```6.3
4. The matrix is

0 -1
1  2

I will use  x  instead of lambda.
The characteristic polynomial is    x^2 -2x + 1  = (x-1)^2.
1  is the only eigenvalue.  Ker (I - A) = the kernel of

1  1
-1 -1

= solutions  of  x+y = 0, i.e.

{  -y  :  y in R}
y

which has basis

-1
1

Since the geometric multiplicity (1) is smaller than
the algebraic multiplicity (2),  there is no eigenbasis.

6. The matrix is

2 3
4 5

with char. polynomial   x^2 - 7x - 2.  By the quadratic formula,
the eigenvalues are  (7 +/- sqrt{57})/2   where  +/-  means
plus or minuis   and  sqrt{57}   means the square of 57.
There will be a one-dimensional eigenspace for each eigenvalue
in this case.

The eigenspace for  (-7 + sqrt{57})/2  is the kernel  of

(7 + sqrt{57})/2 - 2             -3
-4              (7 + sqrt{57})/2 - 5

or

(3 + sqrt{57})/2         -3
-4          (-3 + sqrt{57})/2

(The second row IS a mutliple of the first.)

Then   u =

3
(3 + sqrt{57})/2

is an eigenvector.

Entirely similarly,  v =

3
(3 - sqrt{57})/2

is an eigenvector for the eigenvalue    (-7 - sqrt{57})/2
and u, v  give an eigenbasis.

18. The matrix is

0 0 0 0
0 1 0 1
0 0 0 0
0 0 0 1

The characteristic polynomial is  x^2(x - 1)^2, and the eigenvalues
are  0, 1, both with algebraic multiplicity two.  The eigenspace
for 0  is the kernel of the original matrix, with has basis
e_1, e_3.  The eigenspace for 1 is the kernel of

1 0 0  0
0 0 0 -1
0 0 1  0
0 0 0  0

Since the corresponding equations give  x_1 = -x_4 = x_3 = 0,
the kernel is one-dimensional and has basis  e_2,  the only
eigenvector for the eigenvalue 1.  Since the geometric multiplicity
is less than the algebraic multiplicity, there is no eigenbasis.

6.4
22.  The matrix is

1  3
-4 10

The characteristic polynomial is   x^2 + 11x + 22.  The eigenvalues
happen to be real:

(-11 +/- sqrt{121 - 88})/2 = (-11 =/- sqrt{33})/2

24. The matrix is

0  1 0
0  0 1
5 -7 3

The characteristic polynomial is

x -1   0
det   0  x   -1   =
-5  7  x-3

x(x(x-3) - (-1)(7)) - (-1)(0(x-3) - (-1)(-5)) =
x(x^2 - 3x + 7) +(0 - 5) = x^3 - 3x^2 + 7x - 5.

By inspection,  x = 1 is one root, and we can factor
out  x-1:    we get  (x-1)(x^2 - 2x + 5).
The roots of the quadratic are

(2 +/- sqrt(4-20))/2    or

1 +/- 2i

which, together with 1, give the eigenvalues.

28.  Call the two complex conjugate eigenvalues   a+bi,  a-bi.
Since the real eigenvalue is 2  and the trace and determinant
are  8  and  50, resp.  we know that the sum of the eigenvalues
is  8  and the product is  50.  This gives   2 + 2a = 8  or  a = 3,
and  2(a^2 + b^2) = 50  and so  9 + b^2 = 25,  from which
b = 4 or -4.  This implies that the complex eigenvalues are
3 +/- 4i.
```