1.3.5 Write f= x + 10x,x2 + x as a difference of squares, and f= xí+ 10x1x2 + 30x asa sum of squares. What symmetric matrices correspond to these quadratic forms by f= x' Ax?

Given f=x12+10x1x2+x22. Let the associated matrix of f be A=abba is 2×2 symmetric matrix. Then,…

Answered: Write f= x + 10x1x2 + x3 as a…

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

1.3.5 Write f= x + 10x,x2 + x as a difference of squares, and f= xí+ 10x1x2 + 30x as
a sum of squares. What symmetric matrices correspond to these quadratic forms by f
= x' Ax?

Expert Solution

Step 1

Given $f = x_{1}^{2} + 10 x_{1} x_{2} + x_{2}^{2}$ .

Let the associated matrix of $f$ be $A = [\begin{matrix} a & b \\ b & a \end{matrix}]$ is $2 \times 2$ symmetric matrix. Then,

$\begin{array}{rcl} x^{T} A x & = & f \\ [\begin{matrix} x_{1} & x_{2} \end{matrix}] [\begin{matrix} a & b \\ b & a \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] & = & x_{1}^{2} + 10 x_{1} x_{2} + x_{2}^{2} \\ [\begin{matrix} x_{1} & x_{2} \end{matrix}] [\begin{matrix} a x_{1} + b x_{2} \\ b x_{1} + a x_{2} \end{matrix}] & = & x_{1}^{2} + 10 x_{1} x_{2} + x_{2}^{2} \\ a x_{1}^{2} + 2 b x_{1} x_{2} + a x_{2}^{2} & = & x_{1}^{2} + 10 x_{1} x_{2} + x_{2}^{2} \end{array}$

Comparing the co-efficients from both sides, it results,

$A = [\begin{matrix} 1 & 5 \\ 5 & 1 \end{matrix}]$

Now, let us diagonalize the matrix $A$ orthogonally.

Corresponding characteristic equation is

$\begin{array}{rcl} |\begin{matrix} 1 - k & 5 \\ 5 & 1 - k \end{matrix}| & = & 0 \\ {(1 - k)}^{2} & = & 25 \\ (1 - k) & = & \pm 5 \\ k & = & - 4, 6 \end{array}$

Eigenvectors corresponding to the eigenvalue 6 is $[\begin{matrix} 1 \\ 1 \end{matrix}]$ and that to -4 is $[\begin{matrix} - 1 \\ 1 \end{matrix}]$ . Therefore, $P = [\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}]$ . Then, the diagonal matrix corresponding to the matrix $A$ is

$\begin{array}{rcl} D & = & P^{- 1} A P \\ = & [\begin{matrix} 6 & 0 \\ 0 & - 4 \end{matrix}] \end{array}$

Now, let us consider $x = P y$ where $y = [\begin{matrix} y_{1} \\ y_{2} \end{matrix}]$ . Then

$\begin{array}{rcl} f & = & x_{1}^{2} + 10 x_{1} x_{2} + x_{2}^{2} \\ = & x^{T} A x \\ = & {(P y)}^{T} A (P y) \\ = & y^{T} (P^{T} A P) y \\ = & y^{T} D y \\ = & [\begin{matrix} y_{1} & y_{2} \end{matrix}] [\begin{matrix} 6 & 0 \\ 0 & - 4 \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] \\ = & [\begin{matrix} y_{1} & y_{2} \end{matrix}] [\begin{matrix} 6 y_{1} \\ - 4 y_{2} \end{matrix}] \\ = & 6 y_{1}^{2} - 4 y_{2}^{2} \end{array}$

Hence, $f$ is expressed as a difference of two squares by $f = \begin{array}{rcl} 6 \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} 12 \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 4 \end{array} \begin{array}{rcl} y \end{array} \begin{array}{rcl} 22 \end{array}$

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