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Solution to Linear Algebra Hoffman & Kunze Chapter 3.6


Exercise 3.6.1

Let $n$ be a positive integer and $F$ a field. Let $W$ be the set of all vectors $(x_1,\dots,x_n)$ in $F^n$ such that $x_1+\cdots+x_n=0$.

(a) Prove that $W^0$ consists of all linear functionals $f$ of the form
$$f(x_1,\dots,x_n)=c\sum_{j=1}^nx_j.$$(b) Show that the dual space $W^{*}$ of $W$ can be 'naturally' identified with the linear functionals
$$f(x_1,\dots,x_n)=c_1x_1+\cdots+c_nx_n$$on $F^n$ which satisfy $c_1+\cdots+c_n=0$.

Solution:

(a) Let $g$ be the functional $g(x_1,\dots,x_n)=x_1+\cdots+x_n$. Then $W$ is exactly the kernel of $g$. Thus $\text{dim}(W)=n-1$. Let $\alpha_i=\epsilon_1 - \epsilon_{i+1}$ for $i=1,\dots,n-1$. Then $\{\alpha_1,\dots,\alpha_{n-1}\}$ are linearly independent and are all in $W$ so they must be a basis for $W$. Let $f(x_1,\dots,x_n)=c_1x_1+\cdots+c_nx_n$ be a linear functional. Then $f\in W^0$ $\Rightarrow$ $f(\alpha_1)=\cdots=f(\alpha_n)=0$ $\Rightarrow$ $c_1-c_i=0$ $\forall$ $i=2,\dots,n$ $\Rightarrow$ $\exists$ $c$ such that $c_i=c$ $\forall$ $i$. Thus $f(x_1,\dots,x_n)=c(x_1+\cdots+x_n)$.

(b) Consider the sequence of functions
$$W\rightarrow (F^n)^{*} \rightarrow W^{*}$$where the first function is
$$(c_1,\dots,c_n)\mapsto f_{c_1,\dots,c_n}$$where $f_{c_1,\dots,c_n}(x_1,\dots,x_n)=c_1x_1+\cdots c_nx_n$ and the second function is restriction from $F^n$ to $W$. We know both $W$ and $W^{*}$ have the same dimension. Thus if we show the composition of these two functions is one-to-one then it must be an isomorphism. Suppose $(c_1,\dots,c_n)\in W\mapsto f_{c_1,\dots,c_n}=0\in W^{*}$. Then $\sum c_i=0$ and $\sum c_ix_i=0$ for all $(x_1,\dots,x_n)\in W$. In other words $\sum c_i=0$ and $\sum c_ix_i=0$ for all $(x_1,\dots,x_n)$ such that $\sum x_i=0$.

Let $\{\alpha_1,\dots,\alpha_{n-1}\}$ be the basis for $W$ from part (a). Then $f_{c_1,\dots,c_n}(\alpha_i)=0$ $\forall$ $ i=1,\dots,n-1$; which implies $c_1=c_i$ $\forall$ $i=2,\dots,n$. Thus $\sum c_i=(n-1)c_1$. But $\sum c_i=0$, thus $c_1=0$. Thus $f_{c_1,\dots,c_n}$ is the zero function.

Thus the mapping $W\rightarrow W^{*}$ is a natural isomorphism. We therefore naturally identify each element in $W^{*}$ with a linear functional $f(x_1,\dots,x_n)=c_1x_1+\cdots c_nx_n$ where $\sum c_i=0$.


Exercise 3.6.2

Use Theorem 20 to prove the following. If $W$ is a subspace of a finite-dimensional vector space $V$ and if $\{g_1,\dots,g_r\}$ is any basis for $W^0$, then
$$W=\bigcap_{i=1}^r N_{g_i}.$$ Solution: $\{g_1,\dots,g_r\}$ a basis for $W^0$ $\Rightarrow$ $g_i\in W^0$ $\forall$ $i$ $\Rightarrow$ $g_i(W)=0$ $\forall$ $i$ $\Rightarrow$ $W\subseteq N_{g_i}$ $\forall$ $i$ $\Rightarrow$ $W\subseteq\bigcap_{i=1}^r N_{g_i}$.

