# Selected Exercises #

4. Suppose $m$ and $n$ are positive integers with $m \leq n$, and suppose $\lambda_1, \dots, \lambda_m \in F$. Prove that there exists a polynomial $p \in \mathcal P(\mathbf{F})$ with $\deg p = n$ such that $0 = p(\lambda_1) = \cdots = p(\lambda_m)$ and such that $p$ has no other zeros.

First, we can show that the polynomial $p’(z) = (z - \lambda_1) \cdots (z-\lambda_m)$ with $\deg p’ = m$ has $0 = p’(\lambda_1) = \cdots = p’(\lambda_m)$ and no other zeros. By 4.12, we know that $p’$ has at most $m$ zeros and therefore has no other zeros besides $\lambda_1, \dots, \lambda_m$.

Now, we wish to increase the degree of $p’$ from $m$ to $n$ without adding additional zeros. To accomplish this, we multiply $p’$ by $(z-\lambda_m)^{n-m}$. This gives us $$p(z) = (z-\lambda_m)^{n-m} * p’(z) = (z - \lambda_1) \cdots (z-\lambda_m)^{n-m+1}.$$

It remains to show that $p(z)$ does not have any additional zeros beyond those of $p’$. Similar to 4.14, we can express this as

$$(z-\lambda_1) \cdots (z-\lambda_m)^{m-n+1} = (z-\tau_1) \cdots (z-\tau_n).$$

However, in this case, after dividing by $(z-\lambda_1), \dots, (z-\lambda_m)$, we’re left with

$$(z-\lambda_m)^{n-m} = (z-\tau_{m+1}) \cdots (z-\tau_n).$$

It turns out the same procedure we used above works here too, since each time we divide by $(z-\lambda_m)$, we get a new equality with degree one less.

5. Suppose $m$ is a nonnegative integer, $z_1, …, z_{m+1}$ are distinct elements of $\mathbf{F}$, and $w_1, \dots, w_{m+1} \in \mathbf{F}$. Prove that there exists a unique polynomial $p \in \mathcal P_{m}(\mathbf{F})$ such that $$p(z_j) = w_j$$

for $j = 1, \dots, m+1$.

Define $T: P_m(\mathbf{F}) \rightarrow F^{m+1}$ by $$T(p) = (p(z_1), \dots, p(z_{m+1})).$$

Then, if $T$ is an injective and surjective linear map, we know that there exists a unique polynomial $p \in \mathcal P_{m}(\mathbf{F})$ such that $$p(z_j) = w_j.$$

It’s obvious due to properties of polynomials that $T$ is a linear map, so we skip the proof of this fact.

If $p \in \operatorname{null} T$, $p(z_1) = \cdots = p(z_m) = 0$, so if $p \neq 0$, $p$ would be a degree $m$ polynomial with $m+1$ distinct roots. By 4.12, this is a contradiction, so $p = 0$ and $\operatorname{null} T = \{0\}$.

$T$ is surjective iff $range T = F^{m+1}$. By the Fundamental Theorem of Linear Maps, $$\begin{eqnarray} \dim P_{m} & = & \dim \operatorname{null} T + \dim \operatorname{range} T \\\\ m+1 & = & 0 + \dim \operatorname{range} T \\\\ m+1 & = & \dim \operatorname{range} T \\\\ \dim F^{m+1} & = & \dim \operatorname{range} T \\\\ \end{eqnarray}$$

Hence, $T$ is also surjective.