Eigenvalues and vectors of a Linear Transformation I am kinda lost here. All I did until now was finding eigenvalues and vectors for a matrix but as far as I can understand the question it asks me to find the eigenvalues of a Linear Transformation?
Let V be the vector space of all real polynomials p(x) of degree ≤ n. Define
T : V → V, T (p) = q, q(t) = p(t+1). Determine the eigenvalues and the eigenvectors of T.
I am just looking for the general idea to start the problem. Any help is appreciated.
 A: Hint: You can represent your transformation $T$ as a matrix $A$ w.r.t. to some choice of a basis of polynomials. Find the eigenvectors of $A$ and translate them back to polynomials.
From $Ax = \lambda x$ for some $\lambda \in \mathbb C$ and a nonzero coordinate vector $x$ of a nozero polynomial $p$, you can deduce that the coordinate vector of $Tp$ will be $\lambda x$, i.e. the coordinate vector of $\lambda p$, implying $$Tp = \lambda p,$$ which is the eigenvalue equation for the eigenvalue $\lambda$ of the transformation $T$ with the eigenfunction $p$.
Alternatively, you could also solve the question which polynomials satisfy $\lambda p(t) = p(t+1)$ for some $\lambda$.
A: Linear transformations do not change the dimension of the vector space. They just re-orientate & re-size the vector within the same space.
A vector p = p(x) in this space has a max power (degree) of n.
So its linear transform - here q = q(x) - must have no higher order than n also, though its coefficients must in general change.
(I'm not sure why the variable t is used.)
So let's rewrite the transformation in terms of the space variable.
i.e.                    q(x) = p(x + 1)
This is clearly a linear transformation since whatever coefficients are applied to each power of x, there will be no higher order of x in vector q than in vector p.
Were the transformation something like 
                    q(x) = p(x.x  + 1)   

the q vector would clearly have a higher degree and thus no longer be in the space V.
So our transformation matrix maps a vector of (n + 1) terms to another of the same number of terms. Let's call it T.
We can write this transformation as the matrix equation that relates vectors, p and q :
                   T . p  =  q

This means that matrix T has dimensions (n + 1) x (n + 1).
Let's look at the above relation in more detail :
               T . a . x  = a . x+1

where        a is the row vector holding the coefficients of polynomial p(x)
          x is the column vector holding all the powers of x from 0 to n.
and     x+1 is the column vector holding all the powers of (x+1) from 0 to n.
Cancelling the a from both sides :
             T . x  =  X+1

Simple inspection of the values in each row of vector x+1, i.e. 1, x + 1, x.x + 2.x + 1, etc will quickly suggest to you how to fill in the values of the transformation matrix T.
Your task is then to find eigenvectors and eigenvalues for this matrix.
Over to you.
