In this post I’ll write up the notes I took while trying to follow this “short and simple” linear algebra result which made a few headlines in 2019, sent to me by Juanhe (as so many other cool things are, thank you!)

The general thrust of the result is that simply knowing the eigenvalues of a Hermitian matrix \(A\) and its principal minors, you can construct its eigenvectors.

EDIT, 2019-12-14: the authors of the above paper have reuploaded a much more extensive version replete with multiple proofs and a historical analysis of the identity’s history. The below is now Section 2.5, “Proof using a Cauchy-Binet type formula”; it seems like the simplest of the provided proofs.

Lemma

The lemma asserts that for \(\small n \times n\) Hermitian \(A\) with eigenvalues \(\lambda_i\) and associated eigenvectors \(v_i\) with some eigenvalue \(\lambda_k=0\) and any \(\small n \times n-1\) matrix \(B\), we have:

(∏i=1,i≠knλi(A))∣det⁡(B∣∣vk)∣2=det⁡(B∗AB)\left( \prod_{i=1,i\neq k}^{n}\lambda_i(A) \right) |\det(B \mathbf{||} v_k)|^2 = \det(B^*AB)

where \(B \mathbf{||} v_k\) denotes the \(\small n \times n\) matrix formed by augmenting the \(\small n \times n-1\) matrix \(B\) with the \(\small n \times1\) eigenvector \(v_k\).

Begin by assuming \(A\) is diagonal and that \(k=n\) which by construction means:

A=[λ1λ20⋱0λn−10]=[A′0]A = \begin{bmatrix} \lambda_{1} \\ & \lambda_{2} & & \Large 0 \\ & & \ddots \\ & \Large 0 & & \lambda_{n-1} \\ & & & & 0 \end{bmatrix} = \begin{bmatrix} \Huge A' & \\ & 0 \end{bmatrix} vi=ei,∀iv_i = e_i, \forall i

where \(A’\) is the upper-left diagonal \(\small n − 1 \times n − 1\) submatrix and \(e_i\) is the standard basis vector with a \(1\) in the \(i\)th column and \(0\) elsewhere.

Similarly, write \(B = \begin{bmatrix} B’ \\ X \end{bmatrix} = \begin{bmatrix} \beta_1’ & \beta_2’ & \dots & \beta_{n-1}’ \\ x_1 & x_2 & \dots & x_{n-1} \end{bmatrix}\) where \(B’\) is the upper \(\small n − 1 \times n − 1\) submatrix and \(X\) is the lower \(\small 1 \times n − 1\) row vector, with each \(\beta_i’\) a \(\small n-1 \times 1\) column vector and \(X = \begin{bmatrix}—x_i—\end{bmatrix}\).

We have on the one hand that:

det⁡(B∣∣vn)=det⁡([B′0X1])=0+0+⋯+det⁡B′\det(B\mathbf{||}v_n) = \det( \begin{bmatrix} \Large B' & \Large 0 \\ \\ X & 1 \end{bmatrix} ) = 0 + 0 + \dots + \det B'

by considering the Laplace expansion along the last row (each submatrix but the last contains the \(0\)-vector on the top right, hence each cofactor but the last is also \(0\)), which implies:

⇒(∏i=1n−1λi(A))∣det⁡(B∣∣vn)∣2=(∏i=1n−1λi(A))∣det⁡B′∣2\Rightarrow \left( \prod_{i=1}^{n-1}\lambda_i(A) \right) |\det(B \mathbf{||} v_n)|^2 = \left( \prod_{i=1}^{n-1}\lambda_i(A) \right) |\det B'|^2

On the other hand, note that since the last row and column of \(A\) is identically zero, we have:

