Conditions for this vectors to be linearly dependent

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


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 $\operatorname{rank}(d_1A_1+d_2A_2+\ldots+d_mA_m)<n$.

Tags:

Linear Algebra

Matrices

Vectors

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