Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.

1.Then if all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be?

We already know that if we take $M_{n}(\mathbb{R}) $ instead of $M_{n}(\mathbb{Q}) $ then the answer is $ \rho(n) $. where $ \rho(n) $ is Radon Hurwitz number i.e if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $ .

2.Then if all the non-zero matrices in $N$ are invertible and skew symmetric, what is the maximum the dimension of $N$ can be? you can consider $ n $ is even.

we know that if $ N $ is such subspace of $M_{n}(\mathbb{R}) $ then the answer for $ n=4 $ and $ n= 7 $ are $3$ and $8$ respectively and if $n=6 $ then dim$N$ can't be $5$.

Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.Then if all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be?

We already know that if we take $M_{n}(\mathbb{R}) $ instead of $M_{n}(\mathbb{Q}) $ then the answer is $ \rho(n) $. where $ \rho(n) $ is Radon Hurwitz number i.e if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $ .

Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $. Then if all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be?

We already know that if we take $M_{n}(\mathbb{R}) $ instead of $M_{n}(\mathbb{Q}) $ then the answer is $ \rho(n) $. where $ \rho(n) $ is Radon Hurwitz number i.e if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $ .

Let $M_{n}(\mathbb{Q}) $ denote the $n$ times $n$ matrices over the rational number field. $N$ be a subspace of $M_{n}(\mathbb{Q}) $.

1.Then if all the non-zero matrices in $N$ are invertible, what is the maximum the dimension of $N$ can be?

We already know that if we take $M_{n}(\mathbb{R}) $ instead of $M_{n}(\mathbb{Q}) $ then the answer is $ \rho(n) $. where $ \rho(n) $ is Radon Hurwitz number i.e if $ n = (2a + 1 ) 2^{c+4d} $ where $ 0 \leq c \leq 3 $ then $ \rho(n) = 2^{c} + 8d $ .

2.Then if all the non-zero matrices in $N$ are invertible and skew symmetric, what is the maximum the dimension of $N$ can be? you can consider $ n $ is even.

we know that if $ N $ is such subspace of $M_{n}(\mathbb{R}) $ then the answer for $ n=4 $ and $ n= 7 $ are $3$ and $8$ respectively and if $n=6 $ then dim$N$ can't be $5$.