Can anyone tell me how could i proof this equation:
$ \widehat {Ab} = A^{-T} \hat b A^{-1}$
with the condition: $det(A)=1$
Also, How this equation change when $det(A)\neq1$
This is Exercise 2.4 of book: An Invitation to 3-D Vision-From Images to Geometric Models
by S. Shankar Sastry.
Exercise 2.4 (Skew-symmetric matrices). Given any vector $w = [w_1, w_2, w_3] \in R3$, we know that the matrix $\hat w$ is skew-symmetric; i.e. $\hat w^{T} = -\hat w$. Now for any matrix $A \in R^{3\times3}$ with determinant $det(A) = 1$, show that the following equation holds:
$ A^{T} \hat w A = \widehat {A^{-1}w}$
Then, in particular, if $A$ is a rotation matrix, the above equation holds.
Hint: Both $A^{T} \widehat {(.)}A$ and $\widehat {A^{-1}(.)}$ are linear maps with $w$ as the variable. What do you need in order to prove that two linear maps are the same?
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$\begingroup$For fixed $b$, we have \begin{align*} \widehat {Ab} = A^{-T} \hat b A^{-1} &\Leftrightarrow A^T \widehat{Ab} A = \widehat{b}\\ &\Leftrightarrow y^TA^T \widehat{Ab} Ax = y^T\widehat{b}x \ \text{ for all } x,y\\ &\Leftrightarrow y^TA^T (Ab \times Ax) = y^T(b\times x) \ \text{ for all } x,y\\ &\Leftrightarrow (Ay)\cdot (Ab \times Ax) = y\cdot(b\times x) \ \text{ for all } x,y\\ &\Leftrightarrow \det(Ay,Ab,Ax) = \det(y,b,x) \ \text{ for all } x,y\\ &\Leftrightarrow \det\left(A(y,b,x)\right) = \det(y,b,x) \ \text{ for all } x,y\\ &\Leftrightarrow \det(A)\det(y,b,x) = \det(y,b,x) \ \text{ for all } x,y, \end{align*} and the last statement is true because $\det(A)=1$.
In general, if $\det(A)\neq0$, we have $\det(B)=1$ where $B=\det(A)^{-1/3}A$. Therefore we have $\widehat{Bb} = B^{-T}\widehat{b}B^{-1}$. For any $z=(u,v,w)^T$, since $$ \widehat{z}=\pmatrix{ 0 &-w & v\\ w & 0 &-u\\ -v & u & 0}, $$ we have $\widehat{\lambda z}=\lambda\widehat{z}$ for any scalar $\lambda$. Therefore, by pulling out the factor $\det(A)^{-1/3}$ from $B$ (and $\widehat{Bb}$), we get $$\widehat{Ab} = \det(A)A^{-T}\widehat{b}A^{-1}.$$
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