Question-80

Let \(\displaystyle A\) be a square matrix of order \(2\) such that the sum of each row and the sum of each column is equal to a constant \(\displaystyle k\). Which of the following are eigenvectors of \(\displaystyle A\)?

• \(\displaystyle ( 1,0)\)
• \(\displaystyle ( 0,1)\)
• \(\displaystyle ( 1,1)\)
• \(\displaystyle ( 1,-1)\)

Answer

• \(\displaystyle ( 1,0)\)
• \(\displaystyle ( 0,1)\)
• \(\displaystyle ( 1,1)\)
• \(\displaystyle ( 1,-1)\)

Solution

We have:

\[ \begin{equation*} A=\begin{bmatrix} a & k-a\\ k-a & a \end{bmatrix} \end{equation*} \]

We see that \(\displaystyle ( 1,1)\) is an eigenvector of \(\displaystyle A\) with eigenvalue \(\displaystyle k\) since the sum of the columns of \(\displaystyle A\) is equal to \(\displaystyle ( k,k)\). If we subtract the columns, we get \(\displaystyle ( 2a-k)( 1,-1)\). Hence, \(\displaystyle ( 1,-1)\) is an eigenvector of \(\displaystyle A\) with eigenvalue \(\displaystyle 2a-k\).