1
$\begingroup$

Let $a = (a_1, a_2, a_3)$ and $b = (b_1, b_2, b_3)$ be two orthonormal vectors in $\mathbb{R}^3$ and $a \times b \in \mathbb{R}^3$ their cross product.

I would like to prove by direct calculation that the matrix $A = \begin{bmatrix} a & b & a \times b \end{bmatrix}$ having the vectors $a, b$ and $a \times b$ as its columns has determinant $1$.

We could for example observe that $A$ represents a transformation from an orthonormal basis $\{e_1, e_2, e_3\}$ to an orthonormal basis $\{a, b, a \times b\}$ which could be achieved as a composition of two rotations (one rotating $e_3$ to $a \times b$, and another around the axis $a \times b$ aligning $\{e_1, e_2\}$ with $\{a, b\}$), and hence has determinant $1$. I'm not interested in such approaches.


My attempt:

We have $a\times b = \begin{pmatrix} a_2b_3 - a_3b_2 \\ a_3b_1 - a_1b_3 \\ a_2b_2 - a_2b_1\end{pmatrix}$.

Thus: \begin{align} \det A &= \begin{vmatrix} a_1 & b_1 & a_2b_3 - a_3b_2 \\ a_2 & b_2 & a_3b_1 - a_1b_3 \\ a_3 & b_3 & a_1b_2 - a_2b_1 \\ \end{vmatrix} \\ &= \text{Laplace expansion along the first column}\\ &= a_1 \begin{vmatrix} b_2 & a_3b_1 - a_1b_3 \\ b_3 & a_1b_2 - a_2b_1 \\ \end{vmatrix} - a_2 \begin{vmatrix} b_1 & a_2b_3 - a_3b_2 \\ b_3 & a_1b_2 - a_2b_1 \\ \end{vmatrix} + a_3 \begin{vmatrix} b_1 & a_2b_3 - a_3b_2 \\ b_2 & a_3b_1 - a_1b_3 \\ \end{vmatrix} \\ &= a_1 \big(a_1{b_2}^2 - a_2b_1b_2 - a_3b_1b_3 + a_1{b_3}^2\big) - a_2\big(a_1b_1b_2 - a_2{b_1}^2 - a_2{b_3}^2 + a_3b_2b_3\big) + a_3\big(a_3{b_1}^2 - a_1b_1b_3 - a_2b_2b_3 + a_3{b_2}^2\big) \\ &= \color{red}{{a_1}^2{b_2}^2} - \color{green}{a_1a_2b_1b_2} - a_1a_3b_1b_3 + \color{red}{{a_1}^2{b_3}^2} - \color{green}{a_1a_2b_1b_2} + \color{blue}{{a_2}^2{b_1}^2} + \color{blue}{{a_2}^2{b_3}^2} - a_2a_3b_2b_3 + {a_3}^2{b_1}^2 - a_1a_3b_1b_3 - a_2a_3b_2b_3 + {a_3}^2{b_2}^2 \\ &= \color{red}{{a_1}^2({b_2}^2 + {b_3}^2)} - \color{green}{2a_1a_2b_1b_2} - 2a_1a_3b_1b_3 + \color{blue}{{a_2}^2({b_1}^2+{b_3}^2)} - 2a_2a_3b_2b_3 + {a_3}^2({b_1}^2 + {b_2}^2) \\ \end{align}

Now we could use ${b_1}^2 + {b_2}^2 + {b_3}^2 = 1$ but nothing seems to simplify from this point onward. We also have to use orthogonality at some point. How should we proceed?

$\endgroup$
  • 3
    $\begingroup$ a typo in the first line of your calculation, note the first entry of your $a\times b$ is $a_2b_3-a_3b_2$ $\endgroup$ – Fan Oct 15 '17 at 18:38
  • $\begingroup$ @Fan Thanks. It's really straightforward now. $\endgroup$ – mechanodroid Oct 15 '17 at 19:13
1
$\begingroup$

$$a_1^2(b^2_2+b_3^2) = a_1^2(1-b_1^2) = a_1^2 - a_1^2b_1^2$$

Repeat with the other terms and an $a_1^2+a_2^2+a_3^2 = 1$ will drop out.

Now you have a $\mathbf a \cdot \mathbf b$ appearing.

$\endgroup$
0
$\begingroup$

After fixing the typo in the calculation noticed by @Fan, the statement easily follows.

Using ${b_1}^2 + {b_2}^2 + {b_3}^2 = 1$ we obtain:

\begin{align}\det A &= \color{red}{{a_1}^2({b_2}^2 + {b_3}^2)} - \color{green}{2a_1a_2b_1b_2} - 2a_1a_3b_1b_3 + \color{blue}{{a_2}^2({b_1}^2+{b_3}^2)} - 2a_2a_3b_2b_3 + {a_3}^2({b_1}^2 + {b_2}^2) \\ &= {a_1}^2(1 - {b_1}^2) + {a_2}^2(1 - {b_2}^2) + {a_2}^2(1 - {b_2}^2) - 2a_1a_3b_1b_3 - 2a_2a_3b_2b_3 - 2a_1a_2b_1b_2\\ &= {a_1}^2 + {a_2}^2 + {a_3}^2 + \big({a_1}^2{b_1}^2 + {a_2}^2{b_2}^2 + {a_3}^2{b_3}^2 -2a_1a_3b_1b_3 - 2a_2a_3b_2b_3 - 2a_1a_2b_1b_2\big)\\ &= \|a\|^2 + \langle a, b\rangle^2\\ &= 1 \end{align}

$\endgroup$
  • $\begingroup$ You may also want to try expanding along the last column. It might save more time. $\endgroup$ – Fan Oct 15 '17 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.