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

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

\(A = \left[ {\begin{array}{*{20}{c}}5&2\\0&4\end{array}} \right]\)

The matrix A is triangular; therefore, the diagonal elements are the eigenvalues. So, the eigenvalues of A are 5 and 4

As the eigenvalues are distinct, therefore A is diagonalizable.

Consider a matrix with the same eigenvalues:

\(A = \left[ {\begin{array}{*{20}{c}}3&2\\0&3\end{array}} \right]\)

Apply the characteristic equation:

\(\begin{array}{c}\left( {A - 3I} \right){\bf{x}} = 0\\\left( {\left[ {\begin{array}{*{20}{c}}3&2\\0&3\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}3&0\\0&3\end{array}} \right]} \right)\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\\\left[ {\begin{array}{*{20}{c}}0&2\\0&0\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}0\\0\end{array}} \right]\end{array}\)

So, the eigenvector of the matrix is:

\(\begin{array}{c}{\bf{v}} = \left[ {\begin{array}{*{20}{c}}t\\0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1\\0\end{array}} \right]t\end{array}\)

Since the eigenvector is linearly dependent, therefore matrix A is invertible but not diagonalizable.

So, the matrix A is \(\left[ {\begin{array}{*{20}{c}}3&2\\0&3\end{array}} \right]\).