Question

Which of the following matrices is NOT diagonalizable?

• A. $\left[\begin{matrix}1& 1\\ 1& 2\end{matrix}\right]$
• B. $\left[\begin{matrix}1& 0\\ 3& 2\end{matrix}\right]$
• C. $\left[\begin{matrix}0&-1\\ 1& 0\end{matrix}\right]$
• D. $\left[\begin{matrix}1& 1\\ 0& 1\end{matrix}\right]$

Hint:

Watch out for triangular matrices with repeated values on the diagonal but non-zero entries above/below them; they are often non-diagonalizable.

Answer 1

The Correct Option is D

Step 1: Concept
A matrix is not diagonalizable if the geometric multiplicity of an eigenvalue is less than its algebraic multiplicity, i.e., it does not have enough linearly independent eigenvectors. 

Step 2: Meaning
Consider option (D). It is a Jordan block with eigenvalue \(\lambda=1\) repeated twice, so its algebraic multiplicity is \(2\). 

Step 3: Analysis
Solving \[ (A-I)x=0, \] we get \[ \begin{pmatrix} 0 & amp; 1\\ 0 & amp; 0 \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix} = \begin{pmatrix} 0\\ 0 \end{pmatrix}. \] This implies \[ x_2=0, \] while \(x_1\) is arbitrary. Hence, the eigenvectors are of the form \[ \begin{pmatrix} x_1\\ 0 \end{pmatrix}, \] which gives only one linearly independent eigenvector. 

Step 4: Conclusion
Since the matrix has only one independent eigenvector instead of two, it is defective and therefore not diagonalizable. 

Final Answer: (D)