Question-69
Let \(\displaystyle A\) be a real square matrix of order \(\displaystyle 5\) such that \(\displaystyle A^{2} =0\). Find the number of distinct eigenvalues of \(\displaystyle A\).
Use the eigenvalue equation.
\(1\)
Let \(\displaystyle ( \lambda ,v)\) be an eigenpair of \(\displaystyle A\). Then:
\[ \begin{equation*} \begin{aligned} Av & =\lambda v \end{aligned} \end{equation*} \]
Multiplying both sides by \(\displaystyle A\):
\[ \begin{equation*} \begin{aligned} A^{2} v & =A( \lambda v)\\ & =\lambda ( Av)\\ & =\lambda ^{2} v \end{aligned} \end{equation*} \]
The LHS is the zero vector since \(\displaystyle A^{2} =0\). This gives us:
\[ \begin{equation*} \lambda ^{2} v=0 \end{equation*} \]
Note that the zero on the RHS is the zero vector. \(\displaystyle v\) can’t be zero since its an eigenvector. This forces \(\displaystyle \lambda =0\). Therefore the number of distinct eigenvalues of \(\displaystyle A\) is equal to \(\displaystyle 1\).