1, B+2E and A+2E have the same eigenvalue. 9, 3 ,3.

(λ? - 15λ? +63 λ-8 1 = 0 solution)

For +2e. λ = 9，4x-2y-2z = 0。 -2x+4y-2z = 0。 α 1 = ( 1, 1, 1)'.

λ=3.x+y+z=0.α2=( 1，- 1，0)′α3 =( 1，0，- 1)′。

Note that b+2e = p (- 1) (a+2e) p, and if (a+2e) x = λ x. 。

Then (b+2e) p (-1) x = p (-1) (a+2e) PP (-1) x = λ p (-1) x.

That is, when x is the eigenvector of A+2E about λ. P (- 1) x is the eigenvector with λ of B+2E.

It is not difficult to calculate,

P^(- 1)=

0, 1,- 1

1,0, 0

0,0, 1.

For B+2E, the eigenvectors β 1, β2 and β3 are as follows:

λ=9,β 1=p^(- 1)α 1=(0, 1, 1)′

λ=3,β2=p^(- 1)α2=(- 1, 1,0)′

β3=p^(- 1)α3=( 1, 1,- 1)′.

2，A( 1， 1， 1)′= 3( 1， 1， 1)′。 A(- 1，2- 1)′= 0(- 1，2- 1)′

A(0，- 1， 1)′= 0(0，- 1， 1)′

Make p =

1,- 1,,0

1, 2.- 1

1,- 1, 1.

What about p (- 1) = (1/3) ×

1, 1, 1

-2, 1, 1

-3,0,3

There is p (-1) AP = diag (3,0,0).

A=Pdiag(3,0,0)P^(- 1)=

1, 1, 1

1, 1, 1

1, 1, 1.