PLEASE HELP ME Let A 1 2 1 1 1 5 0 1 0 2 1 4 0 0 1 0 4 3 0

PLEASE HELP ME

Let A = (1 2 1 ?1 ?1 5)( 0 1 0 2 ?1 4)( 0 0 1 0 ?4 3)( 0 0 0 1 ?2 2)( 0 0 0 0 1 1)( 0 0 0 0 0 1)

B) Use the dimension of the appropriate solution space to determine which chain structure it has. Indicate the solution space you used and its dimension.

Solution

Solution: a) Given matrix is A=

The eigenspce of 1 is the null space of

A-I =

This reduces to

There are three free variables x₁, x₄ and x₅ . Here x3 =0

We set x₁ =s, x₄ = t and x₅ = u, the first row of reduction gives

x2 = t+u,

So x is in the eigenspace if

x =

=

+

+

So dimension of its eigenspace 3.

Therefore, geometric multiplicity of \\lambda = 1 is 3(three), and algebraic
multiplicity of \\lambda = 1 is 5(five). Therefore, this matrix is defective as both are not equal.