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Mk Utkarsh
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in Linear Algebra
Oct 4, 2018

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Best answer

We know that MX= ⋋X where X is the eigen vector correspondig to eigen value ⋋

So for ⋋=1,

$\begin{pmatrix} a &b &c \\ b &d &e \\ c &e &f \end{pmatrix} * \begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix} = 1* \begin{bmatrix} 1\\ 1 \\ 1 \end{bmatrix}$

a+b+c=1 --> (1)

b+d+e=1 --> (2)

c+e+f=1 --> (3)

For ⋋=0

$\begin{pmatrix} a &b &c \\ b &d &e \\ c &e &f \end{pmatrix} * \begin{bmatrix} 1\\ -1 \\ 0 \end{bmatrix} = 0* \begin{bmatrix} 1\\ -1 \\ 0 \end{bmatrix}$

a-b=0 => a=b

b-d=0 => b=d

c-e=0 =>c=e

So, a=b=d and c=e

Let us say a=b=d=x and c=e=y

Now the matrix becomes :

$\begin{bmatrix} x & x &y \\ x & x & y\\ y &y & f \end{bmatrix}$

Given eigen values are 1,0,3

Trace of matrix = sum of eigen values

x+x+f=1+0+3 = 4

=>2x+f=4 --> (4)

Also from equ (1)

x+x+y=1 => 2x+y=1 --> (5)

From equ (2)

y+y+f=1 => 2y+f=1 --> (6)

From (4), (5) and (6),

f=1-2y (from (6) )

Putting this in (4)

2x+f=4 => 2x+1-2y=4

=> 2x-2y =3

2x+y=1 (from (5) )

Solving these two we get y= -2/3

f=1-2y (from (6) )

= 1 - 2(-2/3) = 1 +4/3 = 7/3

3f = 7.

The sum of the elements of the main diagonal of a matrix = sum of all eigen values of the matrix

Ref: Theorem 2 here

https://www.adelaide.edu.au/mathslearning/play/seminars/evalue-magic-tricks-handout.pdf

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