eigenvects not functioning as I expected

[markus]

So, for a school project I'm trying to find the eigenvalues of a matrix to see if it is diagonalizable. Our teacher said we should use the function eigenvects.

B:= Matrix(5, 5, {(1, 1) = -9, (1, 2) = 11, (1, 3) = -21, (1, 4) = 63, (1, 5) = -252, (2, 1) = 70, (2, 2) = -69, (2, 3) = 141, (2, 4) = -421, (2, 5) = 1684, (3, 1) = -575, (3, 2) = 575, (3, 3) = -1149, (3, 4) = 3451, (3, 5) = -13801, (4, 1) = 3891, (4, 2) = -3891, (4, 3) = 7782, (4, 4) = -23345, (4, 5) = 93365, (5, 1) = 1024, (5, 2) = -1024, (5, 3) = 2048, (5, 4) = -6144, (5, 5) = 24572});

eigenvects(B);
the result I get is [0, 5, {[0 21/256 -71/128 973/256 1]}]

There are a couple of things I don't understand:
Maple says 0 is an eigenvalue with algebraic multiplicity 5, right? But where are the 5 eigenvectors? [0 21/256 -71/128 973/256 1] is only one. Shouldn't a 5*5 diagonalized matrix have 5 eigenvectors? Using other functions like Eigenvectors(B) gives similarly confusing results. Then they put in the eigenvector in one column and then fills the other with zeroes. Why does it do that? Isn't the definition of an eigenvector that it shouldn't be 0?

Thanks in advance.

[unruh]

Well, the teacher did ask you to show whether or not the matrix is
diagonalizable.
Hae you tried even simpler examples

M= ( 0 1)
( 0 0)
Which has two zero egenvalues. How many eigenvectors does it have?
Or
M=(0 1 0)
(0 0 1)
(0 0 0)
How many eigenvectors does it have?
If it has fewer eigenvectors than its rank, what does that tell you
about its diagonalizability? If it were diagonal, what would the
diagonal elements of the matrix have to be?