Question

Question: Show that if A has n linearly independent eigenvectors, then so does \({A^T}\). [Hint: Use the diagonalization theorem.]

Short answer

It is proved that the matrix \({A^T}\) is diagonalizable and the columns of matrix Q are n linearly independent eigenvectors of \({A^T}\).

Step-by-step solution

Write the diagonalization theorem

If the matrix A has n linearly independent vectors, then by the diagonalization theorem, \(A = PD{P^{ - 1}}\).

The matrix P is invertible and matrix D is diagonalizable.

Check if transpose of A has n linearly independent vector

Using the properties of transpose,

\[\begin{array}{c}{A^T} = {\left( {PD{P^{ - 1}}} \right)^T}\\ = {\left( {{P^{ - 1}}} \right)^T}{D^T}{P^T}\\ = {\left( {{P^T}} \right)^{ - 1}}D{P^T}\\ = QD{Q^{ - 1}}\end{array}\]

Here, \(Q = {\left( {{P^T}} \right)^{ - 1}}\). Thus, the matrix \({A^T}\) is diagonalizable and the columns of matrix Q are n linearly independent eigenvectors of \({A^T}\).