Question

State if the following statements are true and which are false? Justify your answer with a
short proof or a counterexample.

No skew-symmetric matrix is diagonalisable.

Answer 1

Answer on Question #44925 – Math - Linear Algebra

Problem.

State if the following statements are true and which are false? Justify your answer with a short proof or a counterexample.

No skew-symmetric matrix is diagonalisable.

Solution.

The statement is false.

Let $A = \left[ \begin{array}{cc}1 & 2\\ -2 & 1 \end{array} \right]$ and $Q = \left[ \begin{array}{cc} - \frac{i}{2} & \frac{1}{2}\\ \frac{i}{2} & \frac{1}{2} \end{array} \right]$. Then $Q^{-1} = \left[ \begin{array}{cc}i & -i\\ 1 & 1 \end{array} \right]$ (as $\det Q = -\frac{i}{2}$).

Hence

$$D = Q A Q^{-1} = \left[ \begin{array}{cc} - \frac{i}{2} & \frac{1}{2} \\ \frac{i}{2} & \frac{1}{2} \end{array} \right] \left[ \begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array} \right] \left[ \begin{array}{cc} i & -i \\ 1 & 1 \end{array} \right] = \left[ \begin{array}{cc} - \frac{i}{2} & \frac{1}{2} \\ \frac{i}{2} & \frac{1}{2} \end{array} \right] \left[ \begin{array}{cc} 2 + i & 2 - i \\ 1 - 2i & 1 + 2i \end{array} \right] = \left[ \begin{array}{cc} 1 - 2i & 0 \\ 0 & 1 + 2i \end{array} \right].$$

Therefore $A$ is diagonalisable.

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