2

Let $GL_n(\mathbb{F})$ be the set of all invertible $n$-by-$n$ matrices over a field $\mathbb{F}$. For the set $GL_n(\mathbb{F})\cup\{0\}$, it is trivial that it contains $0$ and $1$ dimensional subspaces as its subset.

What's the maximal dimension of subspaces that is subset of $GL_n(\mathbb{F})\cup\{0\}$?

My Progress

When $\mathbb{F}=\mathbb{R}$:

  • It is $1$ when $n$ is odd.
  • It cannot exceed $n$ for all $n$.
  • It is exactly $n$ if and only if $n=1,2,4,8$. (This is from this post.)
Evan_Bradley
  • 456

1 Answers1

2

For $n$ odd $GL_n\cup\{0\}$ does not contain any two-dimensional subspace. Indeed, for any $A,B\in GL_n$, which are linearly independent, i.e. $A$ is not a multiple of $B,$ there exists $\lambda$ such that $A-\lambda B$ is not invertible. Indeed $$A-\lambda B=(AB^{-1}-\lambda I)B$$ The matrix $AB^{-1}$ has a real eigenvalue $\lambda$, as the dimension is odd. Then $AB^{-1}-\lambda I$ is not invertible.

For even $n$ the set $GL_n\cup\{0\}$ contains a two-dimensional subspace. Denote $n=2m.$ Let $A\in GL_{2m},$ have all eignevalues unreal. Then the two-dimensional space generated by $A$ and $I$ is contained in $GL_n.$ For example $A$ could be the direct sum of $m$ matrices of dimension $2\times 2$ $$\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$

Ryszard Szwarc
  • 44,675
  • Will $GL_n\cup{0}$, if $n$ is even, contain even higher dimensional subspaces? Actually I'm hoping that it can contain a subspace of dimension $n$. – Evan_Bradley Apr 02 '22 at 11:27
  • I do not know yet. – Ryszard Szwarc Apr 02 '22 at 11:50
  • Many thanks anyway! – Evan_Bradley Apr 02 '22 at 15:41
  • I have impression that your question is closely related to https://math.stackexchange.com/questions/4418390/a-linear-algebra-problem-about-linear-subspace-of-gl-n/4419151#4419151 – Ryszard Szwarc Apr 03 '22 at 12:46
  • I believe his question is equivalent to my first question. Actually, thanks to your help, I have got a proof that for any even $n$, $GL_n\cup{0}$ has a $2^{n-1}$ dimensional subspace as a subset. But in your question, 祝 said it's true if $n=1,2,4,8$, which I cannot understand. And actually, my second question remains unanswered. – Evan_Bradley Apr 03 '22 at 13:37
  • Alright, it seems that my proof is wrong... – Evan_Bradley Apr 03 '22 at 13:39
  • It seems that an advanced old result called the Hurwitz theorem is involved. Perhaps you can find connections to that. See https://www.google.com/url?sa=t&source=web&rct=j&url=https://student.fau.edu/sclanton/web/presentation.pdf&ved=2ahUKEwiCqorfgvj2AhUqpIsKHWdaCsoQFnoECBIQAQ&usg=AOvVaw2EXlYu4PfqkR_1ZyFgM-e2 – Ryszard Szwarc Apr 03 '22 at 13:45
  • Is this about the Universal Algebra? – Evan_Bradley Apr 03 '22 at 13:50
  • That's not my specialty. I have just googled the term" composition algebra" mentioned in the answer to https://math.stackexchange.com/questions/4418390/a-linear-algebra-problem-about-linear-subspace-of-gl-n/4419151#4419151 – Ryszard Szwarc Apr 03 '22 at 13:57
  • Alright. This is also strange to me, and it requires effort to understand this. Thanks a lot for this great discussion! – Evan_Bradley Apr 03 '22 at 14:00