为记号简便, 用A'表示A的转置, 相应α‘表示α的转置.

首先明确α‘α是一个1*1矩阵, 等同于一个数, 且由α是非零实向量, 有α‘α ≠ 0.

1) A' = (E-(2/(α‘α))αα')' = E'-(2/(α‘α))(αα')' = E-(2/(α‘α))αα'.

(只用了转置的基本性质(X+Y)' = X'+Y', (kX)' = kX', (XY)' = Y'X')

即有A' = A为实对称阵, 因此A相似于对角阵, 进而kE-A也相似于对角阵.

2) A² = (E-(2/(α‘α))αα')² = E²-E(2/(α‘α))αα'-(2/(α‘α))αα'E+((2/(α‘α))αα')²

= E-(4/(α‘α))αα'+(2/(α‘α))²(αα')² = E-(4/(α‘α))αα'+(2/(α‘α))²αα'αα'

= E-(4/(α‘α))αα'+(4/(α‘α)²)α(α'α)α' = E-(4/(α‘α))αα'+(4/(α‘α))αα' = E.

由此可知A的特征值满足x² = 1, 只可能为±1.

kE-A的特征值只可能为k±1, 而k ≠ ±1, 故kE-A没有零特征值.

|kE-A| = kE-A的特征值之积 ≠ 0, 因此kE-A可逆.

3) 设B = E-2αα', 则B' = E'-2(αα')' = E-2αα' = B.

B为正交矩阵当且仅当B'B = E, 即E = B² = (E-2αα')² = E-4αα'+αα'αα'

= E-4αα'+α(α'α)α' = E+(α'α-4)αα', 也即(α'α-4)αα' = 0.

由α非零, 有αα' ≠ 0, 因此B为正交矩阵当且仅当α'α = 4.