Let $A$ be a non-zero linear transformation on a real vector space $V$ of dimension $n$.

Question

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?

1. $\lambda = 1$
2. $\det(A) = |\lambda|^{1/n}$
3. $\lambda$ is the only eigenvalue of $A$
4. 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?

Answer 1

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?

Answer 2

Hit both sides with a vector $v$: you get $A^2v =\lambda A v$ or $A(Av) = \lambda(Av)$. This means that every vector in the image of $A$ is an eigenvector with eigenvalue $\lambda$ and in particular, since it's given that $A$ is nonzero, there must exist at least one nonzero vector in the image. So yes, $\lambda$ must be an eigenvector of $A$.

It doesn't have to be 1, though. Consider for instance

$$A = \left[\begin{array}{cc}2 & 0\\0 & 0\end{array}\right].$$