Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 6A (Vectors and Matrices)

The vector triple product identity gives $n\times (x\times n)=x-(n\cdot x)n$. Decompose $x=x_{\perp }+x_{\parallel }$, where $x_{\parallel }=(n\cdot x)n$ and $x_{\perp }\cdot n=0$. ThenThe linear map acts as the identity on $n^{\perp }$ and as multiplication by $\alpha$ along $n$. Consequently it is invertible exactly when $\alpha \ne 0$, withApplying $\Phi$ to this expression recovers the perpendicular and parallel components of $y$ separately.

At $\alpha =0$, $\Phi$ is the orthogonal projection onto the plane perpendicular to $n$. Its image of a linear map and kernel of a linear map areEvery perpendicular vector is unchanged and every parallel vector is annihilated. The image plane is perpendicular to the kernel axis by the definition of $n^{\perp }$; their dimensions two and one also agree with the rank-nullity theorem.

Expanding the formula from part (a) in components gives $y_i=x_i+(\alpha -1)n_i{n}_j{x}_j$, with repeated indices summed. Hence the matrix of a linear map and its inverse, when $\alpha \ne 0$, have entriesHere ${\delta }_{ij}$ is the Kronecker delta. To check the matrix inverse, write $P=n{n}^T$; $P^2=P$ because $|n|=1$. Thus $A=(I-P)+\alpha P$ and $B=(I-P)+{\alpha }^{-1}P$, whose product is the identity matrix.

For the specified unit vector,For a cyclic matrix with first row $(a,b,c)$, its determinant is $a^3+b^3+c^3-3abc$. Here it givesTherefore $Ax=Cx$ has only the zero solution if $\alpha \ne 1$. At $\alpha =1$, $A=I$ and $A-C$ has kernel consisting of the equal-coordinate vectors. The complete answer isGeometrically, $C^{T}C=I$, $\det C=1$ and $Cn=n$, so $C$ is a rotation matrix about the $n$ axis. Its matrix trace is two, giving $\cos \theta =(\mathrm{tr}C-1)/2=1/2$; its skew part selects angle $-\pi /3$ about the oriented axis $n$. It has no fixed nonzero perpendicular vector. Meanwhile $A$ preserves perpendicular components and scales the axial component by $\alpha$. Comparing these components first forces $x_{\perp }=0$, and then $(\alpha -1)x_{\parallel }=0$, exactly matching the determinant calculation, including the noninvertible case $\alpha =0$.