Question

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)’s and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

Short answer

The eigenvalues are \(\lambda = a - b\) with multiplicity \(n - 1\) and \(\lambda = a + \left( {n - 1} \right)\) with multiplicity \(1\).

Step-by-step solution

Step 1: Write the definition of eigenvalue and eigenvector

Eigenvector and Eigenvalue: An eigenvector of \(n \times n\) matrix \(A\) is a nonzero vector \({\bf{x}}\) such that \(A{\bf{x}} = \lambda {\bf{x}}\) for some scalar \(\lambda \) where scalar \(\lambda \) is called an eigenvalue of \(A\). If there is a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\) then \({\bf{x}}\) is called an eigenvector corresponding to \(\lambda \).

Step 2: Show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\) also find the multiplicities of these eigenvalues

The characteristic polynomial of A is shown below:

\(\begin{aligned}{c}\det \left( {A - \lambda I} \right) = \det \left( {\begin{aligned}{*{20}{c}}{a - \lambda }&b&b&{...}&b\\b&{a - \lambda }&b&{...}&b\\b&b&{a - \lambda }&{...}&b\\:&:&:&:&:\\b&b&b&{...}&{a - \lambda }\end{aligned}} \right)\\ = {\left( {a - \lambda - b} \right)^{n - 1}}\left( {a - \lambda + \left( {n - 1} \right)b} \right)\\ = {\left( {a - b - \lambda } \right)^{n - 1}}\left( {a + \left( {n - 1} \right)b - \lambda } \right)\end{aligned}\)

Thus, the eigenvalues are \(\lambda = a - b\) with multiplicity \(n - 1\) and \(\lambda = a + \left( {n - 1} \right)\) with multiplicity \(1\).