TenLa: an approach based on controllable tensor decomposition and optimized lasso regression for judgement prediction of legal cases

Abstract

With the development of big data and artificial intelligence technology, the computer-assisted judgment of legal cases has become an inevitable trend in the intersection of computer science and law. Judgment prediction methods of legal cases mainly consist of two parts: (1) modeling of legal cases and (2) construction of judgment prediction algorithms. Previous methods for the judgment prediction of legal cases are mainly based on feature models and classification algorithms. Traditional feature models require extensive expert knowledge and manual annotation. They are highly dependent on vocabulary and grammatical information in databases, which are not conducive to the improvement of accuracy and universality of subsequent prediction algorithms. In addition, prediction results obtained by classification algorithms are coarse in granularity and low in accuracy. In general, judgments in similar legal cases are similar. This article proposes a new method for the judgment prediction of legal cases, namely, TenLa, which is based on a controllable algorithm of tensor decomposition and an optimized Lasso regression model. TenLa takes similarities between legal cases as an important indicator of judgment prediction and is mainly divided into three parts: (1) ModTen; we propose a modeling method for legal cases, namely, ModTen, which represents legal cases as three-dimensional tensors. (2) ConTen; we propose a new tensor decomposition algorithm, namely, ConTen, which decomposes tensors obtained by ModTen into core tensors through the intermediary tensor. Core tensors greatly reduce the dimensions of original tensors. (3) OLass; we propose an optimized Lasso regression algorithm, namely, OLass. Core tensors obtained by ConTen are used to train OLass. Specifically, we propose an optimization algorithm for OLass with respect to the intermediary tensor in ConTen; thus, the core tensors obtained by ConTen carry tensor elements and tensor structure information that is most conducive to the improvement of the accuracy of OLass. Experiments show that TenLa has higher accuracy than traditional judgment prediction algorithms.

A Appendix

1.1 A.1 Supplementary material 1

Proofs A.1, A.2, A.3, A.4, A.5, A.6, A.7 and A.8 give the proof process of lemmas 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8 and 4.9, respectively.

Proof A.1

From definition 3.4, we can see that values of corresponding elements in χ and ${\chi _{({n_0})}}$ are equal. According to the definition of Frobenius norm 3.3, the sum of squares of elements in χ and ${\chi _{({n_0})}}$ is equal. That is $\left \| \chi \right \|_F^2 = \left \| {{\chi _{({n_0})}}} \right \|_F^2$.

Proof A.2

Let C = A^TA, $C \in {\mathbb {R}^{J \times J}}$. According to the rule of matrix multiplication, we can obtain that ${C_{ij}} = \sum \limits _{p = 1}^I {A_{ip}^T{A_{pj}}} = \sum \limits _{p = 1}^I {{A_{pi}}{A_{pj}}}$. Based on the definition of the trace norm 3.1, we get that $Trace(C) = \sum \limits _{j = 1}^J {{C_{jj}}}$. That is $Trace(C) = \sum \limits _{j = 1}^J {\sum \limits _{p = 1}^I {A_{pj}^2} }$. From the definition of Frobenius norm 3.3, we obtain that $\left \| A \right \|_F^2 = \sum \limits _{p = 1}^I {\sum \limits _{j = 1}^J {A_{pj}^2} }$. Thus, $\left \| A \right \|_F^2 = Trace({A^T}A)$.

