Question-41

Consider the following matrix:

\[ \begin{equation*} \begin{bmatrix} 0 & 1 & 1 & 1\\ 1 & 0 & 1 & 1\\ 1 & 1 & 0 & 1\\ 1 & 1 & 1 & 0 \end{bmatrix} \end{equation*} \]

What is the largest eigenvalue of the above matrix?

Answer

\(3\)

Solution

Consider:

\[ \begin{equation*} A=\begin{bmatrix} 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1\\ 1 & 1 & 1 & 1 \end{bmatrix} ,\ I=\begin{bmatrix} 1 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{bmatrix} \end{equation*} \]

Then, the matrix we are interested in is \(\displaystyle A-I\). Let \(\displaystyle ( \lambda ,v)\) be an eigenpair of \(\displaystyle A\).

\[ \begin{equation*} \begin{aligned} ( A-I) v & =\lambda v\\ \Longleftrightarrow Av & =( \lambda +1) v \end{aligned} \end{equation*} \]

We have shown \(\displaystyle ( \lambda ,v)\) is an eigenpair of \(\displaystyle A-I\) if and only if \(\displaystyle ( \lambda +1,v)\) is an eigenpair of \(\displaystyle A\). Now:

\[ \begin{equation*} A=\begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix} =uu^{T} ;\ \ \ \ \ \ u=\begin{bmatrix} 1\\ 1\\ 1\\ 1 \end{bmatrix} \end{equation*} \]

Clearly, \(\displaystyle A\) is a matrix of unit rank with \(\displaystyle ( 4,u)\) as an eigenpair. The other kind of eigenpair is \(\displaystyle ( 0,v)\), where \(\displaystyle v\perp u\). We can ignore this since we are only interested in the largest eigenvalue. Therefore, \(\displaystyle ( 3,u)\) is an eigenpair of \(\displaystyle A-I\). By the previous result, \(\displaystyle 3\) happens to be the largest eigenvalue of \(\displaystyle A-I\).