$V$ is a 2 dimensional vector space with basis $v_1$, $v_2$ and T is a linear operator defined by $T(v_1) = v_1 + v_2$ and $T(v_2) = v_1 - v_2$. How does one find its characteristic roots and characteristic vectors without using its matrix representation? The definitions one can use is that $\lambda$ - $T$ must be singular or that there exists a vector $v\neq 0$ such that $T(v) = \lambda v$. But how does one compute using these definitions?
Let's say $v$ is the characteristic vector and $\lambda$ its characteristic root, then $v$ can be represented as $a_1v_1 + a_2v_2$ where $a_1$ and $a_2$ are in $F$. This gives us $T(a_1v_1 + a_2v_2) = \lambda(a_1v_1 + a_2v_2)$. Computing this further using the defintion of $T$ gives us $a_1(v_1 + v_2) + a_2(v_1-v_2) = \lambda(a_1v_1 + a_2v_2)$ 
I'm stuck here. I need to find $\lambda$, $a_1$, $a_2$ but I seem to have only one equation. How to proceed further?
 A: You want to find a solution to $T(v)=\lambda v$.
This means
$$
T(av_1+bv_2)=\lambda(av_1+bv_2)
$$
that is
$$
(a-\lambda a+b)v_1+(a-b-\lambda b)v_2=0
$$
which means
\begin{cases}
a(1-\lambda)+b=0 \\
a-b(1+\lambda)=0
\end{cases}
and you want this linear system in the unknowns $a$ and $b$ to have non trivial solutions.
The matrix of $T$ with respect to the given basis is
$$
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
$$
and you see that the above computation is just a complicated way to arrive at
$$
\det\begin{bmatrix}1-\lambda & 1 \\ 1 & -1-\lambda\end{bmatrix}=0
$$
that is, $\lambda^2-2=0$.
A: The definition of "eigenvalue" and "eigenvector" has nothing to do with matrices so you certainly do not need to use matrices! '
Here you have T defined by T(v1)= v1+ v2 and T(v2)= v1- v2 where v1 and v2 are two basis vectors of this two dimensional vector space.  $\lambda$ is an eigenvalue of T if and only if there exist a non-zero vector, v, such that $Tv= \lambda v$.  But any such vector can be written in terms of the basis vectors, v= av1+ bv2 for some numbers a and b.  Then $Tv= aTv1+ bTv2= a(v1+ v2)+ b(v1- v2)= (a+ b)v1+ (a- b)v2= \lambda(av1+ bv2)$.  So we must have $a+ b= a\lambda$ and $a- b= b\lambda$.
We can rewrite those equations as $(1- \lambda)a+ b= 0$ and $a- (1+ \lambda)b= 0$.  a= b= 0 is an obvious solution to those equations.  $\lambda$ is an eigenvalue if there are other, non-zero solutions.  So lets look for them!  Multiplying the first equation by $1+ \lambda$ makes it $(1- \lambda^2)a+ (1+ \lambda)b= 0$.  Adding to the second equation eliminates b and we have $(2- \lambda^2)a= 0$.  Either a= 0, which gives the "trivial" a= b= 0 solution, or $2- \lambda^2= 0$.  The eigenvalues are $\lambda= \pm\sqrt{2}$.
A: By the way, here is how you would do it using matrices.  Writing the basis vectors v1 as $\begin{bmatrix}1 \\ 0 \end{bmatrix}$,  v2 as $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$, and A as $\begin{bmatrix}a & b \\ c & d \end{bmatrix}$  We have $Av1= \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix} a \\ c \end{bmatrix}= v1+ v2= \begin{bmatrix}1 \\ 1\end{bmatrix}$ so that a= c= 1 and $Av2= \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}0 \\ 1\end{bmatrix}= \begin{bmatrix} b \\ d\end{bmatrix}=  v1- v2= \begin{bmatrix} 1 \\ -1 \end{bmatrix}$ so that b= 1 and d= -1.  That is, $A= \begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$.  The characteristic equation for that matrix is $\left|\begin{array}{cc}1- \lambda & 1 \\ 1 & -1- \lambda \end{array}\right|= (1- \lambda)(-1- \lambda)- 1= \lambda^2- 2= 0$ so that $\lambda= \pm\sqrt{2}$ as before.
