Question

Question: Is \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) an eigenvector of \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\)? If so, find the eigenvalue.

Short answer

Yes, \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\) and the eigenvalue is 0.

Step-by-step solution

Definition of eigenvector

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

Determine whether \(\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the given matrix

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

\(A = \left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\), \({\bf{x}} = \left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\)

According to the definition of eigenvalue, \({\bf{x}} = \left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the matrix \(A\), if \(A{\bf{x}} = \lambda {\bf{x}}\).

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

\(\begin{array}{c}A{\bf{x}} = \left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( 4 \right) + 7\left( { - 3} \right) + 9\left( 1 \right)}\\{ - 4\left( 4 \right) + \left( { - 5} \right)\left( { - 3} \right) + 1\left( 1 \right)}\\{2\left( 4 \right) + 4\left( { - 3} \right) + 4\left( 1 \right)}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{12 - 21 + 9}\\{ - 16 + 15 + 1}\\{8 - 12 + 4}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}\)

The obtained matrix in the form of a given vector can be written as:

\(A{\bf{x}} = 0\left( {\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\)

Because 0 is a multiple of every non-zero number. Hence, the given vector satisfies the condition \(A{\bf{x}} = \lambda {\bf{x}}\).

So, the vector \(\left({\begin{array}{*{20}{c}}4\\{ - 3}\\1\end{array}} \right)\) is the eigenvector of the given matrix \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\).

Determine the eigenvalue 

As the given vector satisfies the condition \(A{\bf{x}} = \lambda {\bf{x}}\). Which implies that \(\lambda \) is the eigenvalue of the given matrix. So, \(\lambda = 0\).

So, 0 is the eigenvalue of the given matrix \(\left( {\begin{array}{*{20}{c}}3&7&9\\{ - 4}&{ - 5}&1\\2&4&4\end{array}} \right)\).