Difference between revisions of "TDSM 8.9"
(Created page with "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}...") |
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| Line 30: | Line 30: | ||
<math>\Rightarrow \left\{ | <math>\Rightarrow \left\{ | ||
\begin{array}{lr} | \begin{array}{lr} | ||
| − | u_1 = | + | u_1 = m_1 & \\ |
| − | lu_1 = | + | lu_1 = m_2 & \\ |
| − | u_3 = | + | u_3 = m_3& \\ |
| − | lu_3 + u_2 = | + | lu_3 + u_2 = m_4 |
\end{array} | \end{array} | ||
\right. | \right. | ||
</math> | </math> | ||
| − | Let <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. | <math>\Rightarrow</math> LU factorization of <math>M</math> not unique. | ||
Latest revision as of 01:47, 12 December 2017
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.
