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Given the matrix A

 

To find real values of X such that A.X=0,                                                 

  1. deduce the forth row with the given rows to come up with a 4×4 matrix
  2. determine p(λ)=det (A- λIn)
  3. 3.        Factorize p(λ) to obtain p(λ)=( λ- λ1)n1. (λ- λ2)n2 …( λ- λk)nk

Where λi, i=1,…..,k.

These might be real or complex, whereby i and the powers ni is referred to as the algebraic multiplicity of the eigenvalue λi

for every eigen value the associated eigenvalues are calculated.

For instance given λi  as the eigenvector, the eigenvectors are represented by a linear system, given as A.X= λiX or (A- λiIn)X)=0

We introduce a forth row to get a 4 by 4 matrix by subtracting two times row one from row two, we get

 

 

 

 

-1      -2      -2      -17

-2      -8      -8      -54

4        7        7      163

0        4        4       20

The next step is to diagonalize the 4 by 4 matrix A and find the diagonal matrix B whereby, B=P-1AP. This is achieved by deducing a characteristic polynomial fA=det (A-tIn)

fA=    λ+1     2            2       17

          2       λ+8         8        54

         -4        -7         λ-7     -63

          0        -4          -4       λ-20

This results to a polynomial of degree n=4

= λ4-18 λ4-95λ2+1156 λ-1120=0

Solving the polynomial simultaneously, we acquire the roots as the possible values of λ by taking arbitrary values from -20 to +20 through scrutinizing and examining the matrix carefully. The following values of λ satisfied the equation, -1, -8, 7, and 20.

Rewriting the polynomial equation in its factorized form

(λ +1)( λ+8)( λ-7)(-20)=0

These values translates to the possible values of x1 ,x2 x3 and x4 respectively

Therefore,

X   =    

 

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