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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 9× in 13 yrs · Read the full method →

Question

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

Technique

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

Solution

Step 1 — Characteristic polynomial

$\det(A-\lambda I)=\det\begin{pmatrix}-\lambda&-i\\i&-\lambda\end{pmatrix}=\lambda^2-(-i)(i)=\lambda^2-(-i^2)=\lambda^2-1$.

Wait, $(-i)(i)=-i^2=-(-1)=1$. So $\det=\lambda^2-1$? Let me recompute: $\det=(-\lambda)(-\lambda)-(-i)(i)=\lambda^2-(-i^2)=\lambda^2-1$ (since $-i^2=1$).

So $\det(A-\lambda I)=\lambda^2-1$.

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

Recompute determinant carefully: $(-\lambda)(-\lambda)-(-i)(i)=\lambda^2-(-i\cdot i)=\lambda^2-(-i^2)$. $i^2=-1$, so $-i^2=1$. $\lambda^2-1=0\Rightarrow\lambda=\pm 1$.

Real eigenvalues ✓.

Step 2 — Eigenvalues

$\lambda_1=1$, $\lambda_2=-1$.

Step 3 — Eigenvector for $\lambda=1$

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

Row 1: $-v_1-iv_2=0\Rightarrow v_1=-iv_2$. Take $v_2=1$: $v_1=-i$.

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

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

Step 4 — Eigenvector for $\lambda=-1$

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

Row 1: $v_1-iv_2=0\Rightarrow v_1=iv_2$. Take $v_2=1$: $v_1=i$.

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

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

Summary

Answer