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Consider $m<n$ positive definite matrices $A_1\dots,A_m\in\mathbb{R}^{n\times n}$ which are linearly independent matrices. This is, that there are no $c_1,\dots,c_n\in\mathbb{R}$ such that $\sum_{i=1}^m c_iA_i = 0$ unless $c_i=0, \forall i=1,\dots,m$.

How can I find vectors $x\in\mathbb{R}^n$ different from $0$ such that the set of vectors $$ A_1x,\dots,A_mx $$ become linearly dependent? This is, that there exists $d_1,\dots,d_n\in\mathbb{R}$ different from all $0$,such that $\sum_{i=1}^m d_iA_ix = 0$.

This question is motivated by an example as the following. Consider the matrices $$ A_i = \begin{bmatrix} M & 0_{3 \times 1}\\ 0_{1\times 3} & a_i \end{bmatrix} $$ where $a_i\in\mathbb{R}$ and $M\in\mathbb{R}^{3\times 3}$, and $a_1,\dots,a_m$ are different from each other. Hence, if I take $$ x = \begin{bmatrix} x' \\ 0 \end{bmatrix} $$ with some $x'\in\mathbb{R}^3$, then $A_ix = Mx', \forall i=1,\dots, m$. Thus, in this case (which is something like a "trivial example") we have that the vectors $A_ix$ are linearly dependent.

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  • $\begingroup$ Just take $x=0$. $\endgroup$ – Dietrich Burde Sep 29 at 15:27
  • $\begingroup$ Thanks. I added the $x\neq 0$ constraint. $\endgroup$ – Rodrigo Aldana Sep 29 at 15:29
  • $\begingroup$ What do you mean by linearly independent matrices? $\endgroup$ – QED Sep 29 at 15:34
  • $\begingroup$ I mean that there is no $c_1,\dots,c_n\in\mathbb{R}$ such that $\sum_{i=1}^m c_iA_i = 0$ unless $c_i=0, \forall i=1,\dots,m$ (this is the classical linearly independent definition right? Or do you suggest me to say something different?). I clarified that in the question too. Thanks. $\endgroup$ – Rodrigo Aldana Sep 29 at 15:37
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Not a full answer, but too long for a comment.

Essentially you are asking if you can find a nontrivial combination of the matrices that is singular. (If you can do that, then the combination $(\sum \alpha_i A_i) x=0$ will have a solution for some $x$.)

This seems to be related: Vector subspace of $M_n(\mathbb{R})$ with invertible matrices

In general, you can not do it for $m<n$.

Without the positive definiteness, a counter-example is trivial: take, in $\mathbb{R}^4$, the identity matrix and the matrix that rotates $x-y$ plane by $90$ degrees and simultaneously rotates the $z-w$ plane by $90$ degrees.

With positive definite symmetric matrices, I did a few experiments in python and here is a counter-example for 2 matrices:

$$ A = \mathrm{diag}(4,3,2,1) $$

$$ R = \begin{pmatrix} -2.00850073 & -1.60957843 & -0.90369738 & -0.50794833\\ 2.38498604 & 0.57307146 & -0.07249097 & 1.52387285\\ -0.57167226 & 0.40667149 & -1.85512324 & 0.44869258\\ -0.21978557 & 1.07929074 & -1.82720749 & -1.99403596 \end{pmatrix} $$ a random matrix, and $B = RAR^{-1}$.

This is the plot of $\det(A + \lambda(B))$ (always positive): enter image description here

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  • $\begingroup$ This is a nice way of putting it. Do you think then that I can pose this as an optimization problem: $$ \min_{\alpha_1,\dots,\alpha_m} \text{det}(\alpha_1A_i+\cdots+\alpha_mA_m) $$ And if it turns out that the value of $\text{det}(\alpha_1A_i+\cdots+\alpha_mA_m)=0$, then I should look for $x$ in the kernel of $\alpha_1A_i+\cdots+\alpha_mA_m$? $\endgroup$ – Rodrigo Aldana Sep 29 at 16:10
  • $\begingroup$ I added a counter-example for $m=2, n=4$, pls check if I'm correct. Yes, but I think this, what you call "optimization problem", could be quite hard in general $\endgroup$ – Peter Franek Sep 29 at 16:11
  • $\begingroup$ Thanks! Just for curiosity, how would you call that, if not an optimization problem? $\endgroup$ – Rodrigo Aldana Sep 29 at 16:12
  • $\begingroup$ I'm no expert on this, but I would use the terminology from the link ("dimension of subspace of real invertible n×n matrices" etc) $\endgroup$ – Peter Franek Sep 29 at 16:16
  • $\begingroup$ By giving a counterexample you mean that there are cases in which linear dependency is not possible right? But there might be other cases in which it is possible. Thanks for providing this nice example! $\endgroup$ – Rodrigo Aldana Sep 30 at 8:11
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We have that

$$d_1A_1x+d_2A_2x+\ldots+d_mA_mx=0 \iff (d_1A_1+d_2A_2+\ldots+d_mA_m)x=0$$

that is $x \in \ker(d_1A_1+d_2A_2+\ldots+d_mA_m)$ then the problem is equivalent to find $d_i$ such that $\operatorname{rank}(d_1A_1+d_2A_2+\ldots+d_mA_m)<n$.

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  • $\begingroup$ But that's not possible for general matrices for some $m,n$. Can you use the positive definiteness somehow? Just curious. $\endgroup$ – Peter Franek Sep 29 at 16:26
  • $\begingroup$ @PeterFranek I can't see a general solution. Since positive definiteness holds, we can exclude trivial solutions but I can't see how use that to find a general solution without any other information on $A_i$. $\endgroup$ – user Sep 29 at 16:31
  • $\begingroup$ Sure, just I'm curious if any counter-example exists (like matrices when this is not possible). Probably yes I think $\endgroup$ – Peter Franek Sep 29 at 16:32
  • $\begingroup$ I think what we are discussing here is very useful to me. Thanks both. This idea of thinking in terms of the rank of $d_1A_1+\dots$ is already very helpful to me. By the way the question is posed right now, this should be marked as the correct answer, unless @PeterFranek is writing another attempt, or want to add something else? Thanks again. $\endgroup$ – Rodrigo Aldana Sep 29 at 16:36
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    $\begingroup$ @PeterFranek I think we need to await for some contribute from expert users in the field. $\endgroup$ – user Sep 29 at 17:00

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