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## Solution to Linear Algebra Hoffman & Kunze Chapter 4.2

#### Exercise 4.2.1

Let $F$ be a subfield of the complex numbers and let $A$ be the following $2\times 2$ matrix over $F$
$$A=\left[\begin{array}{cc}2&1\\-1&3\end{array}\right].$$For each of the following polynomials $f$ over $F$, compute $f(A)$.

(a) $f=x^2-x+2$;
(b) $f=x^3-1$;
(c) $f=x^2-5x+7$;

Solution:

(a) we have
$$A^2=\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]\cdot\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]$$$$=\left[\begin{array}{cc}3&5\\-5&8\end{array}\right]$$Therefore
$$f(A)=A^2-A+2$$$$=\left[\begin{array}{cc}3&5\\-5&8\end{array}\right]-\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]+\left[\begin{array}{cc}2&0\\0&2\end{array}\right]$$$$=\left[\begin{array}{cc}3&4\\-4&7\end{array}\right]$$(b) We have
$$A^2=\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]\cdot\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]\cdot\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]$$$$=\left[\begin{array}{cc}3&5\\-5&8\end{array}\right]\cdot\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]$$$$=\left[\begin{array}{cc}1&18\\-18&19\end{array}\right]$$Therefore
$$f(A)=A^3-1$$$$=\left[\begin{array}{cc}1&18\\-18&19\end{array}\right] – \left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$$$=\left[\begin{array}{cc}0&18\\-18&18\end{array}\right]$$(c) We have
$$f(A)=A^2-5A+7$$$$=\left[\begin{array}{cc}3&5\\-5&8\end{array}\right] – 5\left[\begin{array}{cc}2&1\\-1&3\end{array}\right]+\left[\begin{array}{cc}7&0\\0&7\end{array}\right]$$$$=\left[\begin{array}{cc}0&0\\0&0\end{array}\right].$$

#### Exercise 4.2.2

Let $T$ be the linear operator on $\mathbb R^3$ defined by
$$T(x_1,x_2,x_3)=(x_1,x_3,-2x_2-x_3).$$Let $f$ be the polynomial over $\mathbb R$ defined by $f=-x^3+2$. Find $f(T)$.

Solution: The matrix of $T$ with respect to the standard basis is
$$A=\left[\begin{array}{ccc}1&0&0\\0&0&1\\0&-2&-1\end{array}\right]$$Thus
$$A^2=\left[\begin{array}{ccc}1&0&0\\0&0&1\\0&-2&-1\end{array}\right]\cdot\left[\begin{array}{ccc}1&0&0\\0&0&1\\0&-2&-1\end{array}\right]$$$$=\left[\begin{array}{ccc}1&0&0\\0&-2&-1\\0&2&-1\end{array}\right]$$So
$$A^3=\left[\begin{array}{ccc}1&0&0\\0&-2&-1\\0&2&-1\end{array}\right]\cdot\left[\begin{array}{ccc}1&0&0\\0&0&1\\0&-2&-1\end{array}\right]$$$$=\left[\begin{array}{ccc}1&0&0\\0&2&-1\\0&2&3\end{array}\right]$$Thus
$$-A^3+2=-\left[\begin{array}{ccc}1&0&0\\0&2&-1\\0&2&3\end{array}\right]+\left[\begin{array}{ccc}2&0&0\\0&2&0\\0&0&2\end{array}\right]$$$$=\left[\begin{array}{ccc}-1&0&0\\0&0&1\\0&-2&-1\end{array}\right].$$This corresponds to the transformation$$f(T)(x_1,x_2,x_3)=(-x_1,x_3,-x_2-x_3).$$

#### Exercise 4.2.3

Let $A$ be an $n\times n$ diagonal matrix over the field $F$, i.e., a matrix satisfying $A_{ij}=0$ for $i\not=j$. Let $f$ be the polynomial over $F$ defined by
$$f=(x-A_{11})\cdots(x-A_{nn}).$$What is the matrix $f(A)$?

