以下是答案:

http://www.wolframalpha.com/input/?i=eigenvalues+[[2,+0,+0,+0],+[1,+2,+0,+0],+[0,+1,+2,+0],+[0,+0,+1,+2]]

(我无法建立链接...)

(I cannot make it to a link...)

这样想:

      2    0    0    0
B  =  1    2    0    0
      0    1    2    0
      0    0    1    2.

如果我们从主对角线中减去特征值\ lambda = 2(就像计算特征空间一样)，则得出

If we subtract the eigenvalue \lambda = 2 from the main diagonal (as one does computing eigenspaces), we obtain

              0    0    0    0
(B - 2 I)  =  1    0    0    0
              0    1    0    0
              0    0    1    0.

如果坐标为(x，y，z，w)，那么显然(B-2 I)X = 0得出x = 0(第二行)，y = 0(第三行)，并且z = 0(从最后一行开始).因此，空间由所有点(0，0，0，w)组成，其中w是任意的.也就是说，它是一维的，任何矢量(0、0、0，t)都将用作基本矢量(t非零).

If the coordinates are (x, y, z, w), then, obviously (B - 2 I) X = 0 yields x = 0 (from the second row), y = 0 (from the third row), and z = 0 (from the last row). Hence the space consists of all points (0, 0, 0, w) where w is arbitrary. That is, it is one-dimensional and any vector (0, 0, 0, t) will serve as a basis vector (t nonzero).