Proof A.3

Let $\lambda = \chi { \times _{{n_0}}}P$, ${\mathbb {R}^{{I_1} \times {I_2} \times {\cdots } \times {I_{{n_0} - 1}} \times {J_{{n_0}}} \times {I_{{n_0} + 1}} \times {\cdots } \times {I_N}}}$. From definition 3.6, we get that ${\lambda _{{i_1}{i_2} {\cdots } {i_{{n_0} - 1}}{j_{{n_0}}}{i_{{n_0} + 1}} {\cdots } {i_N}}} = \sum \limits _{i = 1}^{{I_{{n_0}}}} {{\chi _{{i_1}{i_2} {\cdots } {i_{{n_0} - 1}}i{i_{{n_0} + 1}} {\cdots } {i_N}}}{P_{i{j_{{n_0}}}}}}$. According to definition 3.4, we obtain that ${\lambda _{({n_0})}} \in {\mathbb {R}^{({I_1}{I_2} {\cdots } {I_{{n_0} - 1}}{I_{{n_0} + 1}} {\cdots } {I_N}) \times {J_{{n_0}}}}}$. ${\lambda _{({n_0})}}_{({i_1}{i_2} {\cdots } {i_{{n_0} - 1}}{i_{{n_0} + 1}} {\cdots } {i_N}){j_{{n_0}}}} = {\lambda _{{i_1}{i_2} {\cdots } {i_{{n_0} - 1}}{j_{{n_0}}}{i_{{n_0} + 1}} {\cdots } {i_N}}}$. Since ${\chi _{({n_0})}} \in {\mathbb {R}^{({I_1}{I_2} {\cdots } {I_{{n_0} - 1}}{I_{{n_0} + 1}} {\cdots } {I_N}) \times {I_{{n_0}}}}}$, let $A = {\chi _{({n_0})}} \times P$, $A \in {\mathbb {R}^{({I_1}{I_2} {\cdots } {I_{{n_0} - 1}}{I_{{n_0} + 1}} {\cdots } {I_N}) \times {J_{{n_0}}}}}$. From the rule of matrix multiplication, ${A_{({i_1}{i_2} {\cdots } {i_{{n_0} - 1}}{i_{{n_0} + 1}} {\cdots } {i_N}){j_{{n_0}}}}} = \sum \limits _{i = 1}^{{I_{{n_{0}}}}} {{\chi _{({n_{0}})}}_{({i_{1}}{i_{2}} {\cdots } {i_{{n_{0}} - 1}}{i_{{n_{0}} + 1}} {\cdots } {i_{N}})i}{P_{i{j_{{n_{0}}}}}}}$. Since ${\chi _{({n_{0}})}}_{({i_{1}}{i_{2}} {\cdots } {i_{{n_{0}} - 1}}{i_{{n_{0}} + 1}} {\cdots } {i_{N}})i} = {\chi _{{i_1}{i_2} {\cdots } {i_{{n_0} - 1}}i{i_{{n_0} + 1}} {\cdots } {i_N}}}$, ${\lambda _{({n_0})}}_{({i_1}{i_2} {\cdots } {i_{{n_0} - 1}}{i_{{n_0} + 1}} {\cdots } {i_N}){j_{{n_0}}}} = {A_{({i_1}{i_2} {\cdots } {i_{{n_0} - 1}}{i_{{n_0} + 1}} {\cdots } {i_N}){j_{{n_0}}}}}$. ${\lambda _{({n_0})}} = A$. That is ${(\chi { \times _{{n_0}}}P)_{({n_0})}} = {\chi _{({n_0})}} \times P$.

