Question

Find the eigenvalues of the matrices in Exercises 17 and 18.

17. \(A = \left( {\begin{array}{*{20}{c}}0&0&0\\0&2&5\\0&0&{ - 1}\end{array}} \right)\)

Short answer

Eigenvalues of the given matrix are: 0,2 and \( - 1\)

Step-by-step solution

Definition

Eigenvalue: Let \(\lambda \) is a scaler, \(A\) is an \(n \times n\) matrix and \({\bf{x}}\) is an eigenvector corresponding to \(\lambda \), \(\lambda \) is said to an eigenvalue of the matrix \(A\) if there exists a nontrivial solution \({\bf{x}}\) of \(A{\bf{x}} = \lambda {\bf{x}}\).

Theorem 1

If \(A\) is a triangular matrix, the eigenvalues of \(A\) will be the elements of the main diagonal of \(A\).

Find Eigenvalues 

Denote the given matrix by \(A = \left( {\begin{array}{*{20}{c}}0&0&0\\0&2&0\\0&0&{ - 1}\end{array}} \right)\).

It can be observed that it is an upper triangular matrix, as all the entries below the main diagonal are 0.

So, the eigenvalues of the matrix can be found by using theorem 1, according to which the eigenvalues will be the entries of the main diagonal.

The elements of the main diagonal of the matrix \(A\) are 0, 2 and \( - 1\).

So, 0, 2 and \( - 1\) are the eigenvalues of the given matrix.