Chapter 4 Exercise

1. Empty

2. Solution: False. Consider $1=(z^m+1)+(-z^m)\notin \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$. Note that \[(z^m+1)\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}\]and\[-z^m\in \{0\}\cup\{p\in\ca P(\mb F):\deg p=m\},\]it follows that $\{0\}\cup\{p\in\ca P(\mb F):\deg p=m\}$ is not closed under addition. Hence it is not a subspace of $\ca P(\mb F)$.

3. Solution: False. Consider $z=(z^{2}+z)+(-z^2)\notin \{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\}$. Note that \[(z^2+z)\in \{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\}\]and\[-z^2\in \{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\},\]it follows that $\{0\}\cup\{p\in\ca P(\mb F):\deg p\text{~is even}\}$ is not closed under addition. Hence it is not a subspace of $\ca P(\mb F)$.

4. Solution: Define $p\in\ca P(\mb F)$ by \[p(z)=(z-\lambda_1)^{n-m+1}(z-\lambda_2)\cdots(z-\lambda_m).\]Then $p$ is a polynomial of degree $n$ such that $0=p(\lambda_1)=\cdots=p(\lambda_{m})$ and such that $p$ has no other zeros.

5. Solution: See Linear Algebra Done Right Solution Manual Chapter 4 Problem 2.

6. Solution: See Linear Algebra Done Right Solution Manual Chapter 4 Problem 4.

7. Solution: See Linear Algebra Done Right Solution Manual Chapter 4 Problem 5.

8. Solution: First we show that $T$ is a linear map. Then we show $Tp \in\ca P(\R)$ for a basis of $p \in\ca P(\R)$, then by linearity of $T$, we have $Tp \in\ca P(\R)$ for every polynomial $p \in\ca P(\R)$. For any $\lambda\in\R$ and $p,q\in\ca P(\R)$, we have \begin{align*} T(p+q)=&\dfrac{(p+q)-(p+q)(3)}{x-3}=\dfrac{(p+q)-p(3)-q(3)}{x-3}\\ =&\dfrac{p-p(3)}{x-3}+\dfrac{q-q(3)}{x-3}=Tp+Tq, \end{align*} if $x\ne 3$. Similarly, \[T(\lambda p)=\dfrac{(\lambda p)-(\lambda p)(3)}{x-3}=\dfrac{\lambda p-\lambda p(3)}{x-3}=\lambda\dfrac{p-p(3)}{x-3}=\lambda Tp.\]If $x=3$, then $T$ is a composition of the differentiation map and evaluation map. Both of them are linear, hence $T$ is also linear. We can show it directly \[ T(\lambda p+q)=(\lambda p+q)'(3)=(\lambda p’+q’)(3)=\lambda p'(3)+q'(3)=\lambda Tp+Tq. \]Therefore $T$ is a linear map. Let us consider $Tx^n$ for $n\in \mb N^+$, if $x\ne 3$, \[T(x^n)=\frac{x^n-3^n}{x-3}=x^{n-1}+3x^{n-2}+\cdots+3^kx^{n-1-k}+\cdots+3^{n-1}\in\ca P(\R).\] Moreover, if $x=3$, we have $T(x^n)=3^{n-1}n$. Note that when $x=3$, it is true that\[ x^{n-1}+3x^{n-2}+\cdots+3^kx^{n-1-k}+\cdots+3^{n-1}=3^{n-1}n. \]We get \[T(x^n)=x^{n-1}+3x^{n-2}+\cdots+3^kx^{n-1-k}+\cdots+3^{n-1}\in\ca P(\R)\]for $x\in \R$. Similarly, we can show $T(1)=0\in \ca P(\R)$.

Since any polynomial of $\ca P(\R)$ is a linear combination of $1$, $x$, $x^2$, $\cdots$, it follows that $Tp \in\ca P(\R)$ for every polynomial $p \in\ca P(\R)$.

I am not sure the textbook indicates that $1$, $x$, $x^2$, $\cdots$ is a basis of $\ca P(\R)$, so I use some easier arguments such as any polynomial of $\ca P(\R)$ is a linear combination of $1$, $x$, $x^2$, $\cdots$.

