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LU factorization

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matqkks
#1
Nov28-12, 09:54 AM
P: 153
What is the most motivating way to introduce LU factorization of a matrix? I am looking for an example or explanation which has a real impact.
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HallsofIvy
#2
Nov28-12, 07:38 PM
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PF Gold
P: 39,353
One motivation is that it makes solving equations very easy! If A= LU where, of course, L is a "lower diagonal" and U is "upper diagonal", then we can solve Ax= b by writing it as LUx= b so that Ux= y= L[sup]-1[sup]b, which can be done by 'back substitution" and then solving x= U-1y again by back substitution.

For example, if
[tex]A= LU= \begin{bmatrix}2 & 0 \\ 1 & 3\end{bmatrix}\begin{bmatrix}1 & 3 \\ 0 & 2\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix}2 \\ 1\end{bmatrix}[/tex]
We can let Ux= y so the equation becomes
[tex]Ly= \begin{bmatrix}2 & 0 \\ 1 & 3\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix}= \begin{bmatrix}2 & 1 \end{bmatrix}[/tex]

The first row is equivalent to the equation [itex]2y_1= 2[/itex] so we have immediately [itex]y_1= 1[/itex]. With that value, the second equation, [itex]y_1+ 3y_2= 1[/itex] becomes [itex]1+ 3y_2= 1[/itex] so that [itex]3y_2= 0[/itex] and [itex]y_2= 0[/itex].

Then, since we defined y to be Ux, we have
[tex]x= \begin{bmatrix} 1 & 3 \\ 0 & 2 \end{bmatrix}\begin{bmatrix}x_1 \\ x_2\end{bmatrix}= \begin{bmatrix}1 \\ 0 \end{bmatrix}[/tex]

Now, the second row gives the equation [itex]2x_2= 0[/itex] so that [itex]x_2= 0[/itex] and then the top row becomes [itex]x_1+ 3x_2= x_1= 1[/itex].
lurflurf
#3
Nov28-12, 09:03 PM
HW Helper
P: 2,263
Much of linear algebra can be seen as factoring operators. Triangular equations are easy to solve so it is helpful to solve an equation by factoring it into two triangular equations. Another views is to thing about solving an equation by a usual method like row reduction. In doing so we throw away a lot of of the matrix, and as a result if we solve many equations with the same operator we needlessly repeat work. We can by slight modification of row reduction keep this information at almost no cost. If we solve the equation many times we can reuse the information each time. This is the LU decomposition.


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