Proof A.4

Let $C = \upsilon _{({n_0})}^T{\varpi _{({n_0})}}$, $C \in {\mathbb {R}^{{J_{{n_0}}} \times {I_{{n_0}}}}}$. From the rule of matrix multiplication, ${C_{ij}} = \sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {\upsilon {{{~}_{({n_0})}^T}_{ip}}{\varpi _{({n_0})}}_{pj}} = \sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{\upsilon _{({n_0})}}_{pi}{\varpi _{({n_0})}}_{pj}}$. Then ${F_1} = C{B_{{n_0}}}$, ${F_1} \in {\mathbb {R}^{{J_{{n_0}}} \times {J_{{n_0}}}}}$. ${F_1}_{ij} = \sum \limits _{q = 1}^{{I_{{n_0}}}} {{C_{iq}}{B_{{n_0}}}_{qj}} = \sum \limits _{q = 1}^{{I_{{n_0}}}} {\sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{\upsilon _{({n_0})}}_{pi}{\varpi _{({n_0})}}_{pq}{B_{{n_0}}}_{qj}} }$. Based on the definition of the trace norm 3.1, we get that $Trace({F_1}) = \sum \limits _{j = 1}^{{J_{{n_0}}}} {{F_1}_{jj}}$. That is $Trace({F_1}) = \sum \limits _{j = 1}^{{J_{{n_0}}}} {\sum \limits _{q = 1}^{{I_{{n_0}}}} {\sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{\upsilon _{({n_0})}}_{pj}{\varpi _{({n_0})}}_{pq}{B_{{n_0}}}_{qj}} } }$. When q = m, j = n, $\frac {{\partial (\sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{\upsilon _{({n_0})}}_{pn}{\varpi _{({n_0})}}_{pm}{B_{{n_0}}}_{mn}} )}}{{\partial {B_{{n_0}}}_{mn}}} = \sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{\upsilon _{({n_0})}}_{pn}{\varpi _{({n_0})}}_{pm}} = \sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {\varpi {{{~}_{({n_0})}^T}_{mp}}} {\upsilon _{({n_0})}}_{pn}$. When q≠m,j≠n, $\frac {{\partial (\sum \limits _{p = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {{\upsilon _{({n_{0}})}}_{pj}{\varpi _{({n_{0}})}}_{pq}{B_{{n_{0}}}}_{qj}} )}}{{\partial {B_{{n_{0}}}}_{mn}}} = 0$. Therefore, $\frac {{\partial Trace({F_{1}})}}{{\partial {B_{{n_{0}}}}}} = \varpi _{({n_{0}})}^{T}{\upsilon _{({n_{0}})}}$.

Proof A.5

Let $C = B_{{n_{0}}}^{T}\varpi _{({n_{0}})}^{T}$, $C \in {\mathbb {R}^{{J_{{n_{0}}}} \times ({J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}})}}$. ${C_{ij}} = \sum \limits _{p = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{ip}}\varpi {{{~}_{({n_{0}})}^{T}}_{pj}}} = \sum \limits _{p = 1}^{{I_{{n_{0}}}}} {{B_{{n_{0}}}}_{pi}\varpi {{{~}_{({n_{0}})}^{T}}_{pj}}}$. ${F_{2}} = C{\upsilon _{({n_{0}})}}$, ${F_{2}} \in {\mathbb {R}^{{J_{{n_{0}}}} \times {J_{{n_{0}}}}}}$. ${F_{2}}_{ij} = \sum \limits _{q = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {{C_{iq}}{\upsilon _{({n_{0}})}}_{qj}}$. That is ${F_{2}}_{ij} = \sum \limits _{q = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {\sum \limits _{p = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{ip}}\varpi {{{~}_{({n_{0}})}^{T}}_{pq}}{\upsilon _{({n_{0}})}}_{qj}} }$. Based on definition 3.1, $Trace({F_{2}}) = \sum \limits _{j = 1}^{{J_{{n_0}}}} {{F_2}_{jj}}$. $Trace({F_2}) = \sum \limits _{j = 1}^{{J_{{n_0}}}} {\sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {\sum \limits _{p = 1}^{{I_{{n_0}}}} {{B_{{n_0}}}_{pj}\varpi {{{~}_{({n_0})}^T}_{pq}}{\upsilon _{({n_0})}}_{qj}} } }$. When p = m, j = n, $\frac {{\partial (\sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{B_{{n_0}}}_{mn}\varpi {{{~}_{({n_0})}^T}_{mq}}{\upsilon _{({n_0})}}_{qn}} )}}{{\partial {B_{{n_0}}}_{mn}}} = \sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {\varpi {{{~}_{({n_0})}^T}_{mq}}{\upsilon _{({n_0})}}_{qn}}$. When p≠m, j≠n, $\frac {{\partial (\sum \limits _{q = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {{B_{{n_{0}}}}_{pj}\varpi {{{~}_{({n_{0}})}^{T}}_{pq}}{\upsilon _{({n_{0}})}}_{qj}} )}}{{\partial {B_{{n_{0}}}}_{mn}}} = 0$. Thus, $\frac {{\partial Trace({F_{2}})}}{{\partial {B_{{n_{0}}}}}} = \varpi _{({n_{0}})}^{T}{\upsilon _{({n_{0}})}}$.

