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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.


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.

This Post Has 6 Comments

  1. Number 8 is an interesting one. It shows that in order to calculate the derivative of a polynomial p(x) at point x_0, it's enough to
    1. Take the quotient q(x) of division of p(x) by x-x_0: p(x)=q(x)(x-x_0) + r
    2. Evaluate q(x) at the point x_0

    This was unexpected for me. Idk if this fact has any useful applications though.

  2. Problem 11 has a problem.
    The element of a basis of P(F) is not polynomial, but space.

    1. I have no idea what you are talking about. The space $\mathcal P(\mathbf F)$ is the set of all polynomials. You mean elements of $\mathcal P(\mathbf F)$ are spaces?

  3. The another altenative solution for 10 is by deriving the result of 5.
    There is a polynomial p of real coefficients with m+1 distinct values in Pm(R) and p must also be in Pm(C). And since p must be uniquely defined in Pm(C), we have arrived at what is needed to be proven.

  4. Did you choose not to provide a solution to #11? I found one, though it involved proving P(F)/U isomorphic to P_(m-1) (F) by defining a linear transformation T(p)=r where r is the remainder defined by the division algorithm. I was wondering if there was a simpler method.

    1. That is the essential part of this problem.

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