Question

Let F be a field, and let n be a positive integer. For any A ∈ Mn×n(F), the trace of A,
tr(A), is defined to be the sum of the diagonal entries of A; that is, tr(A) = Pn
i=1 A(i, i).
Let W = {A ∈ Mn×n(F) | tr(A) = 0 }.

(b) Find a basis for W, and determine the dimension of W.

Answer 1

What is the definition MNx n(F)?

Answer 2

M nxn (F) is any nxn matrix with elements from F

Answer 3

So we could go with taking each column with a 1 in the appropriate diagonal, but what happens if the number in the diagonal is a 0?

Answer 4

everyone ok with tr(A) and it's definition?

Answer 5

I see what the definitions are, I'm just a little confused by what the question means by the dimension of W. Since W is only a set of matrices, and not a matrix itself, I'm not sure what the dimension of W could be.

Answer 6

dimension of any matrix, is the number of elements in the matrix...at most W will have n elements (numbers in each of the diagonals) but it can have fewer, it could have 0 elements, since 0+0+0+0+0...=0

Answer 7

Well, since W consists of n x n matrices, wouldn't the dimension of W itself be n?

Answer 8

yeah, I guess....first we have to find the basis (i.e a set of vectors that spans W, while being linearly independent

Answer 9

you want a guess, i would say the dimenstion is n - 1

Answer 10

it is a subspace for sure, so the dimension is not n

Answer 11

suppose we were, for example, in
\[\mathbb{R^3}\] then you could use as a basis
\[(1,-1,0), (1,0,1)\]

Answer 12

Am I the only one seeing a math processing error?

Answer 13

\[(1,-1,0), (1,0,-1)\] is what i meant. latex been messing up all day

Answer 14

i got math processing error earlier i have no idea what is going on

Answer 15

dimension of an M nxn matrix is n^2

Answer 16

with numbers in its diagonal dim=n

Answer 17

sorry, not thinking clearly.

Answer 18

Well, it seems to me that the basis for M nxn (F) would also be a basis for W, but it would span W and then lots more, so I'm not sure if it's the best basis to use.