UPSC 2014 Maths Optional Paper 1 Q3c-ii — Step-by-Step Solution

7 marks · Section A

Orthogonal and unitary matrices · Linear Algebra · asked 3× in 14 yrs · Read the full method →

Question

Prove that the eigen values of a unitary matrix have absolute value 1.

Use $U^{*}U=I\Rightarrow\|Uv\|=\|v\|$ (unitary preserves norms); apply to eigenvector.

Strategy. Use the inner-product preservation property of unitary matrices: $\|Uv\|=\|v\|$ for all $v$ .

Let $U$ be unitary, so $U^{*}U=I$ (where $U^{*}$ is the conjugate transpose). Let $\lambda$ be an eigenvalue with eigenvector $v\ne 0$ :

Uv=\lambda v.

Via the unitary property: Using the standard inner product,

\|Uv\|^{2}=\langle Uv,Uv\rangle=\langle v,U^{*}Uv\rangle=\langle v,Iv\rangle=\langle v,v\rangle=\|v\|^{2}.

Via the eigenvalue equation:

\|Uv\|^{2}=\|\lambda v\|^{2}=\langle\lambda v,\lambda v\rangle=\lambda\overline\lambda\langle v,v\rangle=|\lambda|^{2}\|v\|^{2}.

Equating the two expressions:

|\lambda|^{2}\|v\|^{2}=\|v\|^{2}.

Since $\|v\|^{2}>0$ (eigenvector is non-zero), divide:

|\lambda|^{2}=1\;\Longrightarrow\;|\lambda|=1.

\boxed{\;\text{Every eigenvalue of a unitary matrix has absolute value 1.}\;}