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I came across this question recently:

Do there exist $n$ real $n\times n$ matrices $A_1, A_2, \cdots, A_n$, such that for any $n$-dimensional non-zero vector $v$, $A_1 v, A_2 v, \cdots, A_n v$ are always linearly independent?

It seems like a usual linear algebra question, but I can't think of any idea to approach it, and I can't come up with a counterexample either. Is there anything I've missed?

Any help would be appreciated.

Edit: So for $n=2$, this is obvious as we can just take two rotation matrices with different angles. The resulting two vectors will always be on different lines. For $n>2$ however, rotation won't work as there'll be an axis of rotation, on which the vectors are eigenvectors, i.e., on the same line after rotation.

Bumblebee
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Opsimath
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  • Given a none zero vector $v$ is there a matrix $A$ for which $v$ is a cyclic vector? That would answer the question if true, but it seems unlikely. – Paul Aug 04 '23 at 09:36
  • @Widawensen No matter what the axis of rotation is there will be a vector $w$ such that $w$ is an eigenvector. i.e. $Aw = w$, and so $w$ and $Aw$ will not be linearly independent. – Enforce Aug 04 '23 at 09:50
  • For 3 dimensional rotation a selected vector can be on axis, then you can take two other rotations with different axes and it will be rotated. – Widawensen Aug 04 '23 at 09:50
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    @Enforce I suppose however that $w$ is not the vector from the list. The list starts from $A_1 w$., ends on $A_n w$, $...$ $n+1$ vectors would be always linearly dependent in n-dimensional space. – Widawensen Aug 04 '23 at 09:52
  • @Widawensen Yes I misread the question. Although this method still fails as explained in the comments to the answer below. – Enforce Aug 04 '23 at 10:29
  • @Enforce. Yes, my method fails. I see it now. – Widawensen Aug 04 '23 at 10:32

1 Answers1

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Note that $(1)\iff(2)\iff(3)$ in the following:

  1. $A_1v,A_2v,\ldots,A_nv$ are linearly independent whenever $v\ne0$.
  2. $\sum_{i=1}^nc_iA_iv\ne0$ whenever $v\ne0$ and $(c_1,c_2,\ldots,c_n)\in\mathbb R^n\setminus0$.
  3. $\sum_{i=1}^nc_iA_i$ is invertible whenever $(c_1,c_2,\ldots,c_n)\in\mathbb R^n\setminus0$.

So, you are essentially asking whether there exists an $n$-dimensional subspace of $n\times n$ real matrices in which all nonzero members are invertible. This is a non-trivial but solved problem: the answer is “yes” if and only if $n=1,2,4,8$. See here for some references.

David Gao
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user1551
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  • Really we can find such non-zero vector $v$ in 3-dimensional space that $R_x v, R_y v, R_z v$ are linearly dependent? ( operations here are rotations about perpendicular axes) – Widawensen Aug 04 '23 at 10:14
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    @Widawensen Let $v$ be the axis of rotation for $R_x^TR_y$. Then $R_xv-R_yv+0R_zv=0$. – user1551 Aug 04 '23 at 10:16
  • I see. Thank you for the explanation. – Widawensen Aug 04 '23 at 10:19
  • Note on my reversion of an edit: The original answer is correct - there is no reason why the inverses and products of matrices in the space have to be in the space, so this is not just a matter of dimensions of associative real division algebras. – David Gao May 28 '25 at 22:19