Question

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\)? If so, find the eigenvalue.

Short answer

No, \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is not the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\).

Step-by-step solution

Definition of eigenvector

If there exists a non-zero vector \({\bf{x}}\) which satisfies \(A{\bf{x}} = \lambda {\bf{x}}\) for some scaler \(\lambda \), then \({\bf{x}}\) be the eigenvector of an \(n \times n\) matrix \(A\).

Determine whether \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is the eigenvalue of the given matrix

Denote the given matrix by \(A\) and the given vector by \({\bf{x}}\).

\(A = \left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\)

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

Find the product of \(A\), and \({\bf{x}}\).

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1\\{29}\end{array}} \right)\end{array}\)

It can be observed that \(\left( {\begin{array}{*{20}{c}}1\\{29}\end{array}} \right) \ne \lambda \left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\)

So, \(\left( {\begin{array}{*{20}{c}}1\\4\end{array}} \right)\) is not the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}{ - 3}&1\\{ - 3}&8\end{array}} \right)\).