Vector Space-Linear Transformation (Multiple Choice)

Question

I was trying to solve the following problem from a competitive exam paper.

Let $A$ be a nonzero 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 $k=dimV_0 < n$ and suppose for some $\lambda \in \mathbb R$ , $A^2=\lambda A$. Then

1. $\lambda = 1$
2. $detA=|\lambda|^n$
3. $\lambda$ is the only eigen value of $A$
4. There is a non trivial subspace $V_1 \subset V$ such that $Ax=0\forall x \in V_1$

One or more options may be correct.

What I think is that option $4$ is correct as Rank($A$)=$dimV_0=k$ So by Rank-Nullity Theorem $nullity(A)=n-k>0$ Thus nullspace of $A$ is nontrivial. Am I right? Also please help me about the other options. Thnx in advance

Answer 1

For $V = \mathbb{R}^3$, with the canonical basis $(1,0,0),(0,1,0),(0,0,1)$ let $A$ be the map corresponding to the matrix $$ M_A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix} \text{.} $$

Then $V_0 = \textrm{span }\{(0,1,0),(0,0,1)\}$, and $A^2 = 2A$.

This counter-example disproves 1., 2. and 3.

And 4. is indeed true, for the reason you stated.

Answer 2

You have \begin{equation}A^2=\lambda A, \end{equation} so then we must have $A^2-\lambda A=0$ from which it follows that \begin{equation}A(A-\lambda I)=0. \end{equation} Since this equation consists only of distinct linear factors it indicates that $A$ has minimum polynomial $m_A(x)=x(x-\lambda)$, so that $A$ has eigenvalues $0$ and $\lambda$.

This immediately rules out 1,2 and 3 (since for certain choices of $\lambda$ these options will not hold) and in fact confirms 4, since $0$ is eigenvalue of $A$.

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So I will just elaborate a little - the minimum polynomial of $A$ is the monic polynomial of least degree annihilated by $A$ ($m(A)=0$), and every linear factor of the characteristic polynomial is also a factor of the minimum polynomial. So if you only have distinct linear factors in any polynomial annihilated by $A$ then it must be the minimum polynomial. And of course these factors then indicate the eigenvalues of $A$.

Now it could very well be that $\lambda=1$, in that particular case $A$ is a projection, but nothing in the information in the question indicates that $\lambda$ MUST be 1.

Now since $A$ has $0$ as an eigenvalue we must have det$(A)=0$. So again in the special case where $\lambda=0$ we would have that $A$ is nilpotent of index 2 (since it is given that $A$ has a nontrivial image? Perhaps $V_0$ could be trivial?). In such a case 2 and 3 is true, but again nothing in the information given guarantees it.

Option 4 is the only option that MUST be true.