Proof A.6

Let $C = \varpi _{({n_{0}})}^{T}{\varpi _{({n_{0}})}}$, $C \in {\mathbb {R}^{{I_{{n_{0}}}} \times {I_{{n_{0}}}}}}$. Then ${F_{3}} = B_{{n_{0}}}^{T}C{B_{{n_{0}}}}$. Let $D = B_{{n_{0}}}^{T}C$, $D \in {\mathbb {R}^{{J_{{n_{0}}}} \times {I_{{n_{0}}}}}}$. ${F_{3}} = D{B_{{n_{0}}}}$. From the rule of matrix multiplication, ${D_{ij}} = \sum \limits _{p = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{ip}}{C_{pj}}}$. ${F_{3}}_{ij} = \sum \limits _{q = 1}^{{I_{{n_{0}}}}} {{D_{iq}}{B_{{n_{0}}}}_{qj}}$. Based on the definition of the trace norm 3.1, $Trace({F_{3}}) = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {{F_{3}}_{jj}} = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {\sum \limits _{q = 1}^{{I_{{n_{0}}}}} {{D_{jq}}{B_{{n_{0}}}}_{qj}} }$. That is $Trace({F_{3}}) = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {\sum \limits _{q = 1}^{{I_{{n_{0}}}}} {\sum \limits _{p = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{jp}}{C_{pq}}{B_{{n_{0}}}}_{qj}} } }$. When q = m, j = n, $\frac {{\partial (\sum \limits _{p = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{np}}{C_{pm}}} {B_{{n_{0}}}}_{mn})}}{{\partial {B_{{n_{0}}}}_{mn}}} = \sum \limits _{p = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{np}}{C_{pm}}} = \sum \limits _{p = 1}^{{I_{{n_{0}}}}} {C_{mp}^{T}{B_{{n_{0}}}}_{pn}}$. When p = m, j = n, $\frac {{\partial (\sum \limits _{q = 1}^{{I_{{n_{0}}}}} {B{{{~}_{{n_{0}}}^{T}}_{nm}}{C_{mq}}{B_{{n_{0}}}}_{qn}} )}}{{\partial {B_{{n_{0}}}}_{mn}}} = \sum \limits _{q = 1}^{{I_{{n_{0}}}}} {{C_{mq}}{B_{{n_{0}}}}_{qn}}$. In other cases, $\frac {{\partial (B{{{~}_{{n_{0}}}^{T}}_{jp}}{C_{pq}}{B_{{n_{0}}}}_{qj})}}{{\partial {B_{{n_{0}}}}_{mn}}} = 0$. Therefore, $\frac {{\partial Trace({F_{3}})}}{{\partial {B_{{n_{0}}}}}} = C{B_{{n_{0}}}} + {C^{T}}{B_{{n_{0}}}}$. Since $C = \varpi _{({n_{0}})}^{T}{\varpi _{({n_{0}})}}$, C^T = C. $\frac {{\partial Trace({F_{3}})}}{{\partial {B_{{n_{0}}}}}} = 2C{B_{{n_{0}}}}$. Thus, $\frac {{\partial Trace({F_{3}})}}{{\partial {B_{{n_{0}}}}}} = \eta \varpi _{({n_{0}})}^{T}{\varpi _{({n_{0}})}}{B_{{n_{0}}}}$, η = 2.

