Question

Question: A is a \({\bf{4}} \times {\bf{4}}\) matrix with three eigenvalues. One eigenspace is one-dimensional and one of the other eigenspaces is two-dimensional. Is it possible that A is not diagonalizable? Justify your answer.

Short answer

According to diagonalize theorem, matrix A of order \(4 \times 4\) is diagonalizable.

Step-by-step solution

Write the given information

Matrix A is of the order \(4 \times 4\) that has 3 eigenvalues and one eigenspace is one-dimensional and one of the other eigenspaces is two-dimensional.

Find that A is diagonalizable

Consider \(\left\{ {{{\bf{v}}_1}} \right\}\) is the basis of one-dimensional space, \({{\bf{v}}_2}\) and \({{\bf{v}}_3}\) are the basis of two-dimensional space, and let \({{\bf{v}}_4}\) be an eigenvector in the remaining eigenspace.

As per theorem 7, all the eigenvectors are linearly independent.

Therefore, according to diagonalize theorem, matrix A of order \(4 \times 4\) is diagonalizable.