9. Solution: If $f(z)=a_nz^n+\cdots+a_1z+a_0$, where $a_n,\cdots,a_0\in\C$, then \[\overline{f(\bar{z})}=\overline{a_n}z^n+\cdots+\overline{a_1}z+\overline{a_0}.\]That implies $\overline{f(\bar{z})}$ is a polynomial. As the product of polynomials is a polynomial as well, we conclude $q$ is a polynomial.

Now let us show $q$ has only real coefficients. Denote $q(z)$ by \[q(z)=\mu_{2n}z^{2n}+\cdots+\mu_1z+\mu_0.\]Note that $\overline{q(\bar{z})}=\overline{f(\bar{z})\overline{f(z)}}=\overline{f(\bar{z})}f(z)=q(z)$, it follows \[ \overline{\mu_{2n}}z^{2n}+\cdots+\overline{\mu_1}z+\overline{\mu_0}=\mu_{2n}z^{2n}+\cdots+\mu_1z+\mu_0. \]Hence $\overline{\mu_k}=\mu_k$, i.e $\mu_k\in \R$, for $k=0,\cdots,2n$.

Here you can also compute the coefficients $\mu_k$ in terms of $a_i$ and show $\mu_k=\overline{\mu_k}$ or use some calculus methods.

10. Note that $x_0,x_1,\cdots,x_m$ are distinct, we can define the polynomial\[f(x)=\sum_{j=0}^m\frac{(x-x_0)(x-x_1)\cdots(x-x_{j-1})(x-x_{j+1})\cdots(x-x_m)}{(x_j-x_0)(x_j-x_1)\cdots(x_j-x_{j-1})(x_j-x_{j+1})\cdots(x_j-x_m)}p(x_j).\]Then it is obvious that $f(x)\in\ca P_m(\C)$. Moreover, since $x_j$ and $p(x_j)$, $j=0,1,\cdots,m$, are real, it follows that the coefficients of $f(x)$ are real. Hence it suffices to show that $p(x)=f(x)$.

By plugging $x=x_i$ into $f(x)$, we have $f(x_i)=p(x_i)$ since all summands except one are zero (see the link in the remark below),
\]where $\delta_{ij}=0$ if $i\ne j$ and $\delta_{ij}=1$ if $i=j$.

This implies that $f-p$ has $m+1$ distinct zeros. Since $f-p\in\ca P_m(\C)$, it follows from 4.12 that the degree of $f-p$ can not be nonnegative. Hence $f-p$ is the zero polynomial, thus completing the proof.

The polynomial used here is called the Lagrange Interpolating Polynomial . Please see the following link for more detail.

11. By the division algorithm of polynomials in 4.8, we know that for every polynomial $f\in\ca P(\mb F)$ there exist unique polynomials $q$ and $r$ such that \[f=pq+r,\quad\text{and}\quad \deg r<\deg p.\]This implies that $\ca P(\mb F)=U\oplus \ca P_{\deg p-1}(\mb F)$.

Therefore\[\ca P(\mb F)/U\cong\ca P_{\deg p-1}(\mb F).\]It follows that $$\dim\ca P(\mb F)/U=\dim\ca P_{\deg p-1}(\mb F)=\deg p.$$Moreover, a basis of $\ca P(\mb F)/U$ is $1,x,x^2,\cdots,x^{\deg p -1}$.

Here I used the fact that if $V=U\oplus W$, then $V/U\cong W$. Please try to prove as the following alternative solution.

Another solution (explains the solution above more explicitly).

For any given polynomial $f\in\ca P(\mb F)$, let $r(f)$ be the reminder of $f$ divided by $q$. Note that $\deg r(f)<\deg q$we have a map $r:\ca P(\mb F)\to \ca P_{\deg p-1}(\mb F)$. One can check this is a linear map. Moreover, $\mathrm{Null}\, r=U$. By taking polynomials in $\ca P_{\deg p-1}(\mb F)$, we have that $\mathrm{range}\,r=\ca P_{\deg p-1}(\mb F)$.

By 3.91(d), we have that \[\ca P(\mb F)/U=\ca P(\mb F)/\mathrm{Null}\, r\cong \mathrm{range}\,r=\ca P_{\deg p-1}(\mb F).\]Now the problem is solved similarly.

About Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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