Question

Question: Consider an \(n \times n\) matrix A with the property that the row sums all equal the same number s. Show that s is an eigenvalue of A. (Hint: Find an eigenvector.)

Short answer

It is proved that s is an eigenvalue of A.

Step-by-step solution

Assume the matrix A

Let the matrix A is:

\(A = \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \ldots & \cdots & \cdots & \cdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{mn}}}\end{array}} \right)\)

For the elements of all the rows, the given property is:

\(\sum {{a_{1i}}} = \sum {{a_{2i}}} = .....\sum {{a_m}} = s\)

For \(i = 1,2,.....,n\)

Check whether s is the eigenvalue of A

Let the eigenvector be:

\(v = \left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\)

Using the characteristic equation:

\(\begin{array}{c}A{\bf{v}} = \left( {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1n}}}\\{{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2n}}}\\ \ldots & \cdots & \cdots & \cdots \\{{a_{n1}}}&{{a_{n2}}}& \cdots &{{a_{mn}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{\sum {{a_{1i}}} }\\{\sum {{a_{2i}}} }\\ \vdots \\{\sum {{a_{ni}}} }\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}s\\s\\ \vdots \\s\end{array}} \right)\\ = s\left( {\begin{array}{*{20}{c}}1\\1\\ \vdots \\1\end{array}} \right)\\ = s{\bf{v}}\end{array}\)

Therefore, s is an eigenvalue of matrix A.