AB=[A′β1′…A′βn−1′0…0]AB = \begin{bmatrix} A'\beta_1' & \dots & A'\beta_{n-1}' \\ 0 & \dots & 0 \end{bmatrix} ⇒B∗AB=[β1′x1……βn−1′xn−1][A′β1′…A′βn−1′0…0]=[β1′A′β1′…β1′A′βn−1′………βn−1′A′β1…βn−1′A′βn−1]\begin{aligned} \Rightarrow B^*AB & = \begin{bmatrix}\beta_1' & x_1 \\ \dots & \dots \\ \beta_{n-1}' & x_{n-1} \end{bmatrix} \begin{bmatrix} A'\beta_1' & \dots & A'\beta_{n-1}' \\ 0 & \dots & 0 \end{bmatrix} \\ & = \begin{bmatrix} \beta_1'A'\beta_1' & \dots & \beta_1'A'\beta_{n-1}' \\ \dots & \dots & \dots \\ \beta_{n-1}'A'\beta_1 & \dots & \beta_{n-1}'A'\beta_{n-1} \end{bmatrix} \end{aligned}

where the last row of the rightmost matrix being zero eliminates all of the \(x_i\) from the product. This last \(\small n-1 \times n-1\) matrix we recognize as simply the conjugation:

B′∗A′B′B'^* A' B'

and hence:

det⁡(B∗AB)=det⁡(B′∗A′B′)=det⁡(A′)∣det⁡(B′)∣2=(∏i=1n−1λi(A))∣det⁡B′∣2\det(B^* A B) = \det(B'^* A' B') = \det(A') |\det(B')|^2 = \left( \prod_{i=1}^{n-1}\lambda_i(A) \right) |\det B'|^2

as desired.

To complete the lemma, when \(A\) is Hermitian we can find an eigendecomposition \(A = VDV^*\) where the \(V = \begin{bmatrix} v_1 & \dots & v_n \end{bmatrix}\) is orthogonal, and \(D\) is the diagonal matrix of eigenvalues \(\lambda_i\), and moreover we can find such a decomposition where \(\lambda_n\) is the \(0\) eigenvalue. Then:

det⁡(B∗AB)=det⁡(B∗VDV∗B)=(∏i=1n−1λi)∣det⁡(V∗B∣∣en)∣2=(∏i=1n−1λi)∣det⁡(V∗B∣∣V∗vn)∣2=(∏i=1n−1λi)∣det⁡(B∣∣vn)∣2\begin{aligned} \det(B^*AB) = \det(B^*VDV^*B) & = \left( \prod_{i=1}^{n-1}\lambda_i \right) |\det(V^*B \mathbf{||} e_n)|^2 \\ & = \left( \prod_{i=1}^{n-1}\lambda_i \right) |\det(V^*B \mathbf{||} V^*v_n)|^2 \\ & = \left( \prod_{i=1}^{n-1}\lambda_i \right) |\det(B \mathbf{||} v_n)|^2 \end{aligned}

where on the first line we use the diagonal case of the lemma, on the second we use the orthogonality of the rows \(v_i\) of \(V^*\), and on the third we extract \(\det(V^*) = 1\) again by orthogonality.

Main result

The main result applies to Hermitian \(A\), again with eigenvalues/vectors \(\lambda_i, v_i\). Letting \(M_j\) be \(A\) with the \(j\)-th row and column removed, and \(\mu_i\) be its eigenvalues, it says that \(\forall 1\leq i, j\leq n\):

∣vi,j∣2∏k=1,k≠in(λk−λi)=∏k=1n−1(μk−λi)|v_{i,j}|^2 \prod_{k=1, k\neq i}^{n} (\lambda_k - \lambda_i) = \prod_{k=1}^{n-1} (\mu_k - \lambda_i)

It’s actually pretty simple once we already have the Lemma. Fixing \(i, j\) (the paper argues WLOG to consider \(i=n, j=1\)), note that \(A’ = A - \lambda_i I_n\) has eigenvalues \(\lambda_k’ = \lambda_k - \lambda_i, \forall k\) with the same eigenvectors \(v_k’ = v_k\); letting \(M_j’\) be the analogous submatrix similarly \(\mu_k’ = \mu_k - \lambda_i, \forall k\). Substituting in the above, the proof reduces to showing:

