# Can eigenvectors determine their original matrices given the basis of matrices space is small?

Let $$A_1$$, $$A_2$$, and $$A_3$$ be three different mutually orthogonal random hermitian operators, say dimension of $$30\times 30$$. For three random real numbers $$a_1$$, $$a_2$$, $$a_3$$, we have $$(a_1A_1+a_2A_2+a_3A_3)\vec{v}=s\vec{v}.$$ i.e., $$\vec{v}$$ is an eigenvectors of the matrix $$a_1A_1+a_2A_2+a_3A_3$$ with eigenvalue $$s$$. We ask if there exists another set of real numbers $$(a_1',a_2',a_3',s')$$ such that $$(a_1'A_1+a_2'A_2+a_3'A_3)\vec{v}=s'\vec{v}.$$ generically.

I have numerically verified that the $$(a_1,a_2,a_3,s)$$ is unique generically. Now in a more general setting, the number of random hermitian operators of size $$d\times d$$ is $$r$$. We ask if $$(a_i,s)$$ is unique for certain $$\vec{v}$$ when $$r=O(d)$$. Is there any mathematical theorem which guarantte this?

• Of course it's not unique: you can multiply $(a_1, a_2, a_3, s)$ by the same scalar. Generically, that will be all, i.e. the span of $A_1 v$, $A_2 v$, $A_3 v$ and $v$ will have dimension $3$. Oct 25, 2018 at 1:54
• @RobertIsrael Is there any rigorous proof that the span of $A_1v$, $A_2 v$, $A_3v$ and $v$ will have dimension 3? Oct 25, 2018 at 2:37

Given a non-zero vector $$v$$, let $$V(v)=\{ H | \exists \lambda: Hv=\lambda v \}$$ be the space of all Hermitian matrices for which $$v$$ is an eigenvector.
It is clear that this space is conjugate (under a unitary conjugation) to $$V(e_1)$$ where $$e_1=(1,0,\dots,0)$$ is the standard vector. From this it follows that $$V(v)$$ has codimension $$n-1$$ and is a linear subspace of codimension $$n-1$$ in the space $$W$$ of Hermitian matrices.
Problem: Given a subspace $$V$$ of codimension $$k$$ a vector space $$W$$, consider the collection $$C$$ of all $$r$$-tuples $$(A_1,\dots,A_r)$$ in $$W$$ such that their linear span has a non-zero intersection with $$V$$. For a general element of $$C$$ what is the dimension of this intersection?
When $$r\geq k$$, the answer is 1. When $$r, the answer is $$k-r$$.
Applying this to the above example shows that if $$n>4$$, then for a general triple $$(A_1,A_2,A_3)$$ of Hermitian matrices of size $$n$$, a vector $$v$$ which is an eigenvector of a general (real) linear combination $$a_1A_1+a_2A_2+a_3A_3$$ is not an eigenvector any other linear combination except a scalar multiple.