# Given one eigenvector and all eigenvalues of a real, symmetric but unknown matrix, find a set of the remaining eigenvectors or the matrix itself

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.

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.

• I guess n-1 eignevectors is sufficient since there's only one degree of freedom left. But can we go with even fewer?
– fool
Commented Jul 25, 2022 at 4:06
• The counter-example I gave you gives n-2 eigenvectors. Commented Jul 25, 2022 at 12:37
• 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 $A$?
– fool
Commented Jul 25, 2022 at 19:24
• 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. Commented Jul 25, 2022 at 20:43
• 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 $P$. Commented Jul 26, 2022 at 3:13