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Consider the following matrix.$$\begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$$The largest eigenvalue of the above matrix is __________.
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Let $A = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$

Characteristic equation $ \mid A – \lambda I \mid = 0$

$\implies \begin{vmatrix} -\lambda & 1 & 1 & 1\\ 1& -\lambda & 1 & 1\\ 1& 1 & -\lambda & 1 \\1 & 1 & 1 & -\lambda \end{vmatrix} = 0$

Perform the operation, $C_{4} \rightarrow C_{1} + C_{2} + C_{3} + C_{4}$, we get

$\implies \begin{vmatrix} -\lambda & 1 & 1 & 3-\lambda\\ 1& -\lambda & 1 & 3-\lambda\\ 1& 1 & -\lambda & 3-\lambda \\1 & 1 & 1 & 3-\lambda \end{vmatrix} = 0$

$\implies (3 – \lambda) \begin{vmatrix} -\lambda & 1 & 1 & 1 \\ 1& -\lambda & 1 & 1\\ 1& 1 & -\lambda & 1 \\1 & 1 & 1 & 1  \end{vmatrix} = 0$

Perform the operation, $R_{1} \rightarrow R_{1} - R_{4} , R_{2} \rightarrow R_{2} - R_{1}, R_{3} \rightarrow R_{3} - R_{1},$  we get

$\implies (3 – \lambda) \begin{vmatrix} -\lambda - 1 & 0 & 0 & 0 \\ 1 + \lambda & -\lambda-1 & 0 & 0 \\  1 + \lambda & 0 & -\lambda – 1 & 0 \\1 & 1 & 1 & 1  \end{vmatrix} = 0$

$\implies (3-\lambda) (-\lambda – 1)(-\lambda – 1)(-\lambda – 1) = 0$

$\implies \lambda  = -1,-1,-1,3$

$\therefore$ The largest eigenvalue is $3.$

$\textbf{PS:}$  For any matrix $A,$

  • The determinant of $A$ equals the product of its eigenvalues.
  • The trace of $A$ equals the sum of its eigenvalues.
  • The trace of a matrix is defined as the sum of the leading diagonal entries. 
  • A real symmetric matrix has only real eigenvalues.

$$\text(OR)$$

Let $A = \begin{pmatrix} 0 & 1 & 1 & 1\\ 1& 0& 1 & 1\\ 1& 1 & 0 & 1 \\1 & 1 & 1 & 0 \end{pmatrix}$

We can write given matrix as $:3\begin{pmatrix} 0 &\frac{1}{3} &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &0 &\frac{1}{3} &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &0 &\frac{1}{3} \\ \frac{1}{3} &\frac{1}{3} &\frac{1}{3} &0 \end{pmatrix} = 3A$

A Markov matrix is a square matrix with all nonnegative entries, and where the sum of the entries down any column is $1.$ If the entries are all positive, it’s a positive Markov matrix. It is also called a doubly stochastic matrix.

The most important facts about a positive Markov matrix are:

  • $\lambda = 1$ is an eigenvalue.
  • The eigenvector associated with $\lambda = 1$ can be chosen to be strictly positive.
  • All other eigenvalues have a magnitude less than $1.$

So, the correct answer is $3.$

References:

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4 votes

The matrix in question is a 4x4 matrix with all diagonal elements as 0 and all off-diagonal elements as 1. This is a special type of matrix known as a Jacobian matrix for a complete graph with 4 nodes. The eigenvalues of such a matrix can be found without much computation.

Here’s the shortcut trick:

For a complete graph with ‘n’ nodes, the eigenvalues are n-1 (with multiplicity 1) and -1 (with multiplicity n-1). Here, n=4, so the eigenvalues are 3 (with multiplicity 1) and -1 (with multiplicity 3).

This is a well-known property of complete graphs. “Multiplicity” here refers to the number of times the same eigenvalue is repeated.

So, for our 4x4 matrix (which corresponds to a complete graph with 4 nodes), we have:

  • An eigenvalue of 3 (which is 4-1) with multiplicity 1, and

  • An eigenvalue of -1 with multiplicity 3 (which is 4-1).

 

So, the largest eigenvalue of the given matrix is 3.

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1 votes

Easier approach:

Answer: