Linear Algebra

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- λI_n)
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 n_i is referred to as the algebraic multiplicity of the eigenvalue λ_i

for every eigen value the associated eigenvalues are calculated.

For instance given λ_i  as the eigenvector, the eigenvectors are represented by a linear system, given as A.X= λiX or (A- λ_iI_n)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 f_A=det (A-_tI_n)

f_A=    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   =