# Size of GLn

## Main Question or Discussion Point

What is the size of $GL_n(\mathbb{Z}_2)$?

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morphism
Homework Helper
Think in terms of linear independence.

HallsofIvy
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You could also think of this as n by n matrices whose entries must be either 1 or 0. How many entries are then in an n by n matrix? And if there are only two possible values for each?

You could also think of this as n by n matrices whose entries must be either 1 or 0. How many entries are then in an n by n matrix? And if there are only two possible values for each?
Well non-invertible matrices are not in GLn(Z_2), so it's NOT 2^(n^2). Rather it's that minus all the matrices with determinant 0, and I don't know how to count those.

Think in terms of linear independence.
Is it true that an nxn matrix is invertible iff the column space has n dimensions?

$$Z_2= \{0,1\}$$ and $$Z_2$$ has only one unit {1}.
so matrix A which satisfy det A = 1 will be element of $$GL_n ( Z_2)$$

HallsofIvy
Homework Helper
Yes, and the whole question is "how many of them are there"!

In fact, a matrix A, in Z2 is invertible if det(A)= 1 and not invertible if det(A)= 0. Does that imply that exactly half of all n by n matrices in Z2[/sup] are invertible?

morphism
Homework Helper
Is it true that an nxn matrix is invertible iff the column space has n dimensions?
Yes!

In fact, a matrix A, in Z2 is invertible if det(A)= 1 and not invertible if det(A)= 0. Does that imply that exactly half of all n by n matrices in Z2[/sup] are invertible?

No. In fact, this is only true if n=1.

Yes!
I still don't know how to count them. So you have $2^n-1$ choices for the first vector. $2^n-2$ for the second. But how many vectors are linearly independent wrt to the first two?

morphism
Homework Helper
How many linear combinations are there of the first two columns? Remember, there are only two elements in our field.

I think it's
(2^n -1)(2^n - 2) (2^n - 2^2 ) .... (2^n - 2^{n-1}).

the reason is: you just have to make sure all the rows are linearly independent and non-zero.
For the first row there are (2^n - 1) choices.

For the second row there's (2^n-1) minus 1 (the number of rows already used).

For the third row there's (2^n-1) minus 2 (the number of rows already used) minus another 1 for the number of sums of those rows.

For the kth row, there's (2^n-1) minus (k choose 1) minus (k choose 2), etc., minus (k choose (k-1)). You can tell that none of these conditions are redundant because that would imply that some of the previous rows were linearly dependent.

But
k choose 1 + k choose 2 + ... + k choose (k-1) = 2^{k} -1.

?

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