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I'm having trouble proofing following statement from book I'm currently reading.

Let $V=M_n(\mathbb{R})$ be linear space of n by n matrices. i.e. dim $V = n^2$

Let $W$ be a linear subspace of $V$ where $W = \{A \,|\, A=0 \;\text{or A is invertible matrix}\}$

Show the following statements:

  1. For arbritrary n, show $ 1 \le \operatorname{dim}W \le n$
  2. For $n = 1$mod(2), show $\operatorname{dim}W = 1$.
  3. Show $\operatorname{dim}W = n \iff n = 2,4,8$.

So far, I tried to build specific example for $n = 2$

ex)

Let $A_1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$ and $A_2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \\ \end{pmatrix}$
So this will be two basis of $W$ for $n=2$.
But so far have no Idea how to generalize this to higher dimmension, also not sure how to prove that $\operatorname{dim}W$ cannot be bigger than $n$.

Any suggestion would be helpfull. Thanks.

Tim
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mngroon
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  • What have you tried? – CJ Dowd Jul 26 '23 at 10:24
  • I tried to build some examples of these subspaces. but even for n=2 it seems difficult. I'm not sure how to even make such a subspace in more higher dimmensions. So far I have no clue how to tackle this problem. – mngroon Jul 26 '23 at 10:28
  • The lack of quantifiers in the problem statement seems strange. Do you mean that $W$ is a subspace of $V$ such that every matrix in $W$ is either invertible or the zero matrix? And for (3) do you mean to assert that there exists $W$ of dimension $n$ if and only if $n = 2, 4, 8$? – CJ Dowd Jul 26 '23 at 11:01
  • Your $W$ is not a linear subspace. – Sassatelli Giulio Jul 26 '23 at 11:12

  • Do you mean that W is a subspace of V such that every matrix in W is either invertible or the zero matrix?

    yes. every matrix in W is invertible or zero matrix. also, W is linear space so any linear combinations of matrices in W also contained in W.

    do you mean to assert that there exists W of dimension n if and only if n=2,4,8 ?

    thats right. I think assuming det($A \oplus B$) = det(A)*det(B), I think I can make basis for n=4,8 by repeatingly taking direct product of n=2 case basis.(A1 and A2). but not sure why these are only cases

     – mngroon Jul 26 '23 at 11:19

  • Your W is not a linear subspace.

    In general, I agree with you. but the problem is to construct it so it will be linear subspace of V. see example I made for n=2 if it makes sence.. Thanks.

     – mngroon Jul 26 '23 at 11:20
  • You are misreading the book (BTW, what book is it?). The correct statement, is that if $W$ is a non-trivial subspace of $M_n(\mathbb R)$ whose nonzero elements are invertible, then $\dim W=n$ if and only if $n=\color{red}{1},2,4,8$, and $\dim W$ is necessarily equal to $1$ when $n$ is odd. Here are some relevant references. – user1551 Jul 26 '23 at 11:33
  • to user1551, you are right! I misread the book. thanks references!! by the way book Im reading is "数８の神秘", I'm not sure if there is english translation of this book. link to book: https://www.amazon.co.jp/%E6%95%B0%E2%80%9C8-%E3%81%AE%E7%A5%9E%E7%A7%98-8%E3%81%A8%E3%81%84%E3%81%86%E6%95%B0%E3%81%AB%E7%A7%98%E3%82%81%E3%82%89%E3%82%8C%E3%81%9F%E4%B8%8D%E6%80%9D%E8%AD%B0%E3%81%AA%E9%96%A2%E4%BF%82-%E4%BD%90%E4%B9%85%E9%96%93%E4%B8%80%E6%B5%A9/dp/4535888884 – mngroon Jul 26 '23 at 13:05

1 Answers1

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I guess you mean $W \subset \mathrm{GL}_n(\mathbb{R}) \cup \{0\}$ and not $=$ because it is not a vector space. I also assume that $W \neq \{0\}$ or else you could have $\dim(W) = 0 < 1$ contradicting your first point.

If you take $n + 1$ elements $A_1,\ldots,A_{n + 1}$ of $W$, denote by $v_k \in \mathbb{R}^n$ the first row of each $A_k$. It gives you $n + 1$ distinct vectors of $\mathbb{R}^n$ thus this family is not free so we can find real scalars $(\alpha_k)$ not all zeroes such that $\sum_{k = 1}^n \alpha_kv_k = 0$. We deduce that the first row of $B = \sum_{k = 1}^n \alpha_kA_k$ is $0$. We deduce that $B$ is not invertible but $B \in W$ so $B = 0$. Therefore, the family $(A_k)$ is not free. It is true for any family of $n + 1$ elements of $W$ so $\dim(W) \leqslant n$.

If $n$ is odd, take two non-zero elements $A$,$B$ in $W$. $AB^{-1}$ has a real eigenvector $v$ associated to a real eigenvalue $\lambda$ (because $n$ is odd). Let $w = B^{-1}v \neq 0$. We have $Aw = AB^{-1}v = \lambda v = \lambda Bw$ so $A - \lambda B$ is not invertible and belongs to $W$. We deduce that $A - \lambda B = 0$ so $(A,B)$ is not free. Therefore, $\dim(W) = 1$.

For $n = 2$, notice that any complex number $a + ib$ can be represented by $aA_1 + bA_2 \in W \subset \mathrm{M}_2(\mathbb{R})$. Try to do the same kind of parallel with quaternions for $n = 4$ and octonions for $n = 8$ maybe (I am not sure it works).

Cactus
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