Find 3×3 orthogonal matrix satisfying the following conditions.

Question

Let $A$ be 3×3 orthogonal real matrix.

For $\mathbf{u}$ $= \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}, \mathbf{v}$ $= \begin{pmatrix} 0 \\ 0 \\ \sqrt3 \end{pmatrix}, \mathbf{w}$ $= \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$,

$Au=v, Aw=-w$ and $detA=-1$

Find $A$.

I noted that $u, v, w$ are linearly independent. And I write $A\begin{pmatrix} 1 & 0 & 1 \\ 1 & 0 & -1 \\ 1 & \sqrt3 & 0 \end{pmatrix}=\begin{pmatrix} 0 & a & -1 \\ 0 & b & 1 \\ \sqrt3 & c & 0 \end{pmatrix}.$

Then $A=\begin{pmatrix} 0 & a & -1 \\ 0 & b & 1 \\ \sqrt3 & c & 0 \end{pmatrix}\begin{pmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2\sqrt3} & -\frac{1}{2\sqrt3} & \frac{1}{\sqrt3} \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{pmatrix}$

Now, how can I find $a, b, c$?? Am I doing this right?

If there is a better solution, please help me.

Answer 1

Your approach is fine; you can proceed by using the fact that $\det(A)=-1$ and $A$ is orthogonal. From $$-1=\det(A)=\det\begin{pmatrix} 0 & a & -1 \\ 0 & b & 1 \\ \sqrt3 & c & 0 \end{pmatrix}\det\begin{pmatrix} \frac{1}{2} & \frac{1}{2} & 0 \\ -\frac{1}{2\sqrt3} & -\frac{1}{2\sqrt3} & \frac{1}{\sqrt3} \\ \frac{1}{2} & -\frac{1}{2} & 0 \end{pmatrix},$$ you find that $a+b=-2$, so you can substitute $b=-(a+2)$. Now multiply the two matrices and choose $a$ and $c$ so that the matrix is orthogonal.

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Alternatively, the fact that $Au=v$, i.e. $$A\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}0\\0\\\sqrt{3}\end{pmatrix},$$ shows that the first two rows of $A$ are orthogonal to $u$. Because $A$ is an orthogonal matrix, it follows that the third row is orthogonal to the first two, so the third row is a scalar multiple of $u$. Its inner product with $u$ equals $\sqrt{3}$, so the third row must equal $\frac{u}{\sqrt{3}}$.

Can you proceed in this way with $Aw=-w$ to determine $A$?

Answer 2

Hint

As $A$ is an orthogonal matrix it preserves the inner product. Therefore $\Vert Av\Vert = \sqrt 3$. Also $\langle Av, Aw\rangle =\langle v, w\rangle=0$.

Finally use the equation $\det A =-1$.

This provides 3 equations to find $a,b,c$.

Answer 3

Trying to show a more geometric way, which also helps us to understand what is going on with this map:

The map $f\colon \mathbb R^3\to \mathbb R^3, x\mapsto Ax$ is a linear isometry of $\mathbb R^3$. So it preserves angles. Therefore, as $[w]$ is an invariant subspace under $f$, we know that $[w]^\perp$ is also invariant. Restricted to $[w]^\perp$, $f|_{[w]^\perp}$ is a linear isometry with determinant $1$. But there are not many options for this: you always get a rotation in the plane by some angle $\omega \in [0,\pi]$.

Therefore, you know that by changing the basis to another orthonormal basis, you obtain a transformation matrix $$B=\begin{pmatrix}-1&&\\&\cos \omega & -\sin\omega\\&\sin\omega&\cos \omega\end{pmatrix}$$ by the determinant. We may calculate the angle of rotation $\omega$ in the plane $[w]^\perp$ by projecting $u$ and $v$ to $[w]^\perp$ (*). Now you have an orthonormal basis by normalizing $w$ and choosing an orthonormal basis of $[w]^\perp$. Relative to this basis, $f$ has transformation matrix $B$, so now you can obtain $A$ by transformating back to the standard basis.

(*) In this case, as commented by mathcounterexamples, $u$ and $v$ already lie in $[w]^\perp$, so we can just calculate the angle between them. In general, however, the angle between $u$ and $Au$ is not the angle of rotation.