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Let $A \in \mathbb{R}^{n \times n}$ be a real, symmetric, positive semidefinite matrix with $n$ eigenvalues between $[0,1]$

Suppose we have the following information:

  • $\lambda_{min}(A) = 0, \lambda_{max}(A) = 1$
  • $\det(A) = 0$
  • $A\mathbf{1} = 0$ (the one vector of $\mathbb{R}^n$ is an eigenvector associated with the eigenvalue 0)

Since $A$ is symmetric, we know that all eigenvalues are going to be real.

Is it possible to find all the other eigenvalues and eigenvectors?

This sounds like an impossible question because the information presented is minimal, but if there is some result that the distribution of the eigenvalues and eigenvectors are..."uniformly spaced" for this kind of matrix, then perhaps we can find all the other eigenvalue-eigenvector pairs.

Anyone has any ideas how to approach this question?

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2 Answers 2

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There is no way to know what the other eigenvectors or eigenvalues are. Consider a matrix defined as $$A=\lambda_1v_1v_1^T + ... + \lambda_nv_nv_n^T$$ Where $\{v_1,...,v_n\}$ is an orthonormal basis for $\mathbb{R}^n$. Then clearly every $\lambda_i$ is an eigenvalue with eigenvector $v_i$. You can fix $\lambda_1=0$, $v_1=(1,...,1)$ and $\lambda_2=1$, but any choices over the other eigenvalues and eigenvectors on the definition of $A$ will give you a matrix that will hold all the constraints.

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That's not enough information to reconstruct $A$. You can describe explicitly the set of all possible matrices that satisfy $A$ by the following procedure:

  1. Choose vectors $v_2, \dots v_n \in \mathbb{R}^n$ that form an orthonormal basis of $\operatorname{span} \{ \mathbf{1} \}^{\perp}$ and construct a matrix $P$ whose first column is $\frac{1}{\sqrt{n}} \mathbf{1}$ and whose $i$-th column is $v_i$.
  2. Choose a sequence $0 \leq \lambda_2 \leq \lambda_3 \leq \dots \leq \lambda_{n-1}$ of real numbers in $[0,1]$ and set $\lambda_1 = 0, \lambda_n = 1$.

Then the matrix

$$ P \operatorname{diag}(\lambda_1,\lambda_2, \dots, \lambda_{n}) P^{-1} $$

will be a matrix whose eigenvalues are $\lambda_1, \dots, \lambda_n$ corresponding to eigenvectors $\mathbf{1}, v_2, \dots, v_n$ and so it will satisfy the properties requested. Conversely, any matrix that satisfies the properties will be of the form above but there is some redundancy in the description.

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