# a linear transformation matrix

Let $e=(a,b,c)$ be a unit vector in $\mathbb{R}^3$ and let $T$ be the linear transformation on $\mathbb{R}^3$ of rotation by $180^\circ$ about $e$. Find the matrix for $T$ with respect to the standard basis $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$.

The rotation matrix in $\mathbb{R}^3$ by $180^\circ$ is : \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{bmatrix} So rotating $e$ by $180^\circ$ gives : \begin{bmatrix} -a \\ -b \\ c \end{bmatrix} After that how to get the transformation matrix w.r.t the standard basis? P.S.- The answer given involves terms consisting of $a,b,c$.

• Why are you rotating $e$? That’s the axis of the rotation, so it would remain fixed. – amd May 4 '17 at 19:42

The matrix that you gave is for a 180° rotation about $e_3$. A general way to proceed from there would be via a change-of-basis operation, which involves finding an appropriate basis and performing a couple of matrix multiplications and perhaps an inversion, to boot.
Observe, however, that a 180° rotation about a vector $e$ is equivalent to a reflection in the span of $e$, so you can save yourself quite a bit of work by using the formula for reflection of a vector $v$ in a subspace $W$. This is $Rv=2\mathbf\pi_Wv-v$, where $\mathbf\pi_Wv$ is the projection of $v$ onto $W$. In this problem $W$ is the span of $e$, so for this transformation we have $$Tv=2{ee^T\over e^Te}v-v$$ and we know that $\|e\|=\sqrt{e^Te}=1$, so the matrix that you’re looking for is $$2ee^T-I_3.$$ You can easily verify for yourself that $Te=e$ and that if $v$ is orthogonal to $e$, then $Tv=-v$, which is exactly what we want for this rotation.
• Ok. I found that the transformation you mentioned about that reflection $Tv=2\frac{ee^T}{e^Te}v-v$, it's a negative of the Householder transformation. But why that sign is taken negative here? – am_11235... May 5 '17 at 5:31
• @am_11235... You ought to be able to work that out for yourself. Reflection of a vector in a space reverses the orthogonal rejection of the vector from that space, so the reflection of $v$ in $W$ is $\mathbf\pi_Wv-(v-\mathbf\pi_Wv)$. – amd May 5 '17 at 6:00
HINT: The matrix you give, let's call it $M$, is rotation by $180^{\circ}$ about the vector $e_3$. If you can find an orthogonal matrix $T$ that maps $e$ to $e_3$, then $T^{-1}MT$ is the rotation by $180^{\circ}$ about the vector $e$.
• $T$ need not be orthogonal: it’s enough for the images of $e_1$ and $e_2$ to span $e^\perp$. On the other hand, if $T$ is orthogonal, then $T^{-1}=T^T$ and the inversion can be avoided. – amd May 4 '17 at 20:06
• @amd You're absolutely right. I included orthogonality because it's always possible to take $T$ orthogonal, and it simplifies the computations significantly. – Servaes May 6 '17 at 10:01