Solution: The product of $n$ diagonal matrices is again a diagonal matrix, where the $i,i$ element is the product of the entries in the $i,i$ position of the $n$ individual matrices. For each $i$, the $i,i$ terms of $A-A_{ii}$ is zero. Therefore each diagonal entry of $(A-A_{11})\cdots(A-A_{nn})$ is a product of numbers one of which is zero. Therefore $(A-A_{11})\cdots(A-A_{nn})$ is the zero matrix.

#### Exercise 4.2.4

If $f$ and $g$ are independent polynomials over the field $F$ and $h$ is a non-zero polynomial over $F$, show that $fh$ and $gh$ are independent.

Solution: Suppose there are scalars $r,s\in F$ such that $rfh+sgh=$. Then $rfh=-sgh$ and so by Corollary 2 on page 121, it follows that $rf=-sg$ so that $rf+sg=0$. Since $f$ and $g$ are independent, it follows that $r=s=0$. Thus $fh$ and $gh$ are independent.

#### Exercise 4.2.5

If $F$ is a field, show that the product of two non-zero elements of $F^{\infty}$ is non-zero.

Solution: Let $f=(f_0,f_1,f_2,\dots)$ and $g=(g_0,g_1,g_2,\dots)$ be two elements of $F^{\infty}$. Let $n$ be the index of the first non-zero $f_i$ and let $m$ be the index of the first non-zero $g_i$ (could be $n=m=0$). Then what is the $n+m$ coordinate of the product? It is
$$(fg)_{n+m}=\sum_{i=0}^{n+m}f_ig_{n+m-i}=\sum_{i=n}^{n+m}f_ig_{n+m-i}$$and if $i>n$ then $g_{n+m-i}=0$ thus
$$\sum_{i=n}^{n+m}f_ig_{n+m-i}=\sum_{i=n}^{n}f_ig_{n+m-i}=f_ng_m.$$Now we’ve assumed $f_n$ and $g_m$ are non-zero thus $f_ng_m\not=0$ and thus $fg\not=0$ in $F^{\infty}$.

#### Exercise 4.2.6

Let $S$ be a set of non-zero polynomials over a field $F$. If no two elements of $S$ have the same degree, show that $S$ is an independent set in $F[x]$.

Solution: Let $f_1,\dots,f_n\in S$. Suppose deg$(f_i)=d$ and deg$(f_j)<d$ $\forall$ $i\not=j$. We can do this because by assumption all polynomials in the set $\{f_1,\dots,f_n\}$ have different degrees, so one of them must have the largest. Suppose the $d$-th coefficient of $f_i$ is $r\not=0$. Then any linear combination $a_1f_1+\cdots a_nf_n$ has $d$-th coefficient equal to $ra_i$. Thus if this linear combination is zero then necessarily $a_i=0$. We can now apply the same argument to the $f_j$ with the second largest degree to show its coefficient in the linear combination is zero. And then to the third largest, etc. Until we have eventually shown all coefficients in the linear combination are zero. It follows that $\{f_1,\dots,f_n\}$ is a linearly independent subset of $S$ and since it was an arbitrary finite subset of $S$ it follows that $S$ is a linearly independent set.

#### Exercise 4.2.7

If $a$ and $b$ are elements of a field $F$ and $a\not=0$, show that the polynomials $1$, $ax+b$, $(ax+b)^2$, $(ax+b)^3$, $\dots$ form a basis of $F[x]$.

Solution: Let $S=\{1, ax+b, (ax+b)^2, (ax+b)^3,\dots\}$. And let $\langle S\rangle$ be the subspace spanned by $S$. By the previous exercise we know $S$ is a linearly independent set. We must just show $S$ spans the space of all polynomials. Since $1\in S$ and $ax+b\in S$ it follows that $b\cdot 1+\frac1a(a+bx)\in \langle S\rangle$. Thus $x\in\langle S\rangle$. Now we can subtract a multiple of $1$ and a multiple of $x$ from $(a+bx)^2$ to get $a^2x^2\in\langle S\rangle$. Thus $\frac1{a^2}\cdot a^2x^2\in\langle S\rangle$. Thus $x^2\in S$. Continuing in this way we can show that $x^n\in\langle S\rangle$ for all $n$. Since $\{1,x,x^2,\dots\}$ span the space of all polynomials, it follows that $S$ spans the space of all polynomials.

