Question

Linear algebra

4 5 5 (12 points) Consider the symmetric matrix A = 5 4 -5 5 -5 4 The correct characteristic polynomial is 23 – 1222 – 272 +486, but you are still expected to show the steps that lead to this answer. Show details! Hint: show that 9 is one root, and find the others. Find an orthogonal matrix Q that diagonalizes A. Check in writing that AQ = QD, where D is a diagonal matrix. Specify D as part of your solution.

Answer 1

charateristic polynomial of above matrix is

A

$\begin{vmatrix} 4-\lambda &5 &5 \\ 5& 4-\lambda &-5 \\ 5 & -5 &4-\lambda \end{vmatrix}= 4-\lambda \begin{vmatrix} 4-\lambda &-5 \\ -5 & 4-\lambda \end{vmatrix}-5\begin{vmatrix} 5 &-5 \\ 5 & 4-\lambda \end{vmatrix}+5\begin{vmatrix} 5 & 4-\lambda \\ 5 &-5 \end{vmatrix}$

$= (4-\lambda )((4-\lambda )^2-25)-5(5(4-\lambda )+25)+5(-25-(20-5\lambda ))$

$=(4-\lambda )^3-25(4-\lambda )-5(45-5\lambda )+5(-45+5\lambda )$

= 64-3-483 + 123 - 100+ 253-450 + 503

  $=-\lambda ^3+12\lambda ^2+27\lambda -486$

charateristic equation is $-\lambda ^3+12\lambda ^2+27\lambda -486=0$

As 9 is a root so by long division method we have

1-30 -54 13-12-21,486 -१ 3 22 - १ + - 486 31- 27 - 3 27 + + -(५+५86 -54 + ५86 + X (-१) (-3 d -5५) Consider ď²-32-54 ते-१०+6-54 (-१) + 6/-१) (d + 6 ) (-१) So, (-१) (+6)/d-१) 4-6,११

Now we find eigenvector

Now we find eigen vector R=R, -2 (R2) For d=-6 o A-d I = 10 5 5 -5 s 10 e 5 -5 10 Ri= IR; i= 1,2,3 2, + 13 = 0 => 7,5-83 X2 - Xz=0 =) *2 =23 2 1 2. 13 23 1 12 13 13 13 R, <=) R2 2 As X3 is free Let X3= 1 2 V= - R2= R2-2 R, R3 R3-R, 2 -3 3 -3 3 i= 2,3 Ri= -| Ri 2 Rz = R₃-R2 2 1 o

daa A-d I = - 5 5 5 5 -5 -5 -S -5 S i=1,2,3 Ro=-I R 5 1 1 R₂ = R₂ + R, R3= R3 +R, 22-73- o =) a=X2 + x3 % +03 X2 12 23 12 13 :) V2 V₂ Let X2=1 s x=0 co

Let X2=0, X3 = 1 V3 11

we need to find orthogonal matrix Q that diagonalizes A

DEFINITION :- A n x n matrix is orthogonal matrix if $AA^T=I$ where $A^T$ is tranpose of . In particular $A^{-1}=A^T$ i.e a matrix is invertible.

we have to find a orthogonal matrix   that diagonalizes A. In other words we are going to write where is orthogonal matrix and is diagonal.

therefore we normalise $v_1,v_2,v_3$

Vi llvill= 5+12+ (11²+ (0)2 = √3 =( U = V IV, 11 1 I IS V 1₂11=5 (11²+ (11²+0²=52 6) U₂ = V2 11211 O SI SI- r2 V3- (:) (1)²+ o²+ (112 11V31) V2 Uz= V3 "V311 o 15

now our $Q = \begin{bmatrix} \frac{-1}{\sqrt3} & \frac{1}{\sqrt2} &\frac{1}{\sqrt2} \\ \frac{1}{\sqrt3} & \frac{1}{\sqrt2} &0 \\ \frac{1}{\sqrt3} & 0 & \frac{1}{\sqrt2} \end{bmatrix}, D =\begin{bmatrix} 6 &0 &0 \\ 0& 9 &0 \\ 0 & 0 &9 \end{bmatrix}$

$\begin{bmatrix} 4 &5 &5 \\ 5& 4 &-5 \\ 5& -5 &4 \end{bmatrix}\begin{bmatrix} \frac{-1}{\sqrt3} & \frac{1}{\sqrt2} &\frac{1}{\sqrt2} \\ \frac{1}{\sqrt3} & \frac{1}{\sqrt2} &0 \\ \frac{1}{\sqrt3} & 0 & \frac{1}{\sqrt2} \end{bmatrix}=\begin{bmatrix} \frac{-1}{\sqrt3} & \frac{1}{\sqrt2} &\frac{1}{\sqrt2} \\ \frac{1}{\sqrt3} & \frac{1}{\sqrt2} &0 \\ \frac{1}{\sqrt3} & 0 & \frac{1}{\sqrt2} \end{bmatrix}\begin{bmatrix} 6 &0 &0 \\ 0& 9 &0 \\ 0& 0 &9 \end{bmatrix}=\begin{bmatrix} \frac{-6}{\sqrt3} & \frac{9}{\sqrt2} &\frac{9}{\sqrt2} \\ \frac{6}{\sqrt3} &\frac{9}{\sqrt2} &0 \\ \frac{6}{\sqrt3} & 0 & \frac{9}{\sqrt2} \end{bmatrix}$