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Solution to Linear Algebra Hoffman & Kunze Chapter 4.3


Exercise 4.3.1

Use the Lagrange interpolation formula to find a polynomial $f$ with real coefficients such that $f$ has degree $\leq 3$ and $f(-1)=-6$, $f(0)=2$, $f(1)=-2$, $f(2)=6$.

Solution: $t_0=-1$, $t_1=0$, $t_2=1$, $t_3=2$. Therefore
\begin{alignat*}{2}
P_0&=\frac{x(x-1)(x-2)}{(-1)(-2)(-3)} =& {\textstyle \frac{-1}{6}}x(x-1)(x-2)\\
P_1&=\frac{(x+1)(x-1)(x-2)}{(-1)(-2)}=&{\textstyle \frac{1}{2}}(x-1)(x+1)(x-2)\\
P_2&=\frac{(x+1)x(x-2)}{(2)(1)(-1)}=&{\textstyle \frac{-1}{2}}x(x+1)(x-2)\\
P_3&=\frac{(x+1)x(x-1)}{(3)(2)(1)}=&{\textstyle \frac{1}{6}}(x-1)(x+1).
\end{alignat*}Thus
$$f=f(-1)\cdot P_0+f(0)\cdot P_1+f(1)\cdot P_2+f(2)\cdot P_3$$$$=-6P_0+2P_1-2P_2+6P_3$$$$=x(x-1)(x-2)+(x-1)(x+1)(x-2)+x(x+1)(x-2)+x(x-1)(x+1)$$$$=(x^3-3x^2+2x)+(x^3-2x^2-x+2)+(x^3-x^2-2x)+(x^3-x)$$$$=4x^3-6x^2-2x+2.$$Checking:
$$f(-1)=-4-6+2+2=-6$$$$f(0)=2$$$$f(1)=4-6-2+2=-2$$$$f(2)=32-24-4+2=6.$$


Exercise 4.3.2

Let $\alpha,\beta,\gamma,\delta$ be real numbers. We ask when it is possible to find a polynomial $f$ over $\Bbb R$, of degree not more than $2$, such that $f(-1)=\alpha$, $f(1)=\beta$, $f(3)=\gamma$ and $f(0)=\delta$. Prove that this is possible if and only if
$$3\alpha+6\beta-\gamma-8\delta=0.$$Solution: Let $t_0=-1$, $t_1=1$, $t_2=3$. We will apply the Lagrange interpolation formula to get a quadratic satisfying $f(t_0)=\alpha$, $f(t_1)=\beta$, $f(t_2)=\gamma$. Then we will figure out what condition on $\alpha,\beta,\gamma,\delta$ will guarantee that it also satisfies $f(0)=\delta$.
\begin{alignat*}{2}
P_0=&\frac{(x-1)(x-3)}{(-2)(-4)}=&{\textstyle \frac18}(x-1)(x-3)\\
P_1=&\frac{(x+1)(x-3)}{(2)(-2)}=&{\textstyle \frac{-1}{4}}(x+1)(x-3)\\
P_2=&\frac{(x+1)(x-1)}{(4)(2)}=&{\textstyle \frac{1}{8}}(x+1)(x-1)
\end{alignat*}Therefore
$$f=\frac{\alpha}{8}(x-1)(x-3)-\frac{\beta}{4}(x+1)(x-3)+\frac{\delta}{8}(x+1)(x-1)$$$$={\textstyle\frac18}(\alpha x^2-4\alpha x+3\alpha-2\beta x^2+4\beta x+6\beta +\gamma x^2-\gamma).$$Now $f(0)=\gamma$ implies
$$\frac18(3\alpha+6\beta-\gamma)=\delta.$$Simplifying gives
\begin{equation}
3\alpha+6\beta-\gamma-8\delta=0.
\label{f2222f3}
\end{equation}Thus if (\ref{f2222f3}) is satisfied then the four values of $f$ are as required. Since three points determine a quadratic, there cannot be any quadratic other than $f$ that goes through $(-1,\alpha)$, $(1,\beta)$, $(3,\delta)$. Thus this condition is not only sufficient but it is necessary.


Exercise 4.3.3

Let $F$ be the field of real numbers,
\begin{alignat*}{1}
A=&\left[\begin{array}{cccc}
2 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & 0 & 3 & 0\\
0 & 0 & 0 & 1
\end{array}\right]\\
p=&(x-2)(x-3)(x-1).
\end{alignat*}(a) Show that $p(A)=0$.
(b) Let $P_1$, $P_2$, $P_3$ be the Lagrange polynomials for $t_1=2$, $t_2=3$, $t_2=1$. Compute $E_i=P_i(A)$, $i=1,2,3$.
(c) Show that $E_1+E_2+E_3=I$, $E_iE_j=0$ if $i\not=j$, $E_i^2=E_i$.
(d) Show that $A=2E_1+3E_2+E_3$.

