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UPSC 2021 Maths Optional Paper 1 Q4a-ii — Step-by-Step Solution

10 marks · Section A

Eigenvalues and eigenvectors · Linear Algebra · asked 10× in 14 yrs · Read the full method →

Question

Find eigenvalues and corresponding eigenvectors of A=(0−ii0)A=\begin{pmatrix}0&-i\\i&0\end{pmatrix} over the complex number field.

Technique

Standard 2×2 eigenproblem; characteristic polynomial; solve linear systems.

Solution

Step 1 — Characteristic polynomial

det⁡(A−λI)=det⁡(−λ−ii−λ)=λ2−(−i)(i)=λ2−(−i2)=λ2−1\det(A-\lambda I)=\det\begin{pmatrix}-\lambda&-i\\i&-\lambda\end{pmatrix}=\lambda^2-(-i)(i)=\lambda^2-(-i^2)=\lambda^2-1.

Wait, (−i)(i)=−i2=−(−1)=1(-i)(i)=-i^2=-(-1)=1. So det⁡=λ2−1\det=\lambda^2-1? Let me recompute: det⁡=(−λ)(−λ)−(−i)(i)=λ2−(−i2)=λ2−1\det=(-\lambda)(-\lambda)-(-i)(i)=\lambda^2-(-i^2)=\lambda^2-1 (since −i2=1-i^2=1).

So det⁡(A−λI)=λ2−1\det(A-\lambda I)=\lambda^2-1.

Wait, that doesn’t match — AA is Hermitian (A∗=AA^*=A? A∗=(0−ii0)∗=A^*=\begin{pmatrix}0&-i\\i&0\end{pmatrix}^*= transpose conjugate = (0−ii0)\begin{pmatrix}0&-i\\i&0\end{pmatrix} ✓ Hermitian). So eigenvalues should be real.

Recompute determinant carefully: (−λ)(−λ)−(−i)(i)=λ2−(−i⋅i)=λ2−(−i2)(-\lambda)(-\lambda)-(-i)(i)=\lambda^2-(-i\cdot i)=\lambda^2-(-i^2). i2=−1i^2=-1, so −i2=1-i^2=1. λ2−1=0⇒λ=±1\lambda^2-1=0\Rightarrow\lambda=\pm 1.

Real eigenvalues ✓.

Step 2 — Eigenvalues

λ1=1\lambda_1=1, λ2=−1\lambda_2=-1.

Step 3 — Eigenvector for λ=1\lambda=1

(A−I)v=0(A-I)v=0: (−1−ii−1)(v1v2)=0\begin{pmatrix}-1&-i\\i&-1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}=0.

Row 1: −v1−iv2=0⇒v1=−iv2-v_1-iv_2=0\Rightarrow v_1=-iv_2. Take v2=1v_2=1: v1=−iv_1=-i.

Eigenvector: (−i,1)T(-i,1)^T or equivalently (1,i)T(1,i)^T (multiplying by ii).

Check: A(1i)=(−i⋅ii⋅1)=(1i)A\begin{pmatrix}1\\i\end{pmatrix}=\begin{pmatrix}-i\cdot i\\i\cdot 1\end{pmatrix}=\begin{pmatrix}1\\i\end{pmatrix} ✓ (eigenvalue 1).

Step 4 — Eigenvector for λ=−1\lambda=-1

(A+I)v=0(A+I)v=0: (1−ii1)(v1v2)=0\begin{pmatrix}1&-i\\i&1\end{pmatrix}\begin{pmatrix}v_1\\v_2\end{pmatrix}=0.

Row 1: v1−iv2=0⇒v1=iv2v_1-iv_2=0\Rightarrow v_1=iv_2. Take v2=1v_2=1: v1=iv_1=i.

Eigenvector: (i,1)T(i,1)^T.

Check: A(i1)=(−i⋅1i⋅i)=(−i−1)=−(i1)A\begin{pmatrix}i\\1\end{pmatrix}=\begin{pmatrix}-i\cdot 1\\i\cdot i\end{pmatrix}=\begin{pmatrix}-i\\-1\end{pmatrix}=-\begin{pmatrix}i\\1\end{pmatrix} ✓ (eigenvalue −1-1).

Summary

Answer

  λ1=1,  v1=(1,i)T;λ2=−1,  v2=(i,1)T.  \boxed{\;\lambda_1=1,\;v_1=(1,i)^T;\quad\lambda_2=-1,\;v_2=(i,1)^T.\;}
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