Suppose I have a known unit eigenvector $v_1$ and all eigenvalues $\lambda_1, \dots, \lambda_n$ of an unknown but real and symmetric $n\times n$ matrix $A$. In other words, $Av_1 = \lambda_1 v_1$. For now, assume $\lambda_1 > \dots > \lambda_n$.

I have two questions:

  1. Is $A$ be uniquely determined by $v_1$ and $\lambda_1, \dots, \lambda_n$?
  2. How do I prove whether $A$ is determined? Hints and suggestions for an approach would be great! Of course, a direct answer would be equally appreciated.
  3. If the answer to question 1 is yes, how do I find $A$ or a set of $A$'s unit eigenvectors.
  4. If the answer to question 1 is no, what other information do I need to make that a yes?

My intuition suggests the answer to 1 is yes because the three constraints

  1. eigenvectors are pairwise orthogonal,
  2. the matrix $A$ must have eigenvalues as prescribed, and
  3. the first eigenvector must be $v_1$

are strong enough to uniquely identify $A$ in, which lives in a $n(n-1)$ space.


1 Answer 1


No to 1, and here's a counter-example. Suppose $\lambda_1=1, \lambda_2=2,\lambda_3=3$. Set $v_1 = (1, 0, 0)^T$. Two possibilities for the matrix $A$ could be $\mathrm{diag}(1, 2, 3)$ and $\mathrm{diag}(1,3,2)$ (there are infinitely many other possibilities too).

For 4, I will just give a hint. 1 eigenvector is not enough. $n$ eigenvectors is certainly enough (you can use the eigendecomposition of a matrix), but you can get away with fewer.

  • $\begingroup$ I guess n-1 eignevectors is sufficient since there's only one degree of freedom left. But can we go with even fewer? $\endgroup$
    – fool
    Commented Jul 25, 2022 at 4:06
  • $\begingroup$ The counter-example I gave you gives n-2 eigenvectors. $\endgroup$ Commented Jul 25, 2022 at 12:37
  • $\begingroup$ Sorry could you clarify what you mean by "gives $n-2$ eigenvectors"? In your counter example, clearly we need one more eigenvector to determine $A$, requiring a total of $n-1=2$ eigenvectors. How can we only need $n-2=1$ eigenvector to determine $A$? $\endgroup$
    – fool
    Commented Jul 25, 2022 at 19:24
  • $\begingroup$ I'm saying that in the counter-example, n-2 eigenvectors are given and this is insufficient to determine A. Hence, it is not possible in general to determine A from knowledge of n-2 eigenvectors and all eigenvalues of A. $\endgroup$ Commented Jul 25, 2022 at 20:43
  • 1
    $\begingroup$ This was just a coincidence, because I chose some easy matrices to demonstrate the point. If we know only the eigenvalues of A, then we know that $A$ is of the form $P\Lambda P^T$ where the columns of $P$ are the eigenvectors of $A$ (in order) and $\Lambda = \mathrm{diag}(\lambda_1,\ldots, \lambda_n)$. Any orthonormal set of eigenvectors will do. A matrix whose columns are an orthonormal set is called orthogonal. So we find that knowledge of the eigenvalues of $A$ determines $A$ up to an unknown orthogonal matrix $P$. If we know $v_1$, that determines in addition the first column of $P$. $\endgroup$ Commented Jul 26, 2022 at 3:13

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