Solution:

(a) We have
$$(A-2)(A-3)(A-1)$$$$=\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{array}\right]
\left[\begin{array}{cccc}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & -2
\end{array}\right]
\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 0
\end{array}\right]$$$$=
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{array}\right]
\left[\begin{array}{cccc}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]=
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right].
$$

(b) $t_1=2$, $t_2=3$, $t_3=1$.
\begin{alignat*}{1}
P_1=&-(x-3)(x-1)\\
P_2=&{\textstyle\frac12}(x-2)(x-1)\\
P_3=&{\textstyle\frac12}(x-2)(x-3)
\end{alignat*}Thus
$$
E_1=P_1(A)=-(A-3I)(A-I)\\
=-
\left[\begin{array}{cccc}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & -2
\end{array}\right]
\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\\
=
\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\\
$$$$
E_2=P_2(A)=\frac12
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{array}\right]
\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\\
=
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{array}\right]\\
$$$$
E_3=P_3(A)=\frac12
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & -1
\end{array}\right]
\left[\begin{array}{cccc}
-1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & -2
\end{array}\right]\\
=
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right].
$$(c) All three of these facts are basically obvious by visual inspection of the matrices in part (b). $E_1+E_2+E_3=I$ is obvious by inspection. Likewise it is evident by inspection that $E_iE_j=0$ if $i\not=j$. Lastly it is obvious that $E_i^2=E_i$. I'm not sure what there is to prove here.

(d) We have
$$2E_1+3E_2+E_3$$$$=
2\left[\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0
\end{array}\right]
+3
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0
\end{array}\right]
+
\left[\begin{array}{cccc}
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right]
$$$$=
\left[\begin{array}{cccc}
2 & 0 & 0 & 0\\
0 & 2 & 0 & 0\\
0 & 0 & 3 & 0\\
0 & 0 & 0 & 1
\end{array}\right].
$$


Exercise 4.3.4

Let $p=(x-2)(x-3)(x-1)$ and let $T$ be any linear operator on $\mathbb R^4$ such that $p(T)=0$. Let $P_1$, $P_2$, $P_3$ be the Lagrange polynomials of Exercise 3, and let $E_i=P_i(T)$, $i=1,2,3$. Prove that
$$E_1+E_2+E_3=I,\quad E_iE_j=0 \text{ if } i\not=j,$$$$E_i^2=E_i,\quad\text{and}\quad T=2E_1+3E_2+E_3.$$Solution: Recall
\begin{alignat*}{1}
P_1=&-x^2+4x-3\\
P_2=&{\textstyle\frac12}x^2-{\textstyle\frac32}x+1\\
P_2=&{\textstyle\frac12}x^2-{\textstyle\frac52}x+3
\end{alignat*}By definition $P(T)+Q(T)=(P+Q)(T)$ and $P(T)Q(T)=(PQ)(T)$.  Now as polynomials $P_1+P_2+P_3=1$. Thus $$E_1+E_2+E_3=P_1(T)+P_2(T)+P_3(T)=(P_1+P_2+P_3)(T)=I.$$Notice that $P$ divides $P_iP_j$ whenever $i\not=j$. Thus $P_iP_j=PQ$ for some $Q$ (as polynomials). Thus $$E_iE_j=P_i(T)P_j(T)=(P_iP_j)(T)=P(T)Q(T)=0\cdot Q(T)=0.$$ Thus $E_iE_j(T)=0$.

