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Consider the following $M×M$ matrix $$ \mathbf A=\sum_{k=1}^K =a_k \mathbf h_k \mathbf h_k^H,(M≥K) $$ where $a_k$'s are real values and $h_k$'s are $M×1$ randomly generated vectors, e.g., complex Gaussian random vector. Intuitively, since each $\mathbf H_k$ is a random matrix, it is expected that $\mathbf A$ has rank $K$ and its $K$ non-zero eigenvalues are distinct, i.e., $λ_1(A)>⋯>λ_K(A)$ with probability one. I verified it experimentally by generating $\mathbf A$ randomly several times. However, I cannot provide mathematical proof or reasonable explanation.

In fact, I can prove it for $a_k≥0$ for $∀k$. We first assume that $F=[\sqrt{a_1} \mathbf h_1,⋯, \sqrt{a_K} \mathbf h_K]$ has rank $K$ and $K$ distinct non-zero singular values $σ_1(F)>⋯>σ_K(F)$, which is reasonable because $\mathbf h_k$'s are random vectors. Then, we have $\mathbf A=\mathbf F\mathbf F^H$, which has K distinct eigenvalues $σ_1(\mathbf F)>⋯>σ_K^2(\mathbf F)$.

However, I cannot use this approach for $a_k≱0$. Any additional reasonable assumptions can be made for this proof.

Thank you.

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  • $\begingroup$ This question may possibly be approached (for complex-valued Gaussian random vectors) via marginal density of eigenvalues of Wishart matrices, e.g., see "On the marginal distribution of the eigenvalues of Wishart matrices" by Zanella and others. $\endgroup$
    – Aravindh Krishnamoorthy
    Jun 8, 2021 at 20:56

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You could use the fact that within the set of real symmetric matrices, the subset formed by those with a multiple eigenvalue is an algebraic submanifold of codimension $2$. For a reference, have a look to V. I. Arnold, Mathematical methods of classical mechanics. Then you have to prove that an arbitrary submanifold must have zero measure for this distribution.

Something similar is true for complex Hermitian matrices.

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  • $\begingroup$ Probably you mean "subvariety". $\endgroup$
    – YCor
    Jun 7, 2021 at 7:02
  • $\begingroup$ And this subvariety has codimension 1 (in every matrix size $n\ge 2$). It is the vanishing place of the polynomial $M\mapsto \prod_{i\neq j}(\lambda_i-\lambda_j)$, where $\lambda_i$ are eigenvalues, which being symmetric in the $\lambda_i$ is a polynomial with respect to the characteristic polynomial coefficients, and hence with respect to the matrix entries. (Up to sign this is the discriminant of the characteristic polynomial.) $\endgroup$
    – YCor
    Jun 7, 2021 at 7:08
  • $\begingroup$ @Ycor. The situation is different in the space of symmetric matrices, because the subvariety is the locus $D=0$ where $D(M):=\prod_{i<j}(\lambda_j-\lambda_i)^2$ is a non-negative polynomial in the entries. This implies additional identities $\partial D/\partial m_{ij}=0$, which explain the higher codimension. $\endgroup$
    – Denis Serre
    Jun 7, 2021 at 7:12
  • $\begingroup$ Ah, I see, the subvariety has codimension 1, but the set of real points could have lower dimension... Nevertheless for $n=2$, the space of symmetric matrices has dimension 3, and the given subset is the set of scalar matrices, which has codimension 2. I believe it's of codimension 2 in general, because the possible sets of eigenvalues generically has dimension $n-1$, and the orbit then has dimension $\frac{n(n-1)}2-1$, so the union of orbits should have dimension $n+\frac{n(n-1)}2-2$, i.e. codimension 2. $\endgroup$
    – YCor
    Jun 7, 2021 at 7:41
  • $\begingroup$ @Ycor. Yes, it is codim $2$, not $3$. I was too fast. I'll fix it. Of course, the important point it that it is rather small and should be of measure zero. $\endgroup$
    – Denis Serre
    Jun 7, 2021 at 12:25

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