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# Advanced Algebra Beijing Normal University fourth edition after class exercise

mwb_100 注册会员
2023-02-27 08:40

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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< br / > the characteristic value in the A $$n_i$$$\ lambda_i$correspond to the dimensions of the feature subspace. Note that $n_1+n_2+\cdots+n_k=n$, so the above equation can be written as

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< br / > the $m_0 = 0$, $m_k = n$. So, real symmetry

xiaoyudgx 注册会员
2023-02-27 08:40
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The answer part of the reference to GPT, GPT_Pro better solution
Real symmetric molybdenum matrix of order n is n x n square matrix, and every row and every column is the same data. When the n order real symmetric molybdenum matrix is classified according to the contract, it is to classify all n order real symmetric molybdenum matrix, and belong to the same class of real symmetric molybdenum matrix must meet the contract conditions.

Specifically, for each class of real symmetric molybdenum matrix, the following conditions are required between each element: If elements a[i][j] and a[m][n] are elements of the same class of real symmetric molybdenum matrix, then a[i][j] and a[m][n] satisfy that a[i][j] = a[m][n].

Therefore, to calculate the total number of classes for a real symmetric molybdenum matrix of order n, it is necessary to traverse all possible cases, that is, to traverse all possible real symmetric molybdenum matrices. The specific code is as follows:

int count=0; //用于计数总的类数
for (int i=0; i//遍历n×n方阵中的每一行
for (int j=0; j//遍历每一行中的每一个元素
for (int m=0; m//遍历n×n方阵中的每一列
for (int n=0; n//遍历每一列中的每一个元素
if (i==m && j==n) { //若此处元素a[i][j]与a[m][n]处元素相同，则可跳过本次循环
continue;
}
else { //若此处元素a[i][j]与a[m][n]处元素不同，则做如下判断
if (a[i][j] != a[m][n]) { //若a[i][j]不等于a[m][n]，则该实对称钼阵不属于合同，跳出本层循环，重新开始新的循环
break;
}
else if (m==(n-1)) { //若本行最后一个元素a[m][n]与a[i][j]相等，则该实对称钼阵属于合同，故把总数+1
count++;
}
}
}
}
}
}


It can be concluded that there are count classes when n order real symmetric molybdenum matrices are classified by contract.

yazippx 注册会员

Publish Time
2023-02-27 08:40
Update Time
2023-02-27 08:40