## 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  $\int_{I_j}p(x)\,dx=a_j$
[HINT: Make a linear map. If the integral of a continuous function on an interval vanishes, then the function vanishes somewhere on the interval].

### Solution

1. [Vahid] Let $p(x)=\sum_{k=0}^{n-1}e_kx^k$ and $I_i=(c_i,d_i)$ such that $\int_{c_i}^{d_i}p(x)\,dx=a_i$. That is, for each $i=1,2,\dots,n$

$\sum_{k=0}^{n-1}e_k\left(\frac{d_i^{k+1}}{k+1}-\frac{c_i^{k+1}}{k+1}\right)=a_i$.

This corresponds to $n$ equations in $n$ unknowns:

$e_{n-1}\left(\frac{d_1^n}{n}-\frac{c_1^n}{n}\right)+e_{n-2}\left(\frac{d_1^{n-1}}{n-1}-\frac{c_1^{n-1}}{n-1}\right)+\cdots+e_0(d_1-c_1)=a_1$.

$\vdots$

$e_{n-1}\left(\frac{d_n^n}{n}-\frac{c_n^n}{n}\right)+e_{n-2}\left(\frac{d_n^{n-1}}{n-1}-\frac{c_n^{n-1}}{n-1}\right)+\cdots+e_0(d_n-c_n)=a_n$.

This is equivalent to the matrix equation $MX=A$, where $X$ is the vector $(e_{n-1}\, e_{n-2}\,\dots\,e_0)^T$$A$ is the vector $(a_1\,,\,a_2\,,\,\dots\,,\,a_n)^T$ and $M$ is the matrix

$[M]_{ij}=\left(\frac{d_i^{n+1-j}}{n+1-j}-\frac{c_i^{n+1-j}}{n+1-j}\right)$.

A solution to this matrix equation gives a polynomial which satisfies the given condition.

2. Does a solution always exist?? I’m stuck here! It depends on whether the matrix is invertible, etc.

## Summer 2010: Question 1

Suppose that $T:V\rightarrow V$ is a linear map. Suppose that $v_1,v_2,\dots,v_p$ are eigenvectors of $T$ with eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_p$, and that these eigenvectors are all distinct. Prove that the eigenvectors $v_1,v_2,\dots,v_p$ are linearly independent.

### Solution

We use induction on the number of eigenvectors $n$. Let $P(n)$ be the proposition that $n$ of the eigenvectors of $T$ are linearly independent.

$P(1)$ is true because the set $\{v_1\}$ is linearly independent if $v_1\neq 0$ — and eigenvectors are non-zero by definition.

Assume $P(k)$. That is assume if

$\sum_{i=1}^ka_iv_i=\mathbf{0}$ $\Rightarrow c_i=0$  for $i=1,\dots,n$.

Consider $P(k+1)$. Multiply both sides of $\sum_{i=1}^{k+1} a_iv_i$ by $T-\lambda_{k+1}I$:

$(T-\lambda_{k+1}I)(\sum_{i=1}^{k+1}a_iv_i)=(T-\lambda_{k+1}I)\mathbf{0}$

$\Rightarrow \sum_{i=1}^{k+1}a_i(\lambda_i-\lambda_{k+1})v_i=0$

$=a_1(\lambda_1-\lambda_{k+1})v_1+\cdots+a_k(\lambda_k-\lambda_{k+1})v_k+\underbrace{a_{k+1}(\lambda_{k+1}-\lambda_{k+1})v_{k+1}}_{=0}$

$=0$.

By the inductive hypothesis ($P(k)$) we have that $a_i(\lambda_i-\lambda_{k+1})=0$ for each $i=1,\dots,k$. Since $\lambda_i\neq\lambda_{k+1}$, it follows that $a_i=0$ in every case. Therefore we are left with

$\sum_{i=1}^{k+1} a_iv_i=a_{k+1}v_{k+1}=0\Rightarrow a_{k+1}=0$,

also.

Hence by the inductive hypothesis, the eigenvectors $v_1,\dots,v_p$ are linearly independent $\bullet$