设A=（aij)为n阶方阵,且aii>0,aij
1个回答
首先,易见X = (1,1,...,1)'是AX = 0的一组非零解,故r(A) ≤ n-1.
设B是A的左上n-1阶子矩阵,下面证明B可逆,则r(A) ≥ r(B) = n-1,就能完成证明.
实际上,B是所谓严格对角占优阵,满足|a[i,i]| = ∑{1 ≤ j ≤ n,j ≠ i} |a[i,j]| > ∑{1 ≤ j ≤ n-1,j ≠ i} |a[i,j]|.
严格对角占优阵总是可逆的:假设BX = 0有非零解X = (x[1],x[2],...,x[n-1])'.
设|x[k]| = max{|x[1]|,|x[2]|,...,|x[n-1]|} > 0,则由∑{1 ≤ j ≤ n-1} a[k,j]x[j] = 0有:
|a[k,k]|·|x[k]| = |∑{1 ≤ j ≤ n-1,j ≠ i} a[k,j]x[j]|
≤ ∑{1 ≤ j ≤ n-1,j ≠ i} |a[k,j]|·|x[j]|
≤ ∑{1 ≤ j ≤ n-1,j ≠ i} |a[k,j]|·|x[k]|
= |x[k]|·∑{1 ≤ j ≤ n-1,j ≠ i} |a[k,j]|
< |x[k]|·|a[k,k]|,矛盾.
因此BX = 0只有零解,B可逆.
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