Proof A.7

Let $C = \varpi _{({n_{0}})}^{T}$, $C \in {\mathbb {R}^{{I_{{n_{0}}}} \times ({J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}})}}$. From the singular value decomposition, we get that C = PSQ^T. P and Q are orthogonal matrices. S is a diagonal matrix. Since the number of rows is larger than columns in S, there exists U that satisfies SU = E. Let A = QUP^T, CA = PSQ^TQUP^T. Since P and Q are orthogonal matrices, the value of Q^TQ and PP^T is the identity matrix. Since SU = E, CA = E. That is $\varpi _{({n_{0}})}^{T}A = E$. Thus, there exists a matrix A, $A \in {\mathbb {R}^{({J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}) \times {I_{{n_{0}}}}}}$, A satisfies $\varpi _{({n_{0}})}^{T}A = E$.

Proof A.8

According to lemma 4.8. we can obtain that there exists a matrix A, $A \in {\mathbb {R}^{({J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}) \times {I_{{n_0}}}}}$, A satisfies $\varpi _{({n_0})}^TA = E$. E is the identity matrix, $E \in {\mathbb {R}^{{I_{{n_0}}} \times {I_{{n_0}}}}}$. From proof A.7, we get that A = QUP^T. PSQ^T is the singular value decomposition of $\varpi _{({n_0})}^T$. U can be calculated by SU = E. Since ${\varpi _{({n_0})}}{B_{{n_0}}} = {\upsilon _{({n_0})}}$, ${A^T}{\varpi _{({n_0})}}{B_{{n_0}}} = {A^T}{\upsilon _{({n_0})}}$. Since $\varpi _{({n_0})}^TA = {A^T}{\varpi _{({n_0})}} = E$, ${B_{{n_0}}} = {A^T}{\upsilon _{({n_0})}}$. That is ${B_{{n_0}}} = P{U^T}{Q^T}{\upsilon _{({n_0})}}$.

1.2 A.2 Supplementary material 2

Proofs A.9, A.10 and A.11 give the proof process of lemmas 4.11, 4.12 and 4.13, respectively.

Proof A.9

According to the rule of matrix multiplication, ${F_4}_{ij} = \sum \limits _{p = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {P{{{~}_{({n_{0}})}^{T}}_{ip}}} {\widehat \chi _{({n_{0}})pj}}$. From the definition of the trace norm 3.1, $Trace({F_{4}}) = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {{F_{4}}_{jj}} = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {\sum \limits _{p = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {{P_{({n_{0}})}}_{pj}{{\widehat \chi }_{({n_{0}})pj}}} }$. When p = m, j = n, $\frac {{\partial ({P_{({n_{0}})}}_{mn}{{\widehat \chi }_{({n_{0}})mn}})}}{{\partial {{\widehat \chi }_{({n_{0}})mn}}}} = {P_{({n_{0}})}}_{mn}$. When p≠m, j≠n, $\frac {{\partial ({P_{({n_{0}})}}_{pj}{{\widehat \chi }_{({n_{0}})pj}})}}{{\partial {{\widehat \chi }_{({n_{0}})mn}}}} = 0$. Therefore, $\frac {{\partial Trace({F_{4}})}}{{\partial {{\widehat \chi }_{({n_{0}})}}}} = {P_{({n_{0}})}}$.

Proof A.10

According to the rule of matrix multiplication, ${F_{5}}_{ij} = \sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {\widehat \chi _{({n_0})ip}^T{P_{({n_0})}}_{pj}}$. From the definition of the trace norm 3.1, $Trace({F_5}) = \sum \limits _{j = 1}^{{J_{{n_0}}}} {{F_5}_{jj}} = \sum \limits _{j = 1}^{{J_{{n_0}}}} {\sum \limits _{p = 1}^{{J_1}{J_2} {\cdots } {J_{{n_0} - 1}}{J_{{n_0} + 1}} {\cdots } {J_N}} {{{\widehat \chi }_{({n_0})pj}}{P_{({n_0})}}_{pj}} }$. When p = m, j = n, $\frac {{\partial ({{\widehat \chi }_{({n_0})mn}}{P_{({n_0})}}_{mn})}}{{\partial {{\widehat \chi }_{({n_0})mn}}}} = {P_{({n_0})}}_{mn}$. When p≠m, j≠n, $\frac {{\partial ({{\widehat \chi }_{({n_0})pj}}{P_{({n_0})}}_{pj})}}{{\partial {{\widehat \chi }_{({n_0})mn}}}} = 0$. Thus, $\frac {{\partial Trace({F_5})}}{{\partial {{\widehat \chi }_{({n_0})}}}} = {P_{({n_0})}}$.

