← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q3b — Step-by-Step Solution

10 marks · Section A

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

Question

Prove that distinct non-zero eigenvectors of a matrix are linearly independent.

Technique

Induction; eliminate the last eigenvector by forming $(A-\lambda_k I)$ applied to the dependence relation, then invoke the hypothesis and distinctness.

Solution

Step 1 — Precise statement

The intended (and true) statement is: if $\mathbf v_1,\dots,\mathbf v_k$ are non-zero eigenvectors of $A$ belonging to distinct eigenvalues $\lambda_1,\dots,\lambda_k$, then $\mathbf v_1,\dots,\mathbf v_k$ are linearly independent. (Distinct eigenvectors with the same eigenvalue need not be independent, so distinctness of the eigenvalues is what matters.)

Step 2 — Induction on $k$

Base case $k=1$: a single non-zero vector $\mathbf v_1$ is linearly independent.

Inductive step: assume eigenvectors for any $k-1$ distinct eigenvalues are independent. Suppose

Step 3 — Apply $A$ and also multiply by $\lambda_k$

Apply $A$ to (1), using $A\mathbf v_i=\lambda_i\mathbf v_i$:

Multiply (1) by $\lambda_k$:

Subtract (3) from (2); the $\mathbf v_k$ term cancels:

Step 4 — Use the induction hypothesis and conclude

By the inductive hypothesis $\mathbf v_1,\dots,\mathbf v_{k-1}$ are independent, so every coefficient in (4) is zero:

Since the eigenvalues are distinct, $\lambda_i-\lambda_k\ne0$, forcing $c_1=\cdots=c_{k-1}=0$. Substituting back into (1) gives $c_k\mathbf v_k=\mathbf0$, and $\mathbf v_k\ne\mathbf0$ implies $c_k=0$. Hence all $c_i=0$:

Answer