Let $n=\text{dim}(V)$. By Theorem 2, page 71, we know $\text{dim}(N_{g_1})=n-1$. Since the $g_i$'s are linearly independent, $g_2$ is not a multiple of $g_1$, thus by Theorem 20, $N_{g_1}\not\subseteq N_{g_2}$. Thus $\text{dim}(N_{g_1}\cap N_{g_2})\leq n-2$. By Theorem 20 again, $N_{g_3}\not\subseteq N_{g_1}\cap N_{g_2}$ since $g_3$ is not a linear combination of $g_1$ and $g_2$. Thus $$\text{dim}(N_{g_1}\cap N_{g_2}\cap N_{g_3})\leq n-3.$$ By induction $\text{dim}(\bigcap_{i=1}^r N_{g_i})\leq n-r$. Now by Theorem 16, $\text{dim}(W)=n-r$. Thus since $W\subseteq\bigcap_{i=1}^r N_{g_i}$, it follows that $\text{dim}(\bigcap_{i=1}^r N_{g_i})\geq n-r$. Thus it must be that $\text{dim}(\bigcap_{i=1}^r N_{g_i})=n-r$ and it must be that $W=\bigcap_{i=1}^r N_{g_i}$ since we have shown the left hand side is contained in the right hand side and both sides have the same dimension.


Exercise 3.6.3

Let $S$ be a set, $F$ a field, and $V(S;F)$ the space of all functions from $S$ into $F$:
$$(f+g)(x)=f(x)+g(x)$$$$(cf)(x)=cf(x).$$Let $W$ be any $n$-dimensional subspace of $V(S;F)$. Show that there exist points $x_1,\dots,x_n$ in $S$ and functions $f_1,\dots,f_n$ in $W$ such that $f_i(x_j)=\delta_{ij}$.

Solution: I'm not sure using the double dual is really the easiest way to prove this. It can be done rather easily directly by induction on $n$ (in fact see question 121704 on math.stackexchange.com). However, since H\&K clearly want this done with the double dual. At first glance you might try to think of $W$ as a dual on $S$ and $W^{*}$ as the double dual somehow. But that doesn't work since $S$ is just a set. Instead I think you have to consider the double dual of $W$, $W^{**}$ to make it work. I came up with the following solution.

Let $s\in S$. We first show that the function
\begin{alignat*}{1}
\phi_s:&W\rightarrow F\\
&w\mapsto w(s)
\end{alignat*}is a linear functional on $W$ (in other words for each $s$, we have $\phi_s\in W^*$).

Let $w_1,w_2\in W$, $c\in F$. Then $\phi_s(cw_1+w_2)=(cw_1+w_2)(s)$ which by definition equals $cw_1(s)+w_2(s)$ which equals $c\phi_s(w_1)+\phi_s(w_2)$. Thus $\phi_s$ is a linear functional on $W$.

Suppose $\phi_s(w)=0$ for all $s\in S$, $w\in W$. Then $w(s)=0$ $\forall$ $s\in S$, $w\in W$, which implies $\dim(W)=0$. So as long as $n>0$, $\exists$ $s_1\in S$ such that $\phi_{s_1}(w)\not=0$ for some $w\in W$. Equivalently there is an $s_1\in S$ and a $w_1\in W$ such that $w_1(s_1)\not=0$. This means $\phi_{s_1}\not=0$ as elements of $W^*$. It follows that $\langle\phi_{s_1}\rangle$, the subspace of $W^*$ generated by $\phi_{s_1}$, has dimension one. By scaling if necessary, we can further assume $w_1(s_1)=1$.