Proof A.11

According to the rule of matrix multiplication, ${F_6}_{ij} = \sum \limits _{p = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {\widehat \chi _{({n_{0}})ip}^{T}{{\widehat \chi }_{({n_{0}})pj}}}$. From the definition of the trace norm 3.1, $Trace({F_{6}}) = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {{F_{6}}_{jj}} = \sum \limits _{j = 1}^{{J_{{n_{0}}}}} {\sum \limits _{p = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{{n_{0}} - 1}}{J_{{n_{0}} + 1}} {\cdots } {J_{N}}} {\widehat \chi _{({n_{0}})pj}^{2}} }$. When p = m, j = n, $\frac {{\partial (\widehat \chi _{({n_{0}})mn}^{2})}}{{\partial {{\widehat \chi }_{({n_{0}})mn}}}} = 2{\widehat \chi _{({n_{0}})mn}}$. When p≠m, j≠n, $\frac {{\partial (\widehat \chi _{({n_{0}})pj}^{2})}}{{\partial {{\widehat \chi }_{({n_{0}})mn}}}} = 0$. Therefore, $\frac {{\partial Trace({F_{6}})}}{{\partial {{\widehat \chi }_{({n_{0}})}}}} = \eta {\widehat \chi _{({n_{0}})}}$, η = 2.

1.3 A.3 Supplementary material 3

Proofs A.12, A.13, A.14 and A.15 give the proof process of lemmas 4.15, 4.16, 4.17 and 4.18, respectively.

Proof A.12

Let D = AB, $D \in {\mathbb {R}^{m \times p}}$. According to the rule of matrix multiplication, ${D_{ij}} = \sum \limits _{a = 1}^{n} {{A_{ia}}{B_{aj}}}$. Let G = DC, $G \in {\mathbb {R}^{m \times m}}$. Then ${G_{ij}} = \sum \limits _{b = 1}^{p} {{D_{ib}}{C_{bj}}} = \sum \limits _{b = 1}^{p} {\sum \limits _{a = 1}^{n} {{A_{ia}}{B_{ab}}{C_{bj}}} }$. From the definition of the trace norm 3.1, $Trace(G) = \sum \limits _{h = 1}^{m} {{G_{hh}}} = \sum \limits _{h = 1}^{m} {\sum \limits _{b = 1}^{p} {\sum \limits _{a = 1}^{n} {{A_{ha}}{B_{ab}}{C_{bh}}} } }$.

Let D₁ = CA, ${D_{1}} \in {\mathbb {R}^{p \times n}}$. According to the rule of matrix multiplication, ${D_{1ij}} = \sum \limits _{h = 1}^{m} {{C_{ih}}{A_{hj}}}$. Let G₁ = D₁B, ${G_{1}} \in {\mathbb {R}^{p \times p}}$. ${G_{1ij}} = \sum \limits _{a = 1}^{n} {{D_{1ia}}{B_{aj}}} = \sum \limits _{a = 1}^{n} {\sum \limits _{h = 1}^{m} {{C_{ih}}{A_{ha}}{B_{aj}}} }$. From the definition of the trace norm 3.1, $Trace({G_{1}}) = \sum \limits _{b = 1}^{p} {{G_{1bb}}} = \sum \limits _{b = 1}^{p} {\sum \limits _{a = 1}^{n} {\sum \limits _{h = 1}^{m} {{C_{bh}}{A_{ha}}{B_{ab}}} } }$. Thus, Trace(ABC) = Trace(CAB).