#### Exercise 4.2.8

If $F$ is a field and $h$ is a polynomial over $F$ of degree $\geq1$, show that the mapping $f\rightarrow f(h)$ is a one-one linear transformation of $F[x]$ into $F[x]$. Show that this transformation is an isomorphism of $F[x]$ onto $F[x]$ if and only if deg $h=1$.

Solution: Let $G:F[x]\rightarrow F[x]$ be the function $G(f)=f(h)$. Clearly $G$ is a well-defined function from $F[x]$ to $F[x]$. By definition $$G(f+g)=(f+g)(h)=f(h)+g(h)=G(f)+G(g)$$ and for $r\in F$, $$G(rf)=(rf)(h)=r\cdot f(h)=rG(F).$$ Thus $G$ is a linear transformation. Suppose deg $h>1$. Then the coefficient of $x$ in $f(h)$ is zero. Thus if deg $h>1$ then $G$ is not onto. Now suppose deg $h=0$. Then $f(h)$ is a scalar for all $f$. Thus $G$ is not onto. Now suppose deg $h=1$, so that $h(x)=a+bx$. Let $h’= \frac1bx-\frac ab$ and let $G’$ be the corresponding function on $F[x]$, so $G’:F[x]\rightarrow F[x]$ is given by $G(f)=f(\frac1bx-\frac ab)$. Then $G\circ G’$ and $G’\circ G$ both give the identify function on $F[x]$. Thus $G$ is an isomorphism.

Exercise 4.2.9

Let $F$ be a subfield of the complex numbers and let $T$, $D$ be the transformations on $F[x]$ defined by
$$T\left(\sum_{i=0}^nc_ix^i\right)=\sum_{i=0}^n\frac{c_i}{1+i}x^{i+1}$$and
$$D\left(\sum_{i=0}^nc_ix^i\right)=\sum_{i=1}^nic_ix^{i-1}.$$(a) Show that $T$ is a non-singular linear operator on $F[x]$. Show also that $T$ is not invertible.
(b) Show that $D$ is a linear operator on $F[x]$ and find its null space.
(c) Show that $DT=I$, and $TD\not=I$.
(d) Show that $T[(Tf)g]=(Tf)(Tg)-T[f(Tg)]$ for all $f,g$ in $F[x]$.
(e) State and prove a rule for $D$ similar to the one given for $T$ in (d)
(f) Suppose $V$ is a non-zero subspace of $F[x]$ such that $Tf$ belongs to $V$ for each $f$ in $V$. Show that $V$ is not finite-dimensional.
(g) Suppose $V$ is a finite-dimensional subspace of $F[x]$. Prove there is an integer $m\geq0$ such that $D^mf=0$ for each $f$ in $V$.

Solution:

(a) Clearly $T$ is a function from $F[x]$ to $F[x]$. We must show $T$ is linear.
$$T\left(\sum_{i=0}^nc_ix^i+\sum_{i=0}^nc’_ix^i\right)=T\left(\sum_{i=0}^n(c_i+c’_i)x^i\right)$$$$=\sum_{i=0}^n\frac{c_i+c’_i}{i+1}x^{i+1}=\sum_{i=0}^n\left(\frac{c_i}{1+i}x^{i+1}+\frac{c’_i}{1+i}x^{i+1}\right)$$$$=T\left(\sum_{i=0}^nc_ix^i\right)+T\left(\sum_{i=0}^nc’_ix^i\right).$$and
$$T\left(r\sum_{i=0}^nc_ix^i\right)=T\left(\sum_{i=0}^nrc_ix^i\right) =\sum_{i=0}^n\frac{rc_i}{i+1}x^{i+1} =r\sum_{i=0}^n\frac{c_i}{i+1}x^{i+1} =r\cdot T\left(\sum_{i=0}^nc_ix^i\right).$$Thus $T$ is linear.

