# Singular Vectors From Singular Values

In the recent paper <cit.>, Denton et al. provided the eigenvector-eigenvalue identity for Hermitian matrices, and a survey was also given for such identity in the literature. The main aim of this paper is to present the identity related to singular vectors and singular values of a general matrix.

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## 1 Introduction

In the recent paper [1], Denton et al. provided the eigenvector-eigenvalue identity for Hermitian matrices, and a survey was also given for such identity in the literature. The main aim of this paper is to present the identity related to singular vectors and singular values of a general matrix. Indeed, this topic has been studied by Thompson [2]. In particular, the interlacing inequalities for singular values of submstrices are derived in [2].

Let be a matrix with singular values and normed left singular vectors and right singular vectors . denotes the conjugate transpose of . The elements of each left singular vector are denoted and right singular vector . Let be matrix after deleting the j-th row of and be singular values of . Let be matrix after deleting the s-th column of and be singular values of .

Lemma 1[1] Let be a Hermitian matrix with distinct eigenvalues and normed eigenvectors The elements of each eigenvector are denoted . Let be the submatrix of that results from deleting the j-th column and the j-th row, with eigenvalues . Then

 |vij|2=n−1∏k=1(λi(A)−λk(Mj))n∏k=1,k≠i(λi(A)−λk(A)),1≤i,j≤n.

Lemma 2[1] Let be a Hermitian matrix with eigenvalues and normed eigenvectors The elements of each eigenvector are denoted . Let be the submatrix of that results from deleting the j-th column and the j-th row, with eigenvalues . Then

 |vij|2n∏k=1,k≠i(λi(A)−λk(A))=n−1∏k=1(λi(A)−λk(Mj)),1≤i,j≤n.

In this paper, we establish the following theorem. We first present singular vectors from distinct singular values of matrix .

Theorem 1 Let with distinct singular values and be left singular vector matrix of and be right singular vector matrix of . Let be matrix after deleting the j-th row of and be singular values of . Let be matrix after deleting the s-th column of and be singular values of . Then

 |uij|2=n−1∏k=1(σ2i(A)−σ2k(¯Aj))n∏k=1,k≠i(σ2i(A)−σ2k(A)),1≤i,j≤n,
 |vls|2=n−1∏t=1(σ2l(A)−σ2t(^As))n∏t=1,t≠l(σ2l(A)−σ2t(A)),1≤l,s≤n.

Then we deduce singular vectors from singular values of matrix .

Theorem 2 Let with singular values and be left singular vector matrix of and be right singular vector matrix of . Let be matrix after deleting the j-th row of and be singular values of . Let be matrix after deleting the s-th column of and be singular values of . Then

 |uij|2m∏k=1,k≠i(σ2i(A)−σ2k(A))=m−1∏k=1(σ2i(A)−σ2k(¯Aj)),1≤i,j≤m,
 |vls|2n∏t=1,t≠l(σ2l(A)−σ2t(A))=n−1∏t=1(σ2l(A)−σ2t(^As)),1≤l,s≤n.

We note that the identities in Theorems 1 and 2 are not given in the previous results. The organization of this paper is given as follows. In Section 2, we present the proofs of the two main theorems. The concluding remarks are given in Section 3.

## 2 The Proof

In this section, we give the proofs of Theorems 1 and 2.

Proof of Theorem 1.

be , where . (i) Let with Since

 AAH=UHΣΣHU,

by Lemma 1 we have

 |uij|2n∏k=1,k≠i(λi(AAH)−λk(AAH))=n−1∏k=1(λi(AAH)−λk(Sj)),1≤i,j≤n, (1)

where is the submatrix of that results from deleting the th column and the th row, with eigenvalues . It is easy to see that

 Sj=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝a1aH1⋯a1aHj−1a1aHj+1⋯a1aHna2aH1⋯a2aHj−1a2aHj+1⋯a2aHn⋯⋯⋯⋯⋯⋯aj−1aH1⋯aj−1aHj−1aj−1aHj+1⋯aj−1aHnaj+1aH1⋯aj+1aHj−1aj+1aHj+1⋯aj+1aHn⋯⋯⋯⋯⋯⋯anaH1⋯anaHj−1anaHj+1⋯anaHn⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

Let be matrix after deleting the j-th row of and be singular values of , then

 ¯Aj=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝a1⋮aj−1aj+1⋮an⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

and

 σ2k(¯Aj)=λ2k(¯Aj¯AHj). (2)

Since has distinct singular values, by (1) and (2) we have

 |uij|2=n−1∏k=1(σ2i(A)−σ2k(¯Aj))n∏k=1,k≠i(σ2i(A)−σ2k(A)),1≤i,j≤n.