Proof A.13

According to lemma 4.15, we can obtain that $Trace(\varphi {({\widehat \chi ^{(m)}})^{T}}\widehat \chi _{vec}^{(m)T}\mu ) = Trace(\widehat \chi _{vec}^{(m)T}\mu \varphi {({\widehat \chi ^{(m)}})^T})$. Let $A = \mu \varphi {({\widehat \chi ^{(m)}})^T}$, $A \in {\mathbb {R}^{{J_1}{J_2} {\cdots } {J_N}}}$. According to the rule of matrix multiplication, ${A_{ij}} = \sum \limits _{p = 1}^1 {{\mu _{ip}}\varphi ({{\widehat \chi }^{(m)}})_{pj}^T}$. Let $D = \widehat \chi _{vec}^{(m)T}A$, ${D_{ij}} = \sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {\widehat \chi _{veciq}^{(m)T}{A_{qj}}} = \sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {\sum \limits _{p = 1}^1 {\widehat \chi _{veciq}^{(m)T}{\mu _{qp}}\varphi ({{\widehat \chi }^{(m)}})_{pj}^T} }$. From the definition of the trace norm 3.1, $Trace(D) = \sum \limits _{k = 1}^1 {{D_{kk}}} = \sum \limits _{k = 1}^1 {\sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {\sum \limits _{p = 1}^1 {\widehat \chi _{veckq}^{(m)T}{\mu _{qp}}\varphi ({{\widehat \chi }^{(m)}})_{pk}^T} } }$. When q = s, k = t, $\frac {{\partial (\sum \limits _{p = 1}^1 {\widehat \chi _{vecst}^{(m)}{\mu _{sp}}\varphi ({{\widehat \chi }^{(m)}})_{pt}^T} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = \sum \limits _{p = 1}^{1} {{\mu _{sp}}\varphi ({{\widehat \chi }^{(m)}})_{pt}^{T}}$. When q≠s, k≠t, $\frac {{\partial (\sum \limits _{p = 1}^{1} {\widehat \chi _{veckq}^{(m)T}{\mu _{qp}}\varphi ({{\widehat \chi }^{(m)}})_{pk}^{T}} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = 0$. Therefore, $\frac {{\partial {f_{1}}}}{{\partial \widehat \chi _{vec}^{(m)}}} = \mu \varphi {({\widehat \chi ^{(m)}})^{T}}$.

Proof A.14

According to lemma 4.15, we can obtain that $Trace({\mu ^{T}}\widehat \chi _{vec}^{(m)}\varphi ({\widehat \chi ^{(m)}})) = Trace(\varphi ({\widehat \chi ^{(m)}}){\mu ^T}\widehat \chi _{vec}^{(m)})$. Let $A = \varphi ({\widehat \chi ^{(m)}}){\mu ^T}$, According to the rule of matrix multiplication, ${A_{ij}} = \sum \limits _{p = 1}^1 {\varphi {{({{\widehat \chi }^{(m)}})}_{ip}}\mu _{pj}^T}$. Let $D = A\widehat \chi _{vec}^{(m)}$, ${D_{ij}} = \sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {{A_{iq}}\widehat \chi _{vecqj}^{(m)}} = \sum \limits _{q = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{N}}} {\sum \limits _{p = 1}^{1} {\varphi {{({{\widehat \chi }^{(m)}})}_{ip}}\mu _{pq}^{T}\widehat \chi _{vecqj}^{(m)}} }$. From the definition of the trace norm 3.1, $Trace(D) = \sum \limits _{k = 1}^{1} {{D_{kk}}} = \sum \limits _{k = 1}^{1} {\sum \limits _{q = 1}^{{J_{1}}{J_{2}} {\cdots } {J_{N}}} {\sum \limits _{p = 1}^{1} {\varphi {{({{\widehat \chi }^{(m)}})}_{kp}}\mu _{pq}^{T}\widehat \chi _{vecqk}^{(m)}} } }$. When q = s, k = t, $\frac {{\partial (\sum \limits _{p = 1}^{1} {\varphi {{({{\widehat \chi }^{(m)}})}_{tp}}\mu _{ps}^{T}\widehat \chi _{vecst}^{(m)}} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = \sum \limits _{p = 1}^1 {{\mu _{sp}}\varphi ({{\widehat \chi }^{(m)}})_{pt}^T}$. When q≠s, k≠t, $\frac {{\partial (\sum \limits _{p = 1}^1 {\varphi {{({{\widehat \chi }^{(m)}})}_{kp}}\mu _{pq}^T\widehat \chi _{vecqk}^{(m)}} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = 0$. Thus, $\frac {{\partial {f_2}}}{{\partial \widehat \chi _{vec}^{(m)}}} = \mu \varphi {({\widehat \chi ^{(m)}})^T}$.

