Let $V$ be a $n$-dimensional real vector space and let linear operator $ T \in L(V) $ satisfy the equation$ (T^2+I) *(T^2+4I)=0$.

Question

Let $V$ be a $n$-dimensional real vector space and let linear operator $ T \in L(V) $ satisfy the equation $$ (T^2+I) *(T^2+4I)=0$$. Find the eigenvalues for $T$ and prove that $n$ is even. I'm a bit confused here. I tried by defining a function $f(t) =(t^2+1) *(t^2+4)$. Then because of the equation the operator $f(T) $ is nilpotent so its eigenvalue is only 0. Also I know that the eigenvalues of T $ \lambda \in \sigma( T), f(\lambda) \in \sigma(f(T))=f(\sigma(T))$ . So the eigenvalues satsfy the equation $$ (\lambda^2+1) *(\lambda^2+4)=0$$. But then they would have to be complex which cannot be. So T has no eigenvalues? Also the part with the even n I don't even know where to start.

Answer 1

I would agree with your assessment. If $\lambda$ were an eigenvalue, then it would be a real number such that $(\lambda^2 + 1)(\lambda^2 + 4) = 0$, and no such real number exists. Though, it should be pointed out, some people look for eigenvalues not just in their scalar field, but in its algebraic closure. I suspect that you don't, for the purposes of your course.

Now, if $T$ has no eigenvalues, then $T$ must be an operator on an even-dimensional space, since otherwise the characteristic polynomial would be of odd degree, and must have a root somewhere by the intermediate value theorem, and the fact that the polynomial limits to $\pm \infty$ as $x \to \pm \infty$.

Answer 2

Take

$$T=\begin{pmatrix}-4&0&0\\0&-1&1\\0&0&-1\end{pmatrix}$$

Then

$$(T+I)^2(T+4I)=\begin{pmatrix}9&0&0\\0&0&0\\0&0&0\end{pmatrix}\begin{pmatrix}0&0&0\\0&3&1\\0&0&3\end{pmatrix}=0\implies (T+I)^2(T+4I)^2=0$$

yet $\;T\;$ is defined on $\;\Bbb R^3\;$ ...

Maybe some condition is missing...or I made a mistake, of course.