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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 $c_1+\cdots+c_n=0$.


(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
\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


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