Proof A.15

Based on lemma 4.15, we get that $Trace({\mu ^T}\widehat \chi _{vec}^{(m)}\widehat \chi _{vec}^{(m)T}\mu ) = Trace(\widehat \chi _{vec}^{(m)T}\mu {\mu ^T}\widehat \chi _{vec}^{(m)})$. Let D = μμ^T, $A = \widehat \chi _{vec}^{(m)T}D$. According to the rule of matrix multiplication, ${A_{ij}} = \sum \limits _{p = 1}^1 {\widehat \chi _{vecip}^{(m)T}{D_{pj}}}$. Let $H{\text { = }}A\widehat \chi _{vec}^{(m)}$, ${H_{ij}} = \sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {{A_{iq}}\widehat \chi {{{~}_{vec}^{(m)}}_{qj}}} = \sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {\sum \limits _{p = 1}^1 {\widehat \chi _{vecip}^{(m)T}{D_{pq}}\widehat \chi {{{~}_{vec}^{(m)}}_{qj}}} }$. From the definition of the trace norm 3.1, $Trace(H) = \sum \limits _{k = 1}^1 {{H_{kk}}} = \sum \limits _{k = 1}^1 {\sum \limits _{q = 1}^{{J_1}{J_2} {\cdots } {J_N}} {\sum \limits _{p = 1}^1 {\widehat \chi _{veckp}^{(m)T}{D_{pq}}\widehat \chi {{{~}_{vec}^{(m)}}_{qk}}} } }$. When q = s, k = t, $\frac {{\partial (\sum \limits _{p = 1}^1 {\widehat \chi _{vectp}^{(m)T}{D_{ps}}\widehat \chi {{{~}_{vec}^{(m)}}_{st}}} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = \sum \limits _{p = 1}^1 {D_{sp}^T\widehat \chi _{vecpt}^{(m)}}$. When p = s, k = t, $\frac {{\partial (\sum \limits _{p = 1}^{1} {\widehat \chi _{vecst}^{(m)}{D_{sq}}\widehat \chi {{{~}_{vec}^{(m)}}_{qt}}} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = \sum \limits _{p = 1}^{1} {{D_{sq}}\widehat \chi {{{~}_{vec}^{(m)}}_{qt}}}$. In other cases, $\frac {{\partial (\sum \limits _{p = 1}^{1} {\widehat \chi _{veckp}^{(m)T}{D_{pq}}\widehat \chi {{{~}_{vec}^{(m)}}_{qk}}} )}}{{\partial \widehat \chi _{vecst}^{(m)}}} = 0$. Thus, $\frac {{\partial {f_{3}}}}{{\partial \widehat \chi _{vec}^{(m)}}} = D\widehat \chi _{vec}^{(m)} + {D^{T}}\widehat \chi _{vec}^{(m)}$. Since D = D^T, $\frac {{\partial {f_{3}}}}{{\partial \widehat \chi _{vec}^{(m)}}} = \eta \mu {\mu ^{T}}\widehat \chi _{vec}^{(m)}$. η is a constant, η = 2.