If $(I_2 -2Q^T)^{-1}= \left[\begin{smallmatrix}2&1\\1&1\end{smallmatrix}\right]$, find Q

Question

If $(I_2 -2Q^T)^{-1}= \left[\begin{array}{cc}2&1\\1&1\end{array}\right]$, find Q

Cannot figure out how to tackle this problem. Any help is much appreciated.

Answer 1

It's merely a matter of "walking back" or "unravelling" the prescribed operations:

Since

$(I - 2Q^T)^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}, \tag 1$

we have

$I - 2Q^T = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}; \tag 2$

thus

$2Q^T = I - \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 1 & -1 \end{bmatrix}, \tag{3}$

so

$Q^T = \begin{bmatrix} 0 & 1/2 \\ 1/2 & -1/2 \end{bmatrix}; \tag{4}$

finally,

$Q = \begin{bmatrix} 0 & 1/2 \\ 1/2 & -1/2 \end{bmatrix} \tag{5}$

as well.

The problem may of course be viewed in abstracto as

$(I - 2Q^T)^{-1} = A, \tag{6}$

whence

$I - 2Q^T = A^{-1}, \tag 7$

$2Q^T = I - A^{-1}; \tag 8$

$Q^T = \dfrac{1}{2}(I - A^{-1}), \tag 9$

$Q = (\dfrac{1}{2}(I - A^{-1}))^T. \tag{10}$

Answer 2

Let $A:=(I-2Q^T)^{-1}$ then $$A(I-2Q^T)=I\Rightarrow Q^T=A^{-1}(A-I)/2$$ $A$ is given, finds its inverse then these should give you $Q^T$. Lastly take transpose of $Q^T$.