Question

谁帮我解一道线性代数题

Answer 1

这个题目有挑战性, 要有耐心, 不明白的地方就多写出一二行看看结果

A=
0 1 0 0 ... 0 0 0
n 0 2 0 ... 0 0 0
0 n-1 0 3 ... 0 0 0
0 0 n-2 0 ... 0 0 0
... ... ...
0 0 0 0 ... 0 n-1 0
0 0 0 0 ... 2 0 n
0 0 0 0 ... 0 1 0

求x满足 Ax = nx.

解: (A-nE)x = 0.
A-nE =
-n 1 0 0 ... 0 0 0
n -n 2 0 ... 0 0 0
0 n-1 -n 3 ... 0 0 0
0 0 n-2 -n... 0 0 0
... ... ...
0 0 0 0 ...-n n-1 0
0 0 0 0 ... 2 -n n
0 0 0 0 ... 0 1 -n

r(n+1) + r1+r2+...+rn
-n 1 0 0 ... 0 0 0
n -n 2 0 ... 0 0 0
0 n-1 -n 3 ... 0 0 0
0 0 n-2 -n... 0 0 0
... ... ...
0 0 0 0 ...-n n-1 0
0 0 0 0 ... 2 -n n
0 0 0 0 ... 0 0 0

rn + r1+r2+...+r(n-1)
-n 1 0 0 ... 0 0 0
n -n 2 0 ... 0 0 0
0 n-1 -n 3 ... 0 0 0
0 0 n-2 -n... 0 0 0
... ... ...
0 0 0 0 ...-n n-1 0
0 0 0 0 ... 0 -1 n
0 0 0 0 ... 0 0 0

同样继续, 最后得
-n 1 0 0 ... 0 0 0
0 1-n 2 0 ... 0 0 0
0 0 2-n 3 ... 0 0 0
0 0 0 3-n... 0 0 0
... ... ...
0 0 0 0 ...-2 n-1 0
0 0 0 0 ... 0 -1 n
0 0 0 0 ... 0 0 0

r(n-1)+(n-1)rn, rn*(-1)
-n 1 0 0 ... 0 0 0
0 1-n 2 0 ... 0 0 0
0 0 2-n 3 ... 0 0 0
0 0 0 3-n... 0 0 0
... ... ...
0 0 0 0 ...-2 0 2C(n,2)
0 0 0 0 ... 0 1 -C(n,1)
0 0 0 0 ... 0 0 0

依次类推得
1 0 0 0 ... 0 0 -C(n,n)
0 1 0 0 ... 0 0 -C(n,n-1)
0 0 1 0 ... 0 0 -C(n,n-2)
0 0 0 1 ... 0 0 -C(n,n-3)
... ... ...
0 0 0 0 ... 1 0 -C(n,2)
0 0 0 0 ... 0 1 -C(n,1)
0 0 0 0 ... 0 0 0

所以(A-nE)x = 0的基础解系为
(C(n,n),C(n,n-1),C(n,n-2),...,C(n,1), 1)^T

Answer 2

我们假定这样的 x存在 ，Ax = nx， 令 x=(x0,x1,x2,...,xn)'为 nx1向量
则很显然，
A的第一行 和 x相乘得到
x2 = nx1 ---- 1)
第二行和x相乘得到
nx1 +2x2 = nx2 => nx1 = (n-2)x2 --2)
由 1),2)得到x2 = (n-2)x2,所以 x2=0, x1 = 0
由矩阵第三行得到 (n-1)x1 + 3x3 = nx3 所以 x3 =0
这样一层层推导下来，其实最终恐怕还是0