The problems 7 (a),(b) have one line solutions so we did not spent
much time on them:

7 (a) The rank could not exceed the minimum of dimensions of the matrix,
in our case the minimum is 5. Anwer: Maximal possible rank is 5 minimal possible
is 0.

7 (b) Matrix can not have more than 3 pivots and one row is all zeroes hence
the maximal number of pivots is 2. Answer: Maximal rank is 2 minimal is 0.

In problem 8, we just need to check that the inner product of any two vectors
is 0 AND that the length of each vector is 1. The rest of solution was presented
on the board.

Problem 6 is rather routine. We have done many diagonalizations in class. We
just need to do it again. That would be long way to solve the problem.
The short way is below

a) The nul space is defined by the equation
x1+x2+x3+x4=0.
It is a hyperplane with a basis
 v1=(1,-1,0,0) v2=(1,0,-1,0) v3=(1,0,0,-1)
The column space is spanned by the vector (1,2,1,3)

b) The Col(A) is spanned by (1,2,1,3) thus
(1,2,1,3) is an eigenvector! It is eigen values is

lambda=1+2+1+3=7.

The other eigenvalue is 0.

c) The diagonal matrix in diagonalization has diagonal entries: 7,0,0,0.
The matrix of eigenvectors has columns (1,2,1,3), (1,-1,0,0), (1,0,-1,0), (1,0,0,-1).