If suppose I have a square matrix with complex entries with all its eigen value say, $\lambda$ which is real, than what are the possibility of such a matrix. Now, if $\lambda $ is the only real eigen value than by Cayley-Hamilton Theorem we can conclude that $$(A-\lambda I )^n = 0 $$ . Can we also say $$ A^n = \lambda^n I $$.

  • $\begingroup$ It's a nilpotent matrix $\endgroup$ – Tutankhamun Dec 12 '17 at 4:19

You cannot conclude that. Take the matrix

$$ A = \begin{bmatrix} \lambda & 1 \\ 0 & \lambda \end{bmatrix}. $$

Clearly, the only eigenvalue of $A$ is $\lambda$ but $A \ne \lambda I$.

If $A$ is diagonalizable, then indeed you may conclude that $A = \lambda I$. For then $A = P(\lambda I)P^{-1} = \lambda PP^{-1} = \lambda I$. However, if $A$ is not diagonalizable, then it will have $1$'s on the super diagonal of its Jordan form and as the above example shows, we cannot conclude $A = \lambda I$.


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