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Does this matrix have a name?

$$a_{i,j}= \begin{cases} c_{i}c_{j} &\text{if } i \neq j, \\ 0 &\text{if } i=j. \end{cases}$$ where $c_{n}$ are constant real numbers.

This matrix seems too symmetric to be an anonymous commoner matrix.

In any case, what are its eigenvalues?

Rócherz
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athanos lee
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  • Yes this is called a hollow matrix: https://en.wikipedia.org/wiki/Hollow_matrix. Your question is repeated here: https://math.stackexchange.com/questions/79779/properties-of-zero-diagonal-symmetric-matrices – Pavan C. May 02 '25 at 13:22
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    @PavanC. This matrix exhibits more symmetry than any you linked. – athanos lee May 02 '25 at 13:26
  • As commented in the second answer in that post, I don't think much else can be said in general, outside of the interesting case of the tridiagonal laid out there. And as commented in the main post, the eigenvalues will be both positive and negative, except for the null case. – Pavan C. May 02 '25 at 13:40
  • F.W.I.W.: If $c$ is a (column) vector, then $A=cc^{\text T}-\operatorname{diag}(c \odot c)$, where $\operatorname{diag}(c \odot c)$ is fancy notation for a matrix with entries $c_i^2$ along the diagonal and zeroes elsewhere. – Rócherz May 02 '25 at 13:45
  • @Rócherz Yes it is another way to express this matrix. Does this tell us anything about its diagonalization? – athanos lee May 02 '25 at 13:59

1 Answers1

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No idea about the name nor eigenvalues. However, the characteristic polynomial of $A$ can be expressed in terms of $c_k$.

Let $C$ be the diagonal matrix with entries $c_k^2$ and $c$ be the row vector with entries $c_k$, your matrix $A = cc^T - C$, so $$\chi_A(\lambda) \stackrel{\text{def}}{=} \det(\lambda I - A) = \det((\lambda I+C) - cc^T)$$ What's inside the last determinant is a rank-1 update of $\lambda I +C$. By matrix determinant lemma, one has $$\begin{align}\chi_A(\lambda) &= \prod_{k=1}^n(\lambda + c_k^2)\left[1 - \sum_{k=1}^n\frac{c_k^2}{\lambda+c_k^2}\right] \\ &= \lambda P'(\lambda) - (n-1) P(\lambda)\\ &= \lambda^n \left(\frac{P(\lambda)}{\lambda^{n-1}}\right)'\end{align}$$ where $P(\lambda) = \det(\lambda I + C) = \prod\limits_{k=1}^n(\lambda+c_k^2)$.

Finding the eigenvalues of $A$ is more or less equivalent to finding the extremum of $\frac{P(\lambda)}{\lambda^{n-1}}$.

I have no idea how to identify the roots in general.

However, if all $c_k$ are non-zero and distinct, one can reorder them such that $$|c_1|^2 < |c_2|^2 < \cdots < |c_n|^2$$ Since $-|c_k|^2$ are roots of $\frac{P(\lambda)}{\lambda^{n-1}}$, $\chi_A(\lambda)$ has $n-1$ negative roots $\lambda_k$, one in each interval $( -c_{k+1}^2, -c_k^2 )$ for $k = 1,\ldots,n-1$. The remaining root should be positive because all of the eigenvalues of $A$ sum to $0$.

Rócherz
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achille hui
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