Question

Question: Show that if \(\lambda \) is an eigenvalue of A if and only if \(\lambda \) is an eigenvalue of \({A^T}\). (Find out how \(A - \lambda I\) and \({A^T} - \lambda I\) are related.)

Short answer

It is proved that \(\lambda \) must be the eigenvalue of both A and \({A^T}\).

Step-by-step solution

Write the given information

Here, \(\lambda \) is the eigenvalue of a matrix A.

Check for the eigenvalue of \(A\)

Find the relation between \(A - \lambda I\) and \({A^T} - \lambda I\).

\(\begin{array}{c}{\left( {A - \lambda I} \right)^T} = {A^T} - {\left( {\lambda I} \right)^T}\\ = {A^T} - \lambda I\end{array}\)

So, \({\left( {A - \lambda I} \right)^T}\) is invertible if and only if \(A - \lambda I\) is invertible.

According to theorem 6(c), \({A^T} - \lambda I\) is not invertible if only if \(A - \lambda I\) is not invertible.

Therefore, \(\lambda \) must be the eigenvalue of both A and \({A^T}\).