How to get eigenvectors from eigenvalues

I'm not seeing where the eigenvectors are in terms of eigenvalues in the paper that came out recently named "Eigenvectors from Eigenvalues" by the neutrino physicists and Terence Tao:

https://arxiv.org/pdf/1908.03795.pdf

I'm guessing that with as fundamentally groundbreaking as this paper is, some people on here have read it. Please let me know if you know where in the paper the eigenvectors are in terms of eigenvalues.

Edit: To be more specific, I saw the norm squared part, but how do you narrow it down to the actual value of each element of each eigenvector. For any reals, you can just act the transformation on the at most the finite $$2^n$$ possibilities (unless it has infinite dimensions as can be the case in particle physics), where n is the dimension of the vector. But, what about when the values that the norm square is being taken of are complex? Also, if you can address infinite dimensions, that would be appreciated. Meaning, is there a way to get the elements without testing each combination?

• Could you be more specific? -- E.g., corollary 3 is certainly far from groundbreaking Nov 15 '19 at 7:48
• I read it yesterday. Lemma $2$ is the main theorem, although I would definitely recommend reading both proofs, as the second proof contains a "more complete" formula (formula $7$ in the paper). That formula also enables one to recover the phases of the eigenvector elements (up to a global phase factor for each eigenvector). Nov 15 '19 at 7:55
• Lemma 2 gives the norm of the Eigenvector elements in terms of Eigenvalues and Eigenvalues of adjugate matrices. It essentially seems of no practical use for computation of the Eigenvectors.
– user65203
Nov 15 '19 at 8:02
• Right – it gives the norms of the components of the eigenvectors, not the exact values of those components. Nov 15 '19 at 9:14
• I am personally a big fan of formula $7$. When you think about it, it makes sense if $A$ is hermitian and diagonal, with distinct eigenvalues. It also provides a geometric interpretation of the adjugate of a hermitian matrix with exactly one of the eigenvalues being $0$, as a kind of orthogonal projection onto the kernel, times $s_{n-1}$ of the set of eigenvalues (which is nonzero in this case). Nov 15 '19 at 20:11

By formula $$7$$ in that paper, they prove that $$\operatorname{adj}(\lambda_i I_n - A) = \prod_{k\neq i} (\lambda_i - \lambda_k) \operatorname{proj}_{v_i},$$ where $$\operatorname{proj}_{v_i}$$ is the orthogonal projection from $$\mathbb{C}^n$$ onto the complex line $$\mathbb{C}v_i$$. So assuming $$\lambda_i$$ to be different from $$\lambda_k$$, for $$k \neq i$$, which would be the case for instance if all the eigenvalues are distinct, then this implies that the image of the left-hand side is $$\mathbb{C}v_i$$. So taking any non-zero column from the left-hand side and normalizing would give you the eigenvector $$v_i$$, the latter being determined up to a phase factor.

I must admit that, while this does recover the eigenvector $$v_i$$ (up to a phase factor), it involves not only knowing the eigenvalue $$\lambda_i$$, but also computing the adjugate on the left-hand side.

This is one way to answer your question. Another way is that one does recover the modulus squared of any element of each eigenvector by knowing not only the eigenvalues of $$A$$, but also the eigenvalues of each matrix $$A_i$$, for $$1 \leq i \leq n$$, where $$A_i$$ is the $$n-1$$ by $$n-1$$ matrix obtained from $$A$$ by deleting the $$i$$-th row and $$i$$-th column from it. Lemma $$2$$ in that paper does just that, and is (I suppose) the reason for the title of that article.

Let me give an example to show how to derive the eigenvectors from eigenvalues. To be specific, we derive the "norm square of coefficients of eigenvectors from eigenvalues" (link).

Suppose the matrix A is a diagonal matrix A = diag(1, 2, 3). The eigenvalues for A is 1, 2, and 3. That means $$\lambda_1 (A) = 1$$, $$\lambda_2 (A) = 2$$ and $$\lambda_3 (A) = 3$$.

To find the eigenvector $$v_1$$ of the first eigenvalue $$\lambda_1 (A)$$, we need to know the exact values of $$v_{1,1}$$, $$v_{1,2}$$ and $$v_{1,3}$$.

Now use the Lemma2 from the arxiv paper [https://arxiv.org/pdf/1908.03795.pdf].

To solve $$v_{1,1}$$, $$v_{1,1}^2 (1 - 2) (1 - 3) = (1 - 2) (1 - 3)$$, and you know $$v_{1,1} = 1$$ or $$v_{1,1} = -1$$. Note, the $$(1-2) (1-3)$$ on the left is $$( \lambda_1 (A) - \lambda_2 (A) ) \cdot ( \lambda_1 (A) - \lambda_3 (A) )$$, and the $$(1-2) (1-3)$$ on the right is $$( \lambda_1 (A) - \lambda_{k=1} (M_1) ) \cdot ( \lambda_1 (A) - \lambda_{k=2} (M_1) )$$ where $$M_1$$ is $$A$$ after deleting the $$1$$st row and column.

To solve $$v_{1,2}$$, $$v_{1,2}^2 (1 - 2) (1 - 3) = (1 - 1) (1 - 3) = 0$$, and you know $$v_{1,2} = 0$$.

To solve $$v_{1,3}$$, $$v_{1,3}^2 (1 - 2) (1 - 3) = (1 - 1) (1 - 2) = 0$$, and you know $$v_{1,3} = 0$$.

• Thanks for the detailed example! I saw the norm squared part, but how do you narrow it down to the actual value. Between 1 and -1, or any reals, you can just act the transformation on the at most finite $2^n$ possibilities (unless it has infinite dimensions as can be the case in particle physics), where n is the dimension of the vector. But, what about when the values that the norm square is being taken of are complex? Nov 15 '19 at 19:44
• The diagonal case that you are considering here is a bit trivial because here the matrix and the matrix of eigenvalues coincide. You don't need the result of the paper in this case
– lcv
Nov 15 '19 at 23:19
• @lcv you're right. I provided a toy example here. Nov 20 '19 at 21:32
• @user3146, my understanding is that you cannot determine 1 or -1, as this paper is to calculate the "norm square of coefficients of eigenvectors from eigenvalues". Nov 20 '19 at 21:34