Question

Consider the matrix $\mathit{A}\mathbf{=}\left[\begin{array}{ccc}1& 1& -1\\ 3& 2& -5\\ 2& 2& 0\end{array}\right]$ with LDU-factorizationrole="math" localid="1660129574693" $\mathit{A}\mathbf{=}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{0}& \mathbf{0}\\ \mathbf{3}& \mathbf{1}& \mathbf{0}\\ \mathbf{2}& \mathbf{0}& \mathbf{0}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{0}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{0}& \mathbf{2}\end{array}\mathbf{\right]}\mathbf{\left[}\begin{array}{ccc}\mathbf{1}& \mathbf{1}& \mathbf{-}\mathbf{1}\\ \mathbf{0}& \mathbf{1}& \mathbf{2}\\ \mathbf{0}& \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right]}$. Find the LDU-factorization for${\mathit{A}}^{\mathbf{T}}$.

Short answer

The LDU-factorization of $A^T$is $\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ -1& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$.

Step-by-step solution

Determine the LDU-factorization.

Consider the matrix $A=\left[\begin{array}{ccc}1& 1& -1\\ 3& 2& -5\\ 2& 2& 0\end{array}\right]$and LDU-factorization.

$A=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]$

Take transpose both side in the equation role="math" localid="1660129618004" $A=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]$as follows.

$$\begin{gathered} A=\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right] \\ {A}^T={\left(\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\right)}^T \\ {A}^T={\left(\left[\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\right]\right)}^T \\ {A}^T=\left({\left\{\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\right\}}^T\right) \end{gathered}$$

Further, simplify the equation as follows.

role="math" localid="1660129922977" $$\begin{gathered} A^T=\left({\left\{\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]\right\}}^T\right) \\ {A}^T=\left(\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]{\left[\begin{array}{ccc}1& 0& 0\\ 3& 1& 0\\ 2& 0& 0\end{array}\right]}^T\right) \\ {A}^T=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ -1& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 0& 1& 0\\ 0& 0& 0\end{array}\right] \end{gathered}$$

Hence, the LDU-factorization of $A^T$is $\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ -1& 2& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 2\end{array}\right]\left[\begin{array}{ccc}1& 3& 2\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$.