# Bayesian Matrix Learning by Principle Eigenvector for Completing Missing Medical Data
## 3 The Eigenvector of the DTBM  
Since \(B_{ij}^{q}\) indicates the conditional probability value, each element in the DTBM is between [0, 1]. And when a certain condition is given, the sum of its conditional probabilities is equal to 1. Therefore, the sum of each column in the DTBM is 1, which means that the DTBM is a Markov matrix. According to the characteristics of the Markov matrix, the DTBM has a principle eigenvector whose eigenvalue is equal to 1. So, what is this eigenvector?  
The \"0\" in the matrix represents that there is no correlation between the occurrence of independent events set \(X\)\(Y\), which is consistent with [PERSON] theorem's assumption. At this time, the probability of the presence of \(X\) under events \(Y\) is expressed as \(P(X|Y)\), and the likelihood of the occurrence of \(Y\) under events set \(X\) is shown as \(P(Y|X)\).  
According to [PERSON] theorem and the formula of total probability, we get \(p(T)\), \(p(S)\).  
\[p(T|S)p(S) =\left(\begin{array}{cccc}t_{1}|s_{1}&t_{1}|s_{2}&\ldots&t_{1}|s _{m}\\ t_{2}|s_{1}&t_{2}|s_{2}&\ldots&t_{2}|s_{m}\\ \vdots&\vdots&\vdots&\vdots\\ t_{n}|s_{1}&t_{n}|s_{2}&\ldots&t_{n}|s_{m}\end{array}\right)\left(\begin{array} []{c}s_{1}\\ s_{2}\\ \vdots\\ s_{m}\end{array}\right)\] \[=\left(\begin{array}{cccc}\sum\limits_{j=1}^{m}t_{1}|s_{m}\\ \sum\limits_{j=1}^{m}t_{2}|s_{m}\\ \vdots\\ \sum\limits_{j=1}^{m}t_{n}|s_{m}\end{array}\right)=\left(\begin{array}{c}t_{ 1}\\ t_{2}\\ t_{n}\end{array}\right)=p(T)\] \[p(S|T)p(T) =\left(\begin{array}{cccc}s_{1}|t_{1}&s_{1}|t_{2}&\ldots&s_{1}|t _{n}\\ s_{2}|t_{1}&s_{2}|t_{2}&\ldots&s_{2}|t_{n}\\ \vdots&\vdots&\vdots&\vdots\\ s_{m}|t_{1}&s_{n}|t_{2}&\ldots&s_{n}|t_{n}\end{array}\right)\left(\begin{array} []{c}t_{1}\\ t_{2}\\ \vdots\\ t_{n}\end{array}\right)\] \[=\left(\begin{array}{c}\sum\limits_{i=1}^{n}s_{1}|t_{i}\\ \sum\limits_{i=1}^{n}s_{2}|t_{i}\\ \vdots\\ \sum\limits_{i=1}^{n}s_{m}|t_{i}\end{array}\right)=\left(\begin{array}{c}s _{1}\\ s_{2}\\ \vdots\\ s_{m}\end{array}\right)=p(S)\]  
Since \(P(X)\), \(P(Y)\), and \(p(Y|X)\) are used to get \(p(X|Y)\), \(p(X|Y)\) can be expressed by the eigenvalue equation:  
\[\left(\begin{array}{cc}0&p(X|Y)\\ p(Y|X)&0\end{array}\right)\left(\begin{array}{c}p(X)\\ p(Y)\end{array}\right)=\left(\begin{array}{c}p(X)\\ p(Y)\end{array}\right)\] \[\left(\begin{array}{cc}0&P(t|s)\\ P(s|t)&0\end{array}\right)\left(\begin{array}{c}P(t)\\ P(s)\end{array}\right)=\left(\begin{array}{c}\sum\limits_{j=1}^{m}P(t_{1} \Big{|}s_{j})\\ \ldots\\ \sum\limits_{j=1}^{m}P(t_{m}\Big{|}s_{j})\\ \sum\limits_{i=1}^{m}P(s_{1}|t_{i})\\ \ldots\\ \sum\limits_{i=1}^{m}P(s_{n}|t_{i})\end{array}\right)=\left(\begin{array}{c} P(t)\\ P(s)\end{array}\right) \tag{2}\]  
It is not difficult to see that \(\left(\begin{array}{c}P(t)\\ P(s)\end{array}\right)\) is the eigenvector of the DTBM when the eigenvalue equals 1. The eigenvector with an eigenvalue of 1 expresses a stable state of the DTBM.  
After decades of medication research and practice, diabetes complications can be treated with a certain drug, and the proportion of different drugs used for certain complications is stable. Meanwhile, in a relatively stable environment, the proportion of people with various diabetes symptoms also tends to stabilize. This eigenvector represents this stability, where \(p(t)\) and \(p(s)\) can be obtained through data statistics. Therefore, we need to find the missing column vectors in the DTBM based on its eigenvector.