Now suppose $\forall$ $s\in S$ that we have $\phi_s\in\langle\phi_{s_1}\rangle$, the subspace of $W^*$ generated by $\phi_{s_1}$. Then for each $s\in S$ there is a $c(s)\in F$ such that $\phi_s=c(s)\phi_{s_1}$ in $W^*$. Then for each $s\in S$, $w(s)=c(s)w(s_1)$ for all $w\in W$. In particular $w_1(s)=c(s)$ (recall $w_1(s_1)=1$). Let $w\in W$. Let $b=w(s_1)$. Then $w(s)=c(s)w(s_1)=bw_1(s)$ $\forall$ $s\in S$. Notice that $b$ depends on $w$ but does not depend on $s$. Thus $w=bw_1$ as functions on $S$ where $b\in F$ is a fixed constant. Thus $w\in\langle w_1\rangle$, the subspace of $W$ generated by $w_1$. Since $w$ was arbitrary, it follows that $\dim(W)=1$. Thus as long as $\dim(W)\geq 2$ we can find $w_2\in W$ and $s_2\in S$ such that $\langle w_1,w_2\rangle$ (the subspace of $W$ generated by $w_1,w_2$) and $\langle\phi_{s_1},\phi_{s_2}\rangle$ (the subspace of $W^*$ generated by $\{\phi_{s_1},\phi_{s_2}\}$) both have dimension two. Let $W_0=\langle w_1,w_2\rangle$. Then we've shown that $\{\phi_{s_1},\phi_{s_2}\}$ is a basis for $W_0^*$. Therefore there's a dual basis $\{F_1,F_2\}\subseteq W_0^{**}$; so that $F_i(\phi_{s_j})=\delta_{ij}$, $i,j\in\{1,2\}$. By Theorem 17, $\exists$ corresponding $w_1,w_2\in W$ so that $F_i=L_{w_i}$ (in the notation of Theorem 17). Therefore, $$\delta_{ij}=F_i(\phi_{s_j})=L_{w_i}(\phi_{s_j})=\phi_{s_j}(w_i)=w_i(s_j),$$ for $i,j\in\{1,2\}$.

Now suppose $\forall$ $s\in S$ that we have $\phi_s\in\langle\phi_{s_1},\phi_{s_2}\rangle\subseteq W^{*}$. Then $\forall$ $s\in S$, there are constants $c_1(s),c_2(s)\in F$ and we have $w(s)=c_1(s)w(s_1)+c_2(s)w(s_2)$ for all $w\in W$. Similar to the argument in the previous paragraph, this implies $\dim(W)\leq 2$ (for $w\in W$ let $b_1=w(s_1)$ and $b_2=w(s_2)$ and argue as before). Therefore, as long as $\dim(W)\geq3$ we can find $s_3$ so that $\langle\phi_{s_1},\phi_{s_2},\phi_{s_3}\rangle\subseteq W^*$, the subspace of $W^*$ generated by $\phi_{s_1},\phi_{s_2},\phi_{s_3}$, has dimension three. And as before we can find $w_3\in W$ such that $w_i(s_j)=\delta_{ij}$, for $i,j\in\{1,2,3\}$.

Continuing in this way we can find $n$ elements $s_1,\dots,s_n\in S$ such that $\phi_{s_1},\dots,\phi_{s_n}$ are linearly independent in $W^{*}$ and corresponding elements $w_1,\dots,w_n\in W$ such that $w_i(s_j)=\delta_{ij}$. Let $f_i=w_i$ and we are done.

From http://greggrant.org

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.

This Post Has One Comment

  1. Now suppose ∀ s∈S that we have
    ϕs∈⟨ϕs1⟩, the subspace of
    W∗ generated by
    ϕs1. Then for each s∈S there is a c(s)∈F such that
    ϕs=c(s)ϕs1 in
    W∗.

    I don't understand here. The ''s'' in the first sentence satisfies ϕs∈⟨ϕs⟩.But in the second sentence, it become an arbitrary elements in S. I here something wrong?

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