Official PYQ · 2018Q17Eigenvalues and Eigenvectors, LU DecompositionNAT1 mark

GATE CS 2018 Q17 — Eigenvalues and Eigenvectors, LU Decomposition

Consider a matrix

A= uv^T

where

u=\begin{pmatrix}1 \\ 2 \end{pmatrix} , v = \begin{pmatrix}1 \\1 \end{pmatrix}

. Note that

v^T

denotes the transpose of

v

. The largest eigenvalue of

A

is ____

View worked solution

Answer: 3

We are given the vectors:

u = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, \quad v = \begin{pmatrix} 1 \\ 1 \end{pmatrix}

The matrix $A$ is defined as the outer product $A = u v^T$ . Since $A$ is a rank-1 matrix of size $2 \times 2$ , it has one non-zero eigenvalue and one zero eigenvalue.

Method 1: Using properties of outer product

For any column vectors $u$ and $v$ , the non-zero eigenvalue of the rank-1 matrix $A = u v^T$ is given by the inner product $v^T u$ :

\lambda = v^T u = \begin{pmatrix} 1 & 1 \end{pmatrix} \begin{pmatrix} 1 \\ 2 \end{pmatrix} = (1 \times 1) + (1 \times 2) = 3

Thus, the eigenvalues of $A$ are $3$ and $0$ . The largest eigenvalue is $3$ .

Method 2: Direct computation

Let's construct the matrix $A$ explicitly:

A = u v^T = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}

The characteristic equation of $A$ is given by:

\det(A - \lambda I) = 0

\det\begin{pmatrix} 1-\lambda & 1 \\ 2 & 2-\lambda \end{pmatrix} = 0

(1-\lambda)(2-\lambda) - 2 = 0

\lambda^2 - 3\lambda = 0 \implies \lambda(\lambda - 3) = 0

The eigenvalues are $\lambda = 0$ and $\lambda = 3$ .

Therefore, the largest eigenvalue of $A$ is 3.

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Published 10 Oct 2026

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GATE CS 2018 Q17 — Eigenvalues and Eigenvectors, LU Decomposition

Method 1: Using properties of outer product

Method 2: Direct computation

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