← 2022 Paper 1

UPSC 2022 Maths Optional Paper 1 Q4a — Step-by-Step Solution

15 marks · Section A

Eigenvalues and eigenvectors · Linear Algebra · asked 9× in 13 yrs · Read the full method →

Question

Find a linear map T:R2→R2T:\mathbb R^2\to\mathbb R^2 which rotates each vector of R2\mathbb R^2 by an angle θ\theta. Also, prove that for θ=π/2\theta=\pi/2, TT has no eigenvalue in R\mathbb R.

Technique

Standard rotation matrix construction; compute characteristic polynomial and solve.

Solution

Step 1 — Rotation map

A rotation by angle θ\theta takes (x,y)(x,y) to (xcos⁡θ−ysin⁡θ,  xsin⁡θ+ycos⁡θ)(x\cos\theta-y\sin\theta,\;x\sin\theta+y\cos\theta).

In matrix form:

Tθ=(cos⁡θ−sin⁡θsin⁡θcos⁡θ).T_\theta=\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}.

Linearity check. Matrix multiplication is linear ⇒ TθT_\theta is linear ✓.

Effect on basis vectors:

So TθT_\theta is the rotation map.

Step 2 — Eigenvalues of Tπ/2T_{\pi/2}

At θ=π/2\theta=\pi/2: cos⁡(π/2)=0\cos(\pi/2)=0, sin⁡(π/2)=1\sin(\pi/2)=1, so

Tπ/2=(0−110).T_{\pi/2}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}.

Characteristic polynomial:

det⁡(Tπ/2−λI)=det⁡(−λ−11−λ)=λ2+1.\det(T_{\pi/2}-\lambda I)=\det\begin{pmatrix}-\lambda & -1\\ 1 & -\lambda\end{pmatrix}=\lambda^2+1.

Eigenvalue equation: λ2+1=0⇒λ2=−1⇒λ=±i\lambda^2+1=0\Rightarrow\lambda^2=-1\Rightarrow\lambda=\pm i.

Both eigenvalues are purely imaginary, not real.

Answer

  Tπ/2 has eigenvalues ±i, no real eigenvalue.  \boxed{\;T_{\pi/2}\text{ has eigenvalues }\pm i,\text{ no real eigenvalue.}\;}
We post more of this — worked solutions, CSAT trap breakdowns, guide chapters — a few times a week on Telegram. Free, no sign-in. Join

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.