Question

Question: Without calculation, find one eigenvalue and two linearly independent eigenvectors of \(A = \left( {\begin{array}{*{20}{c}}5&5&5\\5&5&5\\5&5&5\end{array}} \right)\). Justify your answer.

Short answer

Eigenvalues of the given matrix is 0, as the determinant of the matrix is 0 so it is non-invertible, which imply \(A{\bf{x}} = 0\) has a nontrivial solution.

Eigenvectors: \(\left( {\begin{array}{*{20}{c}}1\\2\\{ - 3}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)\)

For \(A{\bf{x}} = 0\), the sum of the entries of eigenvectors should be 0 as there is a linearly dependent relation among the columns of the matrix \(A\).

Step-by-step solution

Definitions

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 \(A{\bf{x}} = \lambda {\bf{x}}\).

Eigenvector: For a \(n \times n\) matrix \(A\), whose eigenvalue is \(\lambda \), the set of a subspace of \({\mathbb{R}^n}\) is known as an eigenspace, where the set of the subspace of is the set of all the solutions of \(\left( {A - \lambda I} \right){\bf{x}} = 0\).

Find the Eigenvalues

The given matrix is \(A = \left( {\begin{array}{*{20}{c}}5&5&5\\5&5&5\\5&5&5\end{array}} \right)\).

As per the definition of eigenvalues, if \(A{\bf{x}} = \lambda {\bf{x}}\) has a nontrivial solution, then \(\lambda \) is the eigenvalue, and the matrix \(A\) is non-invertible if and only if \[A{\bf{x}} = 0\] has a nontrivial solution and 0 is the eigenvalue.

Here, a matrix is invertible if its determinant is not 0. So, check, the matrix is invertible or not by checking its determinant.

The determinant of the determinant is 0, as all the rows of the matrix are identical. This implies the matrix is non-invertible. Hence, the eigenvalue will be 0.

Find Eigenvectors 

It is known that 0 is the eigenvalue and \(A{\bf{x}} = 0\).

So, find the eigenvectors for which the eigenvalue is 0, where eigenvectors will be the solutions of \(A{\bf{x}} = 0\), which shows a linearly dependent relation among the columns of the matrix \(A\).

Hence, for any non-zero eigenvector corresponding to \(A{\bf{x}} = 0\), there can be random entries for the eigenvector, but sum of the entries should be 0. Choose 1, 2 and \( - 3\) for the first vector, and choose 1, 0 and \( - 1\) for the second vectors.

\(\begin{array}{c}1 + 2 - 3 = 0\\1 + 0 - 1 = 0\end{array}\)

As the sum of entries is 0 for both vectors, so, \(\left( {\begin{array}{*{20}{c}}1\\2\\{ - 3}\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)\) can be the eigenvectors.