Determine whether a linear transformation has an eigenvalue Let $T$ be a linear transformation from $V$ to $V$ ($V$ is a finite dimensional real vector space).
If $T^2 = 1$, does this imply that $T$ has a real eigenvalue?
 A: Hint.  Represent $T$ by an $n\times n$ matrix $A$.  Then
$$\det(A-I)\det(A+I)=\det(A^2-I)=0\ .$$
A: Here's an outline.  Let $J$ be the Jordan canonical form.  Then $J^2=I$ if and and only if $J$ consists of trivial blocks consisting of $\pm 1.$   
A: Every linear transformation from $V$ to $V$ will have an eigenvalue in $\mathbb{C}$. So let $\lambda$ be a (possibly non-real) eigenvalue of $T$ with associated eigenvector $v$, i.e. $Tv=\lambda v$. Then
$$v=Iv=T^2v=T(Tv)=T(\lambda v)=\lambda(Tv)=\lambda^2v$$
so $\lambda^2=1$. Therefore $\lambda=\pm 1$ is real.
A: Yes. Let $\vec v\in V$ be nonzero and note that
$$
(T+I)(T-I)\vec v=(T^2-I)\vec v=\vec0
$$
It follows that either $T+I$ or $T-I$ has a nontrivial nullspace. Hence one of $\lambda=\pm1$ is an eigenvalue of $T$.
More generally, suppose $T$ is an endomorphism of a vectorspace $V$ with $\dim V=n$. Let $\vec v\in V$ be nonzero and note that the $n+1$ vectors
$$
\vec v, T(\vec v), T^2(\vec v), \dotsc, T^n(\vec v)
$$
are not linearly independent because $\dim V=n$. That is, there exist scalars $\lambda_0,\dotsc,\lambda_n$ such that
$$
\lambda_0\vec v+\lambda_1 T(\vec v)+\lambda_2 T^2(\vec v)+\dotsb+\lambda_n T^n(\vec v)=\vec 0
$$
Let 
$$
p(t)=\lambda_0+\lambda_1t+\lambda_2t^2+\dotsb+\lambda_nt^n
$$
and use The Fundamental Theorem of Algebra to factor this polynomial as
$$
p(t)=a(t-a_1)(t-a_2)\dotsb(t-a_\ell)
$$
where $a\neq 0$.
Now, 
$$
a(T-a_1I)(T-a_2I)\dotsb(T-a_\ell I)\vec v=\vec 0
$$
so that one of $T-a_kI$ has a nontrivial nullspace. Hence $T$ has $\lambda=a_k$ as an eigenvalue.
The advantage to this proof is, of course, that it avoids determinants! For more on determinant-free linear algebra read Axler's famous Down With Determinants!
