Contract Classification Theorem for Real Symmetric Matrices, Any real symmetric matrix $A$of order $n$contracts with the diagonal matrix $\Lambda=\operatorname{diag}(\lambda_1,\lambda_2,\ldots,\lambda_k)$, Where $\lambda_1,\lambda_2,\ldots,\lambda_k$are different eigenvalues of $A$, $k$is the number of different eigenvalues of $A$, and for each eigenvalue $\lambda_i$, The dimension of its corresponding feature subspace in $A$is equal to the number of times it appears in $\Lambda$.
Therefore, two real symmetric matrices of order $n$, $A$and $B$, have the same eigenvalues and the same dimension of the eigensubspace corresponding to each eigenvalue.
Suppose $A$has $k$distinct eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_k$, and for each eigenvalue $\lambda_i$, The dimension of the corresponding feature subspace in $A$is $m_1,m_2,\ldots,m_k$. Then a matrix $P$of $k\times n$can be constructed, where row $i$is a basis of eigenvectors corresponding to eigenvalue $\lambda_i$, such that $P^TAP=\Lambda$.
Since the column vectors of $P$are eigenvectors of $A$, they are linearly independent. And since $A$is a real symmetric matrix, the basis formed by its eigenvectors can be orthogonalized. So you can choose $P$such that $P^TP=I$, that $P$is orthogonal. Therefore, the sufficient and necessary condition of the $A$and $B$contract can be transformed into the existence of the orthogonal matrix $P$, such that $P^TAP$and $P^TBP$have the same eigenvalues and the same dimension of the eigensubspace.
Thus, the number of contract-classified real symmetric matrices is equal to the number of all possible combinations of the product of the number of different eigenvalues and the dimension of the eigensubspace corresponding to each eigenvalue
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