Question

Consider the following matrix.

(0111101111011110)\begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix}

The largest eigenvalue of the above matrix is __________

Official answer and solution
Official answer
3
From
Official GATE 2021 answer key

Solution

Use the all-ones matrix identity or row-sum property of symmetric matrices.

Let AA be the given 4×44 \times 4 matrix:

A=(0111101111011110)A = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix}

Notice that the sum of entries in every row of AA is constant:

0+1+1+1=30 + 1 + 1 + 1 = 3

For any square matrix where all row sums equal a constant cc, the vector of all ones v=(1,1,1,1)Tv = (1, 1, 1, 1)^T is an eigenvector with eigenvalue λ=c\lambda = c. Thus, λ=3\lambda = 3 is an eigenvalue of AA.

To find all eigenvalues, express AA in terms of the all-ones matrix J4J_4 and the identity matrix I4I_4:

A=J4−I4A = J_4 - I_4

The eigenvalues of J4J_4 are 44 (with multiplicity 1) and 00 (with multiplicity 3). Subtracting I4I_4 shifts every eigenvalue by −1-1:

λ(A)=4−1=3,and0−1=−1  (multiplicity 3)\lambda(A) = 4 - 1 = 3, \quad \text{and} \quad 0 - 1 = -1 \;(\text{multiplicity } 3)

The eigenvalues of AA are {3,−1,−1,−1}\{3, -1, -1, -1\}. Therefore, the largest eigenvalue is 3.

Final answer

3

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question
62 of 65, printed as CS Q.52
Section
Computer Science and Information Technology

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