The matrix A6 2 12 4 5 1 11 1 3 1 7 1 3 1 7 1 has two real e

The matrix A=[(6 -2 12 4), (-5 1 -11 -1),( -3 1 -7 -1),( -3 1 -7 -1)] has two real eigenvalues 1<2. Find these eigenvalues, their multiplicities, and the dimensions of their corresponding eigenspaces. please list the eigenvalue, multiplicity, and dimension for each in that order. show steps

Solution

We assume that the vectors given are the rows of A. Then A =

6

-2

12

4

-5

1

-11

-1

-3

1

-7

-1

-3

1

-7

-1

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

The eigenvector of A corresponding to the eigenvalue -1 is the solution to the equation (A+I₄)X = 0.To solve this equation, we will reduce A+I₄ to its RREF, which is

1

0

0

2

0

1

0

-1

0

0

1

-1

0

0

0

0

If X = (x,y,z,w)^T, then the equation (A+I₄)X = 0 is equivalent to x +2w = 0, y-w = 0, z-w = 0 or, x = -2w, y = w, z = w so that X = (-2w,w,w,w)^T=w (-2,1,1,1)^T. Hence, the eigenspace of A corresponding to the eigenvalue -1 is span{ (-2,1,1,1)^T}. Its dimension is 1.

Similarly, the eigenvectors of A corresponding to the eigenvalue 0 are the solution to the equation AX = 0 . To solve this equation, we will reduce A to its RREF, which is

1

0

0

2

0

1

0

-2

0

0

1

-1

0

0

0

0

If X = (x,y,z,w)^T, then the equation AX = 0 is equivalent to x +2w = 0, y-2w = 0, z-w = 0 or, x = -2w, y =2w, z = w so that X = (-2w,2w,w,w)^T=w (-2,2,1,1)^T. Hence, the eigenspace of A corresponding to the eigenvalue 0 is span{ (-2,2,1,1)^T}. Its dimension is 1.

6	-2	12	4
-5	1	-11	-1
-3	1	-7	-1
-3	1	-7	-1