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

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



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.


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