# Linear Algebra Done Right - Chapter 4

# 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.