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?

  • $\begingroup$ 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 $3$. $\endgroup$ Oct 25, 2018 at 1:54
  • $\begingroup$ @RobertIsrael Is there any rigorous proof that the span of $A_1v$, $A_2 v$, $A_3v$ and $v$ will have dimension 3? $\endgroup$
    – Sirui Lu
    Oct 25, 2018 at 2:37

1 Answer 1


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.

Now, we can reduce the problem to the following problem in "linear algebra".

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 $r<k$, 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .