Let $A$ be a non-zero linear transformation on a real vector space $V$ of dimension $n$. Let the subspace $V_0 \subset V$ be the image of $V$ under $A$. Let $k = \dim (V_0) \lt n$ and suppose that for some $\lambda \in \mathbb{R}$, $A^2 = \lambda A$. Then which of the following are true?
- $\lambda = 1$
- $\det(A) = |\lambda|^{1/n}$
- $\lambda$ is the only eigenvalue of $A$
- There is a nontrivial subspace $V_1 \subset V$ such that $Ax = 0$ for all $x \in V_1$
I am able to say that 4 is true as $\operatorname{rank}(A) < n$. 2 is false by using the determinant function on both sides. Using the argument used for 4, 0 is also an eigenvalue. So 3 is also false.
But I want to know what we can say about $\lambda$. Is is an eigenvalue? Is it equal to 1?
Best Answer
If $\lambda\ne0$ then the polynomial with simple roots $x^2-\lambda x=x(x-\lambda)$ annihilates $A$ and clearly $A\ne \lambda I_n$ and $A\ne0$ so $0$ and $\lambda$ are eigenvalues of $A$ and the multiplicity of $\lambda$ is $k$. If $\lambda=0$ then $A$ is nilpotent and $A$ in it's Jordan canonical form has a Jordan block with size $k$. Can you know now the correct options?