Question

Question: Construct a nondiagonal \({\bf{2}} \times {\bf{2}}\) matrix that is diagonalizable but not invertible.

Short answer

The required matrix is \(\left[ {\begin{array}{*{20}{c}}2&{10}\\1&5\end{array}} \right]\).

Step-by-step solution

Write the matrix which is not invertible

Let the matrix of the order \(2 \times 2\) be:

\(A = \left[ {\begin{array}{*{20}{c}}2&{10}\\1&5\end{array}} \right]\)

The determinant of matrix A is:

\(\begin{array}{c}\det A = 5 \times 2 - 10 \times 1\\ = 10 - 10\\ = 0\end{array}\)

Therefore, the matrix is not invertible.

Find the eigenvalues of A

Find the eigenvalues of A.

\(\begin{array}{c}\left| {A - \lambda I} \right| = 0\\\left| {\begin{array}{*{20}{c}}{2 - \lambda }&{10}\\1&{5 - \lambda }\end{array}} \right| = 0\\\left( {2 - \lambda } \right)\left( {5 - \lambda } \right) - 10 = 0\\10 - 2\lambda - 5\lambda + {\lambda ^2} - 10 = 0\\{\lambda ^2} - 7\lambda = 0\\\lambda \left( {\lambda - 7} \right) = 0\\\lambda = 0,7\end{array}\)

As the eigenvalues are different, therefore A is diagonalizable.

Thus, the required matrix is \(\left[ {\begin{array}{*{20}{c}}2&{10}\\1&5\end{array}} \right]\).