Since $F$ has characterisitc zero we can find $a,b\in F$, such that $a\not=b$. Consider $a$ and $b$ as constant polynomials in $F$. Then $T(a)=T(b)=0$. Thus $T$ is not one-to-one. Thus $T$ is not invertible.

(b) Clearly $D$ is a function from $F[x]$ to $F[x]$. We must show $D$ is linear.
$$D\left(\sum_{i=0}^nc_ix^i+\sum_{i=0}^nc’_ix^i\right)=D\left(\sum_{i=0}^n(c_i+c’_i)x^i\right)$$$$=\sum_{i=1}^ni(c_i+c’_i)x^{i-1}=\sum_{i=1}^n(ic_ix^{i-1}+ic’_ix^{i-1})$$$$=D\left(\sum_{i=0}^nc_ix^i\right)+D\left(\sum_{i=0}^nc’_ix^i\right).$$and
$$D\left(r\sum_{i=0}^nc_ix^i\right)=D\left(\sum_{i=0}^nrc_ix^i\right) =\sum_{i=1}^nrc_ix^{i-1} =r\sum_{i=1}^nc_ix^{i-1} =r\cdot D\left(\sum_{i=0}^nc_ix^i\right).$$Thus $D$ is linear.

Suppose $f(x)=\sum_{i=0}^nc_ix^i$ is in the null space of $D$. Then
$$D(f)=\sum_{i=1}^nic_ix^{i-1}=0.$$ A polynomial is zero if and only if every coefficient is zero. Thus it must be that $0=c_1=c_2=c_3=\cdots$. So it must be that $f(x)=c_0$ a constant polynomial. Thus the null space of $D$ contains the constant polynomials. Since $D(f)=0$ for all constant polynomials, the null space of $D$ consists of exatly the constant polynomials.

(c)Note that$$D\left(T\left(\sum_{i=0}^nc_ix^i\right)\right)=D\left(\sum_{i=0}^n\frac{c_i}{1+i}x^{i+1}\right).$$The first non-zero term of this sum is the linear term $c_0x$. Thus when we apply $D$ the sum still starts at zero:
$$=\sum_{i=0}^n(i+1)\frac{c_i}{1+i}x^{i+1-1}=\sum_{i=0}^nc_ix^i.$$Thus
$$D\left(T\left(\sum_{i=0}^nc_ix^i\right)\right)=\sum_{i=0}^nc_ix^i.$$Thus $DT=I$.

Let $f(x)=1$. Then $TD(f)=T(D(f))=T(0)=0$. Thus $TD(1)\not=1$. Thus $TD\not=I$.

(d) This follows rather easily from part (e). And likewise (e) follows rather easily from (d). Thus one can derive (d) straight from the definition of $T$ and then derive (e) from it, or one can derive (e) straight from the definition of $D$ and then derive (d) from it. I’ve chosen to do the latter.