∣vi,j∣2∏k=1,k≠inλk′=∏k=1n−1μk′|v_{i,j}|^2 \prod_{k=1, k\neq i}^{n} \lambda_k' = \prod_{k=1}^{n-1} \mu_k'

Our new matrix \(A’\) has \(\lambda_k’ = 0\) so we may apply the Lemma with \(B\) the \(\small n \times n-1\) matrix:

bu,v={1u<j,u=v1u>j,u=v+10otherwise⟶B=[11⋱000…000⋱11]b_{u,v} = \begin{cases} 1 & u < j, u = v \\ 1 & u > j, u = v + 1 \\ 0 & \text{otherwise} \end{cases} \qquad \longrightarrow \qquad B = \begin{bmatrix} 1 \\ & 1 \\ & & \ddots \\ 0 & 0 & 0 & \dots & 0 & 0 & 0 \\ & & & & \ddots \\ & & & & & 1 \\ & & & & & & 1\\ \end{bmatrix}

that is, \(B\) is \(I_{n-1}\) with the zero vector \(0_{1, n-1}\) inserted at the \(j\)-th row.

Note \(|\det(B || v_i)|^2\) is precisely \(|v_{i,j}|^2\) again by considering the Laplace expansion along the \(j\)-th row. And:

det⁡(B∗A′B)=det⁡(B∗[a1,1′…a1,j−1′a1,j+1′…a1,n′a2,1′…a2,j−1′a2,j+1′…a2,n′………………an,1′…an,j−1′an,j+1′…an,n′])=det⁡([a1,1′…a1,j−1′a1,j+1′…a1,n′………………aj−1,1′…aj−1,j−1′aj−1,j+1′…aj−1,n′aj+1,1′…aj+1,j−1′aj+1,j+1′…aj+1,n′………………an,1′…an,j−1′an,j+1′…an,n′])=det⁡(Mj′)=∏k=1n−1μk′\begin{aligned} \det(B^*A'B) & = \det\left(B^* \begin{bmatrix} a'_{1,1} & \dots & a'_{1,j-1} & a'_{1,j+1} & \dots & a'_{1,n} \\ a'_{2,1} & \dots & a'_{2,j-1} & a'_{2,j+1} & \dots & a'_{2,n} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a'_{n,1} & \dots & a'_{n,j-1} & a'_{n,j+1} & \dots & a'_{n,n} \\ \end{bmatrix}\right) \\ \\ & = \det\left(\begin{bmatrix} a'_{1,1} & \dots & a'_{1,j-1} & a'_{1,j+1} & \dots & a'_{1,n} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a'_{j-1,1} & \dots & a'_{j-1,j-1} & a'_{j-1,j+1} & \dots & a'_{j-1,n} \\ a'_{j+1,1} & \dots & a'_{j+1,j-1} & a'_{j+1,j+1} & \dots & a'_{j+1,n} \\ \dots & \dots & \dots & \dots & \dots & \dots \\ a'_{n,1} & \dots & a'_{n,j-1} & a'_{n,j+1} & \dots & a'_{n,n} \\ \end{bmatrix}\right) \\ \\ & = \det(M'_j) = \prod_{k=1}^{n-1} \mu_k' \end{aligned}

so that the Lemma immediately gives our result:

(∏k=1,k≠inλk(A′))∣det⁡(B∣∣vi)∣2=det⁡(B∗A′B)⇒(∏k=1,k≠inλk′)∣vi,j∣2=∏k=1n−1μk′\begin{aligned} & \left( \prod_{k=1,k\neq i}^{n}\lambda_k(A') \right) |\det(B \mathbf{||} v_i)|^2 = \det(B^*A'B) \\ \Rightarrow & \left( \prod_{k=1,k\neq i}^{n}\lambda_k' \right) |v_{i,j}|^2 = \prod_{k=1}^{n-1} \mu_k' \end{aligned}