怎么证明矩阵特征值的和等于矩阵的迹_

2025-04-06 23:36:33
推荐回答(3个)
回答1:

矩阵的特征多项式xE-A,把行列式展开,是一个n次多项式,由根系关系可得;特征值的和就等于多项式得根得和,是第n-1次项的系数,是a11+a22+`````+ann。总之,把那个行列式展开,比较系数即可。

设A是n阶方阵,如果数λ和n维非零列向量x使关系式Ax=λx成立,那么这样的数λ称为矩阵A特征值,非零向量x称为A的对应于特征值λ的特征向量。式Ax=λx也可写成( A-λE)X=0。这是n个未知数n个方程的齐次线性方程组,它有非零解的充分必要条件是系数行列式| A-λE|=0。

扩展资料:

矩阵的迹性质

(1)设有N阶矩阵A,那么矩阵A的迹(用tr(A)表示)就等于A的特征值的总和,也即矩阵A的主对角线元素的总和。

1、迹是所有对角元素的和

2、迹是所有特征值的和

3、某些时候也利用tr(AB)=tr(BA)来求迹

4、tr(mA+nB)=m tr(A)+n tr(B)

(2)奇异值分解(Singular value decomposition )

奇异值分解非常有用,对于矩阵A(p*q),存在U(p*p),V(q*q),B(p*q)(由对角阵与增广行或列组成),满足A = U*B*V

U和V中分别是A的奇异向量,而B是A的奇异值。AA'的特征向量组成U,特征值组成B'B,A'A的特征向量组成V,特征值(与AA'相同)组成BB'。因此,奇异值分解和特征值问题紧密联系。如果A是复矩阵,B中的奇异值仍然是实数。

回答2:

相似矩阵的对角线元素的和相等,
以特征值为对角线元素的矩阵与原矩阵相似,所以矩阵特征值的和等于矩阵的迹 。

回答3:

看图

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