The matrix A 2 0 0 0 2 2 0 2 0 2 0 2 0 2 2 4 has two distin

The matrix A = [2 0 0 0 -2 -2 0 -2 0 2 0 2 0 2 -2 4] has two distinct eigenvalues lambda_1

Solution

The characteristic equation of A is det (A- I₄) = 0 or, ⁴ -4 ³+4 ²= 0 or, ² (²-4 +4) = 0 or, ²(-2)²= 0. Thus, A has 2 distinct eigenvalues, ₁ = 0 and ₂ = 2, of multiplicity 2 each.

The eigenvectors of A corresponding to the eigenvalue are solutions to the equation (A- I₄)X = 0. Thus, the eigenvectors of A corresponding to the eigenvalue 2 are solutions to the equation (A- 2I₄)X = 0.We will reduce A- 2I₄ to its RREF as under:

Interchange the 1st row and the 2nd row

Multiply the 1st row by -1/2

Interchange the 2nd row and the 3rd row

Multiply the 2nd row by ½

Add -2 times the 2nd row to the 4th row

Add -2 times the 2nd row to the 1st row

Then the RREF of A-2I₄ is

1

0

2

-1

0

1

-1

1

0

0

0

0

0

0

0

0

  If X = (x,y,z,w)^T , then the above equation is equivalent to x +2z –w = 0 and y-z+w = 0. Hence x = -2z+w and y = z-w so that X = (-2z+w, z-w, z,w)^T = z(-2,1,1,0)^T+w(1,-1,0,1)^T. Thus, the eigenbasis of A corresponding to the eigenvalue 2 is { (-2,1,1,0)^T, (1,-1,0,1)^T}.

The eigenvectors of A corresponding to the eigenvalue 0 are solutions to the equation AX = 0. We will reduce A to its RREF as under:

Multiply the 1st row by ½

Add 2 times the 1st row to the 2nd row

Multiply the 2nd row by -1/2

Add -2 times the 2nd row to the 3rd row

Add -2 times the 2nd row to the 4th row

Nterchange the 3rd row and the 4th row

Multiply the 3rd row by -1/2

Then the RREF of a is

1

0

0

0

0

1

0

1

0

0

1

-1

0

0

0

0

If X = (x,y,z,w)^T , then the above equation is equivalent to x = 0, y+w = 0, z-w = 0 i.e. x = 0, y = -w and z = w so that X = ( 0,-w, w,w)^T = w(0,-1,1,1)^T. Thus, the eigenbasis of A corresponding to the eigenvalue 0 is    {(0,-1,1,1)^T}.

1	0	2	-1
0	1	-1	1
0	0	0	0
0	0	0	0