# Thread: Verify this proof using determinants

1. ## Verify this proof using determinants

Hello,

Please tell me what you think of the following;

v = [v1 , v2, v3]

Skew(v) = $\left[ \begin{array}{ccc} 0 & -v3 & v2 \\ v3 & 0 & -v1 \\ -v2 & v1 & 0 \end{array} \right]$

$R\epsilon SO(3)$; $v \epsilon \Re^3$

Prove that $(R)Skew(v)R^T=Skew(Rv)$

$det[(R) Skew(v) (R^T)] = det[Skew(Rv)]$

$det(R) det[Skew(v)] det(R^T) = det[Skew(Rv)]$

Since by definition $det[SO(3)] = 1$;
Since by definition of square matrices $det(R^T) = det(R)$

$det[Skew(v)] = det [Skew(Rv)]$

Since by definition the determinant of a Skew matrix is 0 for nxn matrices where n is odd

$0 = 0$

Proofs with 0=0 always make me a little nervous...

2. Originally Posted by tarnishd
Hello,

Please tell me what you think of the following;

v = [v1 , v2, v3]

Skew(v) = $\left[ \begin{array}{ccc} 0 & -v3 & v2 \\ v3 & 0 & -v1 \\ -v2 & v1 & 0 \end{array} \right]$

$R\epsilon SO(3)$; $v \epsilon \Re^3$

Prove that $(R)Skew(v)R^T=Skew(Rv)$

$det[(R) Skew(v) (R^T)] = det[Skew(Rv)]$

$det(R) det[Skew(v)] det(R^T) = det[Skew(Rv)]$

Since by definition $det[SO(3)] = 1$;
Since by definition of square matrices $det(R^T) = det(R)$

$det[Skew(v)] = det [Skew(Rv)]$

Since by definition the determinant of a Skew matrix is 0 for nxn matrices where n is odd

$0 = 0$

Proofs with 0=0 always make me a little nervous...
Presumably you're hoping that from the correct conclusion 0 = 0 you can work back to the statement $R\,\text{Skew}(v)R^{\textsc t}=\text{Skew}(Rv)$. But this fails at the final step. You can work back all the way to $\det[R\,\text{Skew}(v)R^{\textsc t}] = \det[\text{Skew}(Rv)]$ (which is correct because both sides are zero). But just because two matrices have the same determinant it doesn't follow that the two matrices are equal. So you cannot conclude that $R\,\text{Skew}(v)R^{\textsc t}=\text{Skew}(Rv)$.

To prove this relation, notice that for a vector $x = (x_1,x_2,x_3)^{\textsc t}$, Skew(v)x is just the cross product of v with x:

$\text{Skew}(v)x = \begin{bmatrix}0&-v_3&v_2\\ v_3&0&-v_1\\ -v_2&v_1&0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3\end{bmatrix} = \begin{bmatrix}-v_3x_2 + v_2x_3\\ v_3x_1 -v_1x_3\\ -v_2x_1 + v_1x_2\end{bmatrix} = v\times x$.

Therefore $(R\,\text{Skew}(v)R^{\textsc t})x = R\bigl(\text{Skew}(v)(R^{\textsc t}x)\bigr) = R(v\times R^{\textsc t}x)$. But the image of a cross product under a rotation is just what you get by taking the cross product of the rotated vectors. So $R(v\times R^{\textsc t}x) = Rv\times RR^{\textsc t}x = Rv\times x = \text{Skew}(Rv)x$, and it follows that $R\,\text{Skew}(v)R^{\textsc t}=\text{Skew}(Rv)$.