an invertable matrix [math]\Rightarrow x_4x_1 - x_2x_3 = 0[/math]

[math]\iff (lu_3+u_2)u_1 - lu_1u_3 = 0[/math]

[math]\iff u_2 u_1 = 0[/math]

[math]\iff u_1 = 0[/math] or [math]u_2 = 0[/math] (proved)