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UPSC Maths 2013 Paper 1 Q2c-i — Solution

8 marks · Section A

Question

Let $A$ be a Hermitian matrix having all distinct eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_n$. If $X_1,X_2,\ldots,X_n$ are corresponding eigenvectors then show that the $n\times n$ matrix $C$ whose $k^{\text{th}}$ column consists of the vector $X_k$ is non singular.

Technique

Standard “Hermitian + distinct eigenvalues $\Rightarrow$ orthogonal eigenvectors $\Rightarrow$ basis” chain.

Solution

Strategy. Show the eigenvectors are pairwise orthogonal (Hermitian + distinct eigenvalues $\Rightarrow$ orthogonal eigenvectors), hence linearly independent, hence $C$ has full column rank, hence is non-singular.

Step 1 — Orthogonality of eigenvectors for distinct eigenvalues

Let $i\ne j$. Then $\lambda_i\ne\lambda_j$ by hypothesis.

Compute $\langle AX_i,X_j\rangle$ two ways (using the standard inner product on $\mathbb{C}^{n}$):

Way 1: $AX_i=\lambda_i X_i$, so $\langle AX_i,X_j\rangle=\langle\lambda_i X_i,X_j\rangle=\lambda_i\langle X_i,X_j\rangle$.

Way 2: Use the Hermitian property $A^{*}=A$. Then $\langle AX_i,X_j\rangle=\langle X_i,A^{*}X_j\rangle=\langle X_i,AX_j\rangle=\langle X_i,\lambda_j X_j\rangle=\overline{\lambda_j}\langle X_i,X_j\rangle$.

Hermitian $\Rightarrow$ eigenvalues real (standard), so $\overline{\lambda_j}=\lambda_j$. Thus

Since $\lambda_i\ne\lambda_j$, we conclude $\langle X_i,X_j\rangle=0$.

Step 2 — Orthogonal non-zero vectors are linearly independent

Eigenvectors are non-zero by definition. Suppose

for scalars $c_k$. Inner-product both sides with $X_k$:

Since $\|X_k\|^{2}>0$ (eigenvector non-zero), $c_k=0$. Holds for each $k$, so all $c_k=0$.

Step 3 — $C$ is non-singular

The columns of $C$ are linearly independent (Step 2), so $C$ has rank $n$, so $C$ is non-singular (invertible).

Answer