We prove the next part in general. Let
\begin{alignat*}{1}
f_1=&x-a\\
f_2=&x-b\\
f_3=&x-c
\end{alignat*}Thus
\begin{alignat*}{1}
P_1=&\frac{f_2f_3}{(a-b)(a-c)}\\
P_2=&\frac{f_1f_3}{(b-a)(b-c)}\\
P_3=&\frac{f_1f_2}{(c-a)(c-b)}\\
\end{alignat*}Let $d=2c-b-a$. Then it follows by simply multiplying it out that
$$\frac{df_3+f_3^2}{(c-a)(c-b)}=\frac{f_1f_2}{(c-a)(c-b)}-1.$$Which is equivalent to
\begin{equation}
\frac{df_3+f_3^2}{(c-a)(c-b)}=P_3-1.
\label{fff2303r}
\end{equation}This equation is true as polynomials. We now evaluate things at $T$.
$$f_1(T)f_2(T)f_3(T)=0$$$$\Longrightarrow f_1(T)f_2(T)f_3^2(T)=0$$$$\Longrightarrow\frac{f_1(T)f_2(T)}{(c-a)(c-b)}\cdot\frac{f_3^2(T)}{(c-a)(c-b)}=0.$$Since $f_1(T)f_2(T)f_3(T)=0$, this implies
$$\frac{f_1(T)f_2(T)}{(c-a)(c-b)}\cdot\frac{df_3(T)+f_3^2(T)}{(c-a)(c-b)}=0.$$Equivalently
$$P_3(T)\cdot\frac{df_3(T)+f_3^2(T)}{(c-a)(c-b)}=0.$$By (\ref{fff2303r}) this implies
$$P_3(T)(P_3(T)-1)=0.$$Thus
$$P_3^2(T)=P_3(T).$$Thus $E_3^2=E_3$. Since $a,b,c$ were general, the same follows for $E_1$ and $E_2$.

It remains to show $T=2E_1+3E_2+E_3$. We first note that as polynomials
$$2P_1+3P_2+P_3$$$$=(-2x^2+8x-6)+({\textstyle\frac32}x^2-{\textstyle\frac92}x+3)+({\textstyle\frac12}x^2-{\textstyle\frac52}x+3)=x.$$Plugging in $T$ we get
$$2P_1(T)+3P_2(T)+P_3(T)=T.$$Thus
$$2E_1+3E_2+E_3=T.$$


Exercise 4.3.5

Let $n$ be a positive integer and $F$ a field. Suppose $T$ is an $n\times n$ matrix over $F$ and $P$ is an invertible $n\times n$ matrix over $F$. If $f$ is any polynomial over $F$, prove that
$$f(P^{-1}TP)=P^{-1}f(T)P.$$Solution: First note that $(P^{-1}xP)^n=P^{-1}x^n(T)P.$ This is obvious by inspection, it follows basically from the fact that multiplication is associative and $P^{-1}P=I$.

The general result now follows
$$P^{-1}f(T)P=P^{-1}(a_0+a_1T+a_2T^2+\cdots+a_nT^n)P$$$$=P^{-1}a_0P+P^{-1}a_1TP+P^{-1}a_2T^2P+\cdots+P^{-1}a_nT^nP$$$$=a_0+a_1(P^{-1}TP)+a_2(P^{-1}TP)^2+\cdots+a_n(P^{-1}TP)^n=f(P^{-1}TP).$$


Exercise 4.3.6

Let $F$ be a field. We have considered certain special linear functionals on $F[x]$ obtained via `evaluation at $t$':
$$L(f)=f(t).$$Such functionals are not only linear but also have the property that $L(fg)=L(f)L(g)$. Prove that if $L$ is any linear functional on $F[x]$ such that
$$L(fg)=L(f)L(g)$$for all $f$ and $g$, then either $L=0$ or there is a $t$ in $F$ such that $L(f)=f(t)$ for all $f$.

Solution: Let $L$ be a non-zero linear transformation. First note that $L(1)\not=0$ since otherwise $$L(f)=L(f\cdot 1)=L(f)L(1)=L(f)\cdot 0=0$$ for all $f$. Next note that $L(1)=L(1\cdot 1)=L(1)L(1)$ $\Rightarrow$ $L(1)=1$. It follows that $$L(a)=L(a\cdot1)=aL(1)=a$$ for all $a\in F$. Now let $t=L(x)$. Let $f(x)=a_0+a_1x+\cdots+a_nx^n$. Then
$$L(f)=L(a_0+a_1x+\cdots+a_nx^n)$$$$=L(a_0)+L(a_1)L(x)+L(a_2)L(x^2)\cdots+L(a_n)L(x^n)$$$$=a_0+a_1L(x)+a_2L(x)^2+\cdots+a_nL(x)^n$$$$=a_0+a_1t+a_2t^2+\cdots+a_nt^n=f(t).$$

From http://greggrant.org

Linearity

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This Post Has 2 Comments

  1. I do not understand this line in Exercise 4.3.6 solution:
    L(a) = L(a*1) = aL(1) = a
    So L(a*1) = L(a)L(1), but why is that aL(1)? Isn't there an assumption of L(a) = a, the very thing it is trying to prove?

    1. Because $L$ is also linear.

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