Science:Math Exam Resources/Courses/MATH307/April 2005/Question 01 (a)

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MATH307 April 2005

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Question 01 (a)
Consider the symmetric matrix $B = (\begin{matrix} 2 & - 2 & 0 & 4 \\ - 2 & 3 & 2 & 1 \\ 0 & 2 & 3 & 3 \\ 4 & 1 & 3 & 2 \end{matrix})$ Find the $L D U$ decomposition of B.

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Hint
Science:Math Exam Resources/Courses/MATH307/April 2005/Question 01 (a)/Hint 1

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Solution
We want to find a lower triangular matrix, a diagonal matrix, and an upper triangular matrix, which we call $L, D, U$ respectively such that when multiplied together, we get $B = L D U$ . First we perform row operations on $B$ until $B$ becomes an upper triangular matrix, keeping track of the the row operations performed: $\begin{aligned} B = (\begin{matrix} 2 & - 2 & 0 & 4 \\ - 2 & 3 & 2 & 1 \\ 0 & 2 & 3 & 3 \\ 4 & 1 & 3 & 2 \end{matrix}) \to (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 2 & 3 & 3 \\ 4 & 1 & 3 & 2 \end{matrix}) \to (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 2 & 3 & 3 \\ 0 & 5 & 3 & - 6 \end{matrix}) \\ \to (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & - 1 & - 7 \\ 0 & 5 & 3 & - 6 \end{matrix}) \to (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & - 1 & - 7 \\ 0 & 0 & - 7 & - 31 \end{matrix}) \to (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & - 1 & - 7 \\ 0 & 0 & 0 & 18 \end{matrix}) . \end{aligned}$ The row operations performed are: $E_{1}$ : row1 + row2 $E_{2}$ : -2row1 + row4 $E_{3}$ : -2row2 + row3 $E_{4}$ : -5row2 + row4 $E_{5}$ : -7row3 + row4 These row operations translate into the following 5 elementary matrices $E_{i}$ as described above: $\begin{aligned} E_{1} = (\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}), E_{2} = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ - 2 & 0 & 0 & 1 \end{matrix}), E_{3} = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & - 2 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) \\ E_{4} = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & - 5 & 0 & 1 \end{matrix}), E_{5} = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & - 7 & 1 \end{matrix}) . \end{aligned}$ Hence, we have $\begin{aligned} (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & - 1 & - 7 \\ 0 & 0 & 0 & 18 \end{matrix}) = E_{5} E_{4} E_{3} E_{2} E_{1} B = (\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ - 2 & - 2 & 1 & 0 \\ 7 & 9 & - 7 & 1 \end{matrix}) B . \end{aligned}$ Let $E_{L} = E_{5} E_{4} E_{3} E_{2} E_{1}$ describe the matrix product of elementary matrices required to take $B$ to an upper triangular matrix. Similarly, we perform 3 more row operations to make the diagonal entries of $B$ be all ones: $E_{6}$ : 1/2row1 $E_{7}$ : -1row3 $E_{8}$ : 1/18*row4 and we let $E_{D} = E_{8} E_{7} E_{6}$ to describe the matrix product of elementary matrices required to make the diagonal entries of $B$ be all ones. That is, we have our desired upper triangular matrix $U$ after multiplying $E_{D}, E_{L}, B$ : $E_{D} E_{L} B = E_{D} (\begin{matrix} 2 & - 2 & 0 & 4 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & - 1 & - 7 \\ 0 & 0 & 0 & 18 \end{matrix}) = (\begin{matrix} 1 & - 1 & 0 & 2 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1 \end{matrix}) = U$ In summary, $\begin{aligned} E_{L} = (\begin{matrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ - 2 & - 2 & 1 & 0 \\ 7 & 9 & - 7 & 1 \end{matrix}), E_{D} = (\begin{matrix} 1 / 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 1 / 18 \end{matrix}), \\ U = (\begin{matrix} 1 & - 1 & 0 & 2 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1 \end{matrix}) \end{aligned}$ Since $E_{D} E_{L} B = U$ , we can invert $E_{D}$ and $E_{L}$ to attain $D$ and $L$ respectively: $\begin{aligned} E_{D} E_{L} B = U \Rightarrow B = E_{L}^{- 1} E_{D}^{- 1} U \\ \Rightarrow D = E_{D}^{- 1} = (\begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 18 \end{matrix}) \\ \Rightarrow L = E_{L}^{- 1} = (\begin{matrix} 1 & 0 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 2 & 5 & 7 & 1 \end{matrix}) \end{aligned}$ . Hence, we have $\begin{aligned} B = L D U = (\begin{matrix} 1 & 0 & 0 & 0 \\ - 1 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 2 & 5 & 7 & 1 \end{matrix}) (\begin{matrix} 2 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & - 1 & 0 \\ 0 & 0 & 0 & 18 \end{matrix}) (\begin{matrix} 1 & - 1 & 0 & 2 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 1 & 7 \\ 0 & 0 & 0 & 1 \end{matrix}) . \end{aligned}$

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Question 01 (a)

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