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Just a nice little problem I saw. The solution is not difficult but here I present a different one which I like.

Let $\mathbb{P}_{1,001}$ be the vector space of polynomials of degree at most 1,000. Let $T:\mathbb{P}_{1,001}\rightarrow \mathbb{P}_{1,001}$ be the linear map defined by: $\displaystyle T\{p(x)\}=2p'(x)-p(x)$.

Find the eigenvalues and eigenfunctions of the linear map $T$.

## Summer 2011: Question 3

Suppose that $n$ is a positive integer. Pick $n$ intervals $I_1,I_2,\dots,I_n$ on the real number line. Assume that any pair of these intervals are disjoint. Pick $n$ real numbers $a_1,a_2,\dots,a_n$.

1. Using ideas of linear algebra, prove that there is a polynomial function $p(x)$ of degree at most $n-1$, with real number coefficients, so that $\int_{I_j}p(x)\,dx=a_j$
2. How many such polynomial functions are there? Justify your answer?
[HINT: Make a linear map. If the integral of a continuous function on an interval vanishes, then the function vanishes somewhere on the interval].

An essence of mathematics is to cut through the gloss of each topic and exhibit the core features of the construct.

We all encountered vectors as a teenager. We may have been told that a vector is a physical quantity with magnitude in direction. This is to distinguish it from a scalar – which has magnitude alone. It begs the question – what has direction without magnitude? Anyway we can think of those familiar arrow vectors in the plane, think of all possible magnitudes and directions, and think of them as some big, infinite set $V$. One basic thing about these vectors is that we can add them together – by the parallelogram law and more than that when we add them together we get another vector. So there is a way to add vectors $u$, $v$ together, and also $u+v\in V$. This means $+$ is a closed binary relation. Furthermore it is doesn’t matter in what order we add the vectors: $u+v=v+u$, the addition is commutative. Also we can take a vector $v$ and double it, or treble it or indeed multiply it by any real number $k\in \mathbb{R}$ – and again we’ll get another vector $kv\in V$ (with $1v=v$). We don’t care about multiplying two vectors – at the moment. There is special zero vector, $\mathbf{0}$, such that for any vector $v$, $v+\mathbf{0}=v$. Every vector $v$ has a negative, namely $-v$, such that $v+(-v)=\mathbf{0}$. Also the addition and scalar multiplication satisfies some nice algebraic laws; those of  associativity and distributivity. 