Clarification of Sherman-Morrison formula for: $(A + \alpha bc^T)^{-1}$

Question

Sherman-Morrison formula (e.g., cf. in the matrix cookbook, sec. 3.2.4)

$$(A + bc^T)^{-1} = A^{-1} - \frac{A^{-1} bc^T A^{-1}}{1 + c^T A^{-1} b} \ .$$

Question/Clarification: If $$(A + \alpha bc^T)^{-1} = A^{-1} - \alpha \frac{A^{-1} bc^T A^{-1}}{1 + \alpha^2 c^T A^{-1} b}$$

Is it true? If yes, then how? If incorrect, please suggest the correct one.

Answer 1

If $\alpha \in \mathbb{R}$, then

view the problem as

$$(A+(\alpha b)c^T)^{-1}$$

that is whenever we see a $b$, we replace it by $\alpha b$, then we obtain

$$(A + (\alpha b)c^T)^{-1} = A^{-1} - \frac{A^{-1} (\alpha b)c^T A^{-1}}{1 + \alpha c^T A^{-1} b} =A^{-1} - \alpha \frac{A^{-1} bc^T A^{-1}}{1 + \alpha \cdot c^T A^{-1} b}$$

where all the inverse terms are assumed to exist.

Edit:

A numerical example:

octave:3> A = eye(2)
A =

Diagonal Matrix

   1   0
   0   1

octave:4> b = [1 ; 0]
b =

   1
   0

octave:5> c = b
c =

   1
   0

octave:6> alpha = 2
alpha =  2
octave:7> inv(A+alpha*b*c')-inv(A)+alpha * inv(A)*b*c' * inv(A)/(1+alpha * c' * inv(A)*b)
ans =

  -1.1102e-16   0.0000e+00
   0.0000e+00   0.0000e+00

octave:8> inv(A+alpha*b*c')-inv(A)+alpha * inv(A)*b*c' * inv(A)/(1+alpha*alpha * c' * inv(A)*b)
ans =

  -0.26667   0.00000
   0.00000   0.00000

Answer 2

You can use the Alternative verification. Moreover, consider $b$ and $c$ as $1\times 1$ matrices (i.e. single numbers). Then: $$(A + bc^T)^{-1} = A^{-1} - \frac{A^{-1} bc^T A^{-1}}{1 + c^T A^{-1} b} \ \Rightarrow (1+bc)^{-1}=1-\frac{bc}{1+cb};$$ Similarly: $$(A + \alpha bc^T)^{-1} = A^{-1} - \alpha \frac{A^{-1} bc^T A^{-1}}{1 + \color{red}{\alpha} c^T A^{-1} b} \Rightarrow \\ (1 + \alpha bc)^{-1} = 1 - \alpha \frac{bc}{1 + \alpha cb}$$ Or you can denote: $\alpha b=d$, where $\alpha$ is a scalar, $b,d$ are the vectors.