Answered: Find 2 by 2 symmetric matrices S = ST…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Find 2 by 2 symmetric matrices S = ST with these properties:
(a) Sis not invertible.
(b) S is invertible but cannot be factored into L U (row exchanges needed).
(c) S can be factored into LDLT but not into LLT (because of negative D).

Expert Solution

Step 1

Given: Let S be a $2 \times 2$ symmetric matrix.

To Find :

a) S is not invertible.

b) S is invertible but cannot be factored into $L U$ (row exchanges needed).

c) S can be factored into $L D L^{T}$ but not into $L L^{T}$ (because of negative D).

Step 2

Let $S = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]$ and $S^{T} = [\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}]$

Since $S = S^{T}$ , therefore S is a symmetric matrix.

Now determinant of S is calculated as follows :

$\begin{array}{rcl} d e t S & = & |\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}| \\ = & 1 \times 0 - 0 \times 0 \\ = & 0 \end{array}$

Hence S is not invertible

Let $S = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ and $S^{T} = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$

Since $S = S^{T}$ , therefore S is a symmetric matrix.

Then, let L= lower triangular matrix and U=Upper triangular matrix

$L = (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix})$ , $U = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix})$

Now,

$\begin{array}{rcl} L U & = & (\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}) (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) \\ = & (\begin{matrix} 0 \times 0 + 0 \times 0 & 0 \times 1 + 0 \times 0 \\ 1 \times 0 + 0 \times 0 & 1 \times 1 + 0 \times 0 \end{matrix}) \\ = & (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) \\ \neq & (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) \\ = & S \end{array}$

$\Rightarrow L U \neq S$

$\begin{array}{rcl} d e t S & = & |\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}| \\ = & 1 \times 1 - 0 \times 0 \\ = & 1 - 0 \\ = & 1 \end{array}$

But $d e t S \neq 0$ , hence S is invertible.

Therefore, S is invertible but cannot be factored into $L U$ .

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