0
$\begingroup$

$$ \begin{array}{l}{\text { The regular Matrix } A \in \mathbb{R}^{n \times n} \text { with the LU decomposition } A=L R \text { gets an extra column }} \\ {\text { anr row so }} \\ {\qquad \widehat{A}=\left(\begin{array}{cc}{A} & {b} \\ {c^{T}} & {d}\end{array}\right), \quad b, c \in \mathbb{R}^{n}, \quad d \in \mathbb{R}}\end{array} $$ $$ \begin{array}{l}{\text { a) State the LU-Decomposition of } \widehat{A} \text { with a matrix } \widehat{L} \text { of the form }} \\ {\qquad \widehat{L}=\left(\begin{array}{cc}{L} & {0} \\ {x^{T}} & {1}\end{array}\right), \quad x \in \mathbb{R}^{n}}\end{array} $$ $$ \begin{array}{l}{\text { b) Show that } \widehat{A} \text { is only regular if } d-c^{T} A^{-1} b \neq 0 .} \\ {\text { c) In singular care state a solution } z \in \mathbb{R}^{n+1} \backslash\{0\} \text { of the homogeneously }} \\ {\text { equation } \widehat{A} z=0}\end{array} $$

Could someone give me advice on how to solve that? I have literally no clue how to start that one

$\endgroup$
1
$\begingroup$

HINT

For (a), assume $\hat{U}$ in $\hat{A}=\hat{L}\hat{U}$ of the form $$ \hat{U}=\begin{bmatrix}U&u\\0&\upsilon\end{bmatrix}. $$ From $$ \begin{bmatrix} A&b\\c^T&d \end{bmatrix} = \begin{bmatrix} L & 0 \\ x^T & 1 \end{bmatrix} \begin{bmatrix}U&u\\0&\upsilon\end{bmatrix} = \begin{bmatrix} LU & Lu \\ x^TU & x^Tu+\upsilon \end{bmatrix}, $$ we have $$ A=LU, \quad b=Lu, \quad c=U^Tx, \quad d=x^Tu+\upsilon. $$ Since $A=LU=LR$, this gives $U=R$. Can you go ahead and solve the remaining three equations for $x$, $u$, and $\upsilon$?

For (b), it is clear that $\hat{A}$ is singular if and only if $\upsilon=0$. When does this happen?

For (c), use the previously obtained LU factorization or, better, directly the definition. The matrix $\hat{A}$ is singular if and only if there is a nonzero $\hat{x}:=[x^T,\xi]^T$ such that $\hat{A}\hat{x}=0$, that is, $$ Ax+\xi b=0, \quad c^Tx+d\xi=0. $$ Clearly, $\xi\neq 0$. Otherwise, $A$ would need to be singular. Given $\xi\neq 0$, we can get the corresponding $x$ from the first equation because $A$ is nonsingular. Obviously, $\xi=0$ leads to the zero solution of the homogeneous equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.