(ii) Let with Since

 AHA=VHΣHΣV,

by Lemma 1 we have

 |vij|2n∏k=1,k≠i(λi(AHA)−λk(AHA))=n−1∏k=1(λi(AHA)−λk(Hj)),1≤i,j≤n, (3)

where is the submatrix of that results from deleting the th column and the th row, with eigenvalues . It is easy to see that

 Hj=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝bH1b1⋯bH1bj−1bH1bj+1⋯bH1bnbH2b1⋯bH2bj−1bH2bj+1⋯bH2bn⋯⋯⋯⋯⋯⋯bHj−1b1⋯bHj−1bj−1bHj−1bj+1⋯bHj−1bnbHj+1b1⋯bHj+1bj−1bHj+1bj+1⋯bHj+1bn⋯⋯⋯⋯⋯⋯bHnb1⋯bHnbj−1bHnbj+1⋯bHnbn⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

Let be matrix after deleting the j-th row of and be singular values of , then

 ^Aj=(b1,…,bj−1,bj+1,…,bn)

and

 σ2k(^Aj)=λ2k(^AHj^Aj). (4)

Since has distinct singular values, by (3) and (4) we have

 |vls|2=n−1∏t=1(σ2l(A)−σ2t(^As))n∏t=1,t≠l(σ2l(A)−σ2t(A)),1≤l,s≤n.

This completes the proof.

By using the similar trick, we can show the results of Theorem 2.

Proof of Theorem 2. Let singular value decompositions of be , where . (i) Let with Since

 AAH=UHΣΣHU,

by Lemma 1 we have

 |uij|2m∏k=1,k≠i(λi(AAH)−λk(AAH))=m−1∏k=1(λi(AAH)−λk(~Sj)),1≤i,j≤m, (5)

where is the submatrix of that results from deleting the th column and the th row, with eigenvalues . It is easy to see that

 ~Sj=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝a1aH1⋯a1aHj−1a1aHj+1⋯a1aHma2aH1⋯a2aHj−1a2aHj+1⋯a2aHm⋯⋯⋯⋯⋯⋯aj−1aH1⋯aj−1aHj−1aj−1aHj+1⋯aj−1aHmaj+1aH1⋯aj+1aHj−1aj+1aHj+1⋯aj+1aHm⋯⋯⋯⋯⋯⋯amaH1⋯amaHj−1amaHj+1⋯amaHm⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

Let be matrix after deleting the j-th row of and be singular values of , then

 ~Aj=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝a1⋮aj−1aj+1⋮am⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠

and

 σ2k(~Aj)=λ2k(~Aj~AHj). (6)

By (5) and (6) we have

 |uij|2m∏k=1,k≠i(σ2i(A)−σ2k(A))=m−1∏k=1(σ2i(A)−σ2k(¯Aj)),1≤i,j≤m.

(ii) Let with Since

 AHA=VHΣHΣV,

by Lemma 1 we have

 |vij|2n∏k=1,k≠i(λi(AHA)−λk(AHA))=n−1∏k=1(λi(AHA)−λk(Hj)),1≤i,j≤n, (7)

where is the submatrix of that results from deleting the th column and the th row, with eigenvalues . It is easy to see that

 Hj=⎛⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜⎝bH1b1⋯bH1bj−1bH1bj+1⋯bH1bnbH2b1⋯bH2bj−1bH2bj+1⋯bH2bn⋯⋯⋯⋯⋯⋯bHj−1b1⋯bHj−1bj−1bHj−1bj+1⋯bHj−1bnbHj+1b1⋯bHj+1bj−1bHj+1bj+1⋯bHj+1bn⋯⋯⋯⋯⋯⋯bHnb1⋯bHnbj−1bHnbj+1⋯bHnbn⎞⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟⎠.

Let be matrix after deleting the j-th row of and be singular values of , then

 ^Aj=(b1,…,bj−1,bj+1,…,bn)

and

 σ2k(^Aj)=λ2k(^AHj^Aj). (8)

By (7) and (8) we have

 |vls|2=n−1∏t=1(σ2l(A)−σ2t(^As))n∏t=1,t≠l(σ2l(A)−σ2t(A)),1≤l,s≤n.

This completes the proof.

## 3 Concluding Remarks

In [1], Denton et al. revisited the identity of eigenvalue-eigenvector for Hermitian matrices. This identity has been discovered by many researchers in the literature. In this paper, we derived similar identity for singular vectors and singular values of general matrices. However, we find these results have not been studied previously. As a future research work, it may be useful to apply this new identity formula to matrix perturbation problems and the calculation of arbitrary singular values and singular vector instead of all singular values and singular vectors.