In part (e) below the product formula is proven straight from the definition. So we will use it here to prove this part. In particular, we apply the product formula from part (e) to $(Tf)(Tg)$
$$D[(Tf)(Tg)]=(Tg)D(Tf)+(Tf)D(Tg).$$By part (c) $DT=I$ so this is equivalent to
$$D[(Tf)(Tg)]=f(Tg)+(Tf)g.$$Thus
$$D[(Tf)(Tg)]-f(Tg)=(Tf)g.$$Now apply $T$ to both sides
$$T\left(\rule{0mm}{4mm}D[(Tf)(Tg)]-f(Tg)\right)=T((Tf)g).$$Since $T$ is a linear transformation this is equivalent to
$$T\left(\rule{0mm}{4mm}D[(Tf)(Tg)]\right)-T[f(Tg)]=T((Tf)g).$$We showed in part (c) that $TD\not=I$, however if $f$ has constant term equal to zero then in fact $T(D(f))$ does equal $f$. Now $Tf$ and $Tg$ have constant term equal to zero, so $(Tf)(Tg)$ has constant term zero, thus
$$T\left(\rule{0mm}{4mm}D[(Tf)(Tg)]\right)=(Tf)(Tg).$$Thus
$$(Tf)(Tg)-T[f(Tg)]=T((Tf)g).$$(e) I believe they are after the product formula here:
\begin{equation}
D(fg)=fD(g)+gD(f).
\label{abbdwefe}
\end{equation}We prove this by brute force appealing just to the definition and to the product formula for polynomials. Let $f(x)=\sum_{i=0}^nc_ix^i$ and $g(x)=\sum_{i=0}^md_ix^i$. Then using the product formula (4-8) on page 121 we have
$$D(fg)=D\left(\sum_{i=0}^nc_ix^i\sum_{i=0}^md_ix^i\right) =D\left(\sum_{i=0}^{n+m}\left(\sum_{j=0}^ic_jd_{i-j}\right)x^i\right)$$And using the linearity of the differentiation operator $D$ this equals
\begin{equation}
\sum_{i=0}^{n+m}i\left(\sum_{j=0}^ic_jd_{i-j}\right)x^{i-1}.
\label{j203r2}
\end{equation}Now we write down the sum for the right hand side of (\ref{abbdwefe}):
$$fD(g)+gD(f)$$\begin{equation}
=\sum_{i=0}^nc_ix^i\sum_{i=1}^mid_ix^{i-1} + \sum_{i=1}^nic_ix^{i-1}\sum_{i=0}^md_ix^i.
\label{f320r}
\end{equation}Consider
$$x\cdot\sum_{i=0}^nc_ix^i\sum_{i=1}^mid_ix^{i-1}.$$This equals
$$\sum_{i=0}^nc_ix^i\sum_{i=1}^mid_ix^i$$and since $0d_0=0$ we can write this as
$$\sum_{i=0}^nc_ix^i\sum_{i=0}^mid_ix^i.$$It’s straightforward to apply (4-8) page 121 to this product. In (4-8) we let $f_i=c_i$ and $g_i=id_i$ and it equals
$$\sum_{i=0}^{m+n}\left(\sum_{j=0}^i(i-j)c_jd_{i-j}\right)x^i.$$ The constant terms is zero thus we can write it as
$$\sum_{i=1}^{m+n}\left(\sum_{j=0}^i(i-j)c_jd_{i-j}\right)x^i.$$And thus the sum
$$\sum_{i=0}^nc_ix^i\sum_{i=1}^mid_ix^{i-1}$$equals
$$\sum_{i=1}^{m+n}\left(\sum_{j=0}^i(i-j)c_jd_{i-j}\right)x^{i-1}.$$Similary the second sum is
$$\sum_{i=1}^{n+m}\left(\sum_{j=0}^ijc_jd_{i-j}\right)x^i.$$Thus (\ref{f320r}) does equal (\ref{j203r2}).

(f) Suppose $V$ is finite dimensional. Let $\{b_1,\cdots,b_n\}$ be a basis for $V$. Let $d=\max_{i=1,\cdots n} \deg(b_i)$. It follows from Theorem 1(v) and induction that the degree of a linear combination of polynomials is no larger than the max of the degress of the individual polynomials involved in the linear combination. Thus no element of $V$ has degree greater than $d$. Now let $f\in V$ be any non-zero element. Let $d’=\deg(f)$. Then $Tf$ has degree $d’+1$, $T^2f$ has degree $d’+2$, etc. Thus for some $n$, $\deg(T^nf)>d$. If $T^nf\in V$ then this is a contradiction. Thus if $T^nf\in V$ for all $f\in V$ it must be that $V$ is not finite dimensional.

(g) Let $\{b_1,\cdots,b_n\}$ be a basis for $V$. Let $d=\max_{i=1,\cdots n} \deg(b_i)$. For any $f\in F[x]$, we know $\deg(Df)<\deg(f)$. Thus $D^{d+1}b_i=0$ for all $i=1,\dots,n$. Since $D^{d+1}(b_i)=0$ for all elements of the basis $\{b_1,\dots,b_n\}$ it follows that $D^{d+1}(f)=0$ for all $f\in V$.