(a) Find a linear map T : R² → R² which rotates each vector of R² by an angle θ. Also, prove that for θ = (π)/2, T has no…

Part (a). A rotation through θ sends e₁=(1,0) to (cosθ,sinθ) and e₂=(0,1) to (−sinθ,cosθ). Hence T(x,y)=(x cosθ−y sinθ, x sinθ+y cosθ), so its matrix is [[cosθ,−sinθ],[sinθ,cosθ]]. To verify linearity, for vectors (x₁,y₁),(x₂,y₂) and scalar λ, T((x₁,y₁)+(x₂,y₂))=T(x₁+x₂,y₁+y₂)=T(x₁,y₁)+T(x₂,y₂), and T(λx,λy)=λT(x,y), because both coordinates are linear combinations of x and y. For θ=π/2, T(x,y)=(−y,x), with matrix A=[[0,−1],[1,0]]. The characteristic equation is det(A−λI)=λ²+1=0, giving λ=±i. Since no root is real, T has no eigenvalue in R.

Part (b). If a=0, the equation becomes x²(y²−1)=0, so the locus is the y-axis x=0 together with the lines y=±1. Assume now a≠0. Since x²y²≥0, we must have x²≥a²; hence |x|≥|a| and x≠0. The curve can be written as y²=1−a²/x²=(x²−a²)/x². It is symmetric about both axes because only x² and y² occur. It meets the x-axis where y=0, giving x=±a, and has no y-intercept. The region |x|<|a| is empty. As |x|→∞, y²→1, so y=±1 are horizontal asymptotes; there is no vertical asymptote because y remains finite. For |x|>|a|, 0<y²<1, so the branches lie between the asymptotes and begin at (±a,0). Differentiating implicitly, 2xy²+2x²yy′=2x, so, for x≠0, y²+x yy′=1 and y′=(1−y²)/(xy). Using y²=1−a²/x² gives y′=a²/(x³y). At (a,0) and (−a,0) the derivative is infinite; the tangents are the vertical lines x=a and x=−a. Also, from y²=(x²−a²)/x², on x>|a| the quantity y² increases from 0 to 1, while on x<−|a| it decreases from 1 to 0 as x moves toward −|a|; this fixes the opening of the arcs. Thus, for a>0, four branches appear: from (a,0) they go into the first and fourth quadrants and approach y=1 and y=−1 as x→∞; from (−a,0) they go into the second and third quadrants and approach y=±1 as x→−∞. For a<0 the same picture occurs with intercepts at x=±|a|. The sketch is two symmetric pairs of open arcs, one on each side of the y-axis, bounded by the horizontal asymptotes.

Part (c). Let p=(u,v,w)≠(0,0,0) and A=diag(a,b,c). A generator of the cone lying in the plane pᵀr=0 has direction vector r=(l,m,n) satisfying rᵀAr=0 and pᵀr=0. Choose an orthonormal basis e₁,e₂ of the plane P=p⊥, and write a unit generator direction as r=αe₁+βe₂; α and β are its direction cosines in P. Substituting this parametric form into rᵀAr=0 gives the homogeneous quadratic B₁₁α²+2B₁₂αβ+B₂₂β²=0, where Bᵢⱼ=eᵢᵀAeⱼ. This is the elimination of the scale factor from the two defining equations. Putting t=α/β, the two generators correspond to the two roots t₁,t₂ of B₁₁t²+2B₁₂t+B₂₂=0; if B₁₁=0, use the reciprocal parameter and the same final condition follows. Their direction vectors in the orthonormal basis are (t₁,1) and (t₂,1); if their full direction cosines are (l₁,m₁,n₁) and (l₂,m₂,n₂), perpendicularity means l₁l₂+m₁m₂+n₁n₂=0, which in this basis is t₁t₂+1=0. By Vieta, t₁t₂=B₂₂/B₁₁; hence perpendicularity is equivalent to B₁₁+B₂₂=0. But B₁₁+B₂₂ is the trace of the restriction of A to P. If n=p/‖p‖ is the unit normal to P, then e₁,e₂,n form an orthonormal basis of R³, so B₁₁+B₂₂=tr(A)−nᵀAn=(a+b+c)−(au²+bv²+cw²)/(u²+v²+w²). Here tr(A)=a+b+c, and nᵀAn=(au²+bv²+cw²)/(u²+v²+w²) because n=(u,v,w)/‖p‖. Setting this equal to zero gives (a+b+c)(u²+v²+w²)−(au²+bv²+cw²)=0, which simplifies to (b+c)u²+(c+a)v²+(a+b)w²=0. This is the required condition.