The LU Factorization

Let $\mathbf{A} = \begin{bmatrix} 1 & 4 & 5 \\ 4 & 18 & 26 \\ 3 & 16 & 30 \end{bmatrix}$.

Let $\mathbf{A}$ and $\overrightarrow {b}$ be the matrices

Determine all values of $\xi$ for which $\mathbf{A} = \begin{bmatrix} \xi & 2 & 0 \\ 1 & \xi & 1 \\ 0 & 1 & \xi \end{bmatrix}$ fails to have an LU factorization.

If $\mathbf{A}$ is a nonsingular matrix that possesses an LU factorization, prove that the pivot that emerges after $(k+1)$ stages of standard Gaussian elimination using only Type III operations is given by

where $\mathbf{A}_{k}$ and $\mathbf{A}_{k+1} = \begin{pmatrix} \mathbf{A}_{k} & \overrightarrow {b} \\ \overrightarrow {c}^{T} & a_{k+1,k+1} \end{pmatrix}$ are the leading principal submatrices of orders $k$ and $k+1$, respectively. Use this to deduce that all pivots must be nonzero when an LU factorization for $\mathbf{A}$ exists.

If $\mathbf{A}$ is a matrix that contains only integer entries and all of its pivots are $1$, explain why $\mathbf{A}^{-1}$ must also be an integer matrix.

Consider the tridiagonal matrix $\mathbf{T} = \begin{bmatrix} \beta_{1} & \gamma_{1} & 0 & 0 \\ \alpha_{1} & \beta_{2} & \gamma_{2} & 0 \\ 0 & \alpha_{2} & \beta_{3} & \gamma_{3} \\ 0 & 0 & \alpha_{3} & \beta_{4} \end{bmatrix}$.

$\mathbf{A}_{n \times n}$ is called a band matrix if $a_{ij} = 0$ whenever $\left|i-j\right| > w$ for some positive integer $w$, called the bandwidth. In other words, the nonzero entries of $\mathbf{A}$ are constrained to be in a band of $w$ diagonal lines above and below the main diagonal. For example, tridiagonal matrices have bandwidth one, and diagonal matrices have bandwidth zero. If $\mathbf{A}$ is a nonsingular matrix with bandwidth $w$, and if $\mathbf{A}$ has an LU factorization $\mathbf{A} = \mathbf{L}\mathbf{U}$, then $\mathbf{L}$ inherits the lower band structure of $\mathbf{A}$, and $\mathbf{U}$ inherits the upper band structure in the sense that $\mathbf{L}$ has "lower bandwidth" $w$, and $\mathbf{U}$ has "upper bandwidth" $w$. Illustrate why this is true by using a generic $5 \times 5$ matrix with a bandwidth of $w = 2$.

Recall the LDU factorization: factoring the diagonal pivots out of the upper-triangular factor $\mathbf{U}$ of an LU factorization $\mathbf{A} = \mathbf{L}\mathbf{U}$ produces $\mathbf{A} = \mathbf{L}\mathbf{D}\mathbf{U}'$, where $\mathbf{D} = \diag (u_{11}, u_{22}, \ldots , u_{nn})$ is the diagonal matrix of pivots and $\mathbf{U}'$ is a rightmost upper-triangular matrix with $1$'s on its diagonal (renaming $\mathbf{U}'$ as $\mathbf{U}$ hereafter, so both $\mathbf{L}$ and $\mathbf{U}$ have unit diagonals in the LDU factorization).

Recall: a symmetric matrix $\mathbf{A}$ possessing an LU factorization in which each pivot is positive is said to be positive definite, and such a matrix can be uniquely factored as $\mathbf{A} = \mathbf{R}^{T}\mathbf{R}$, where $\mathbf{R}$ is an upper-triangular matrix with positive diagonal entries; this is the Cholesky factorization of $\mathbf{A}$, and $\mathbf{R}$ is the Cholesky factor of $\mathbf{A}$.

Explain why $\mathbf{A} = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 8 & 12 \\ 3 & 12 & 27 \end{bmatrix}$ is positive definite, and then find the Cholesky factor $\mathbf{R}$.