08. 쌍대공간Dual Space

Proof. (1) For a functional of the form $f(x,y,z)=ax+by+cz$, it suffices to solve for the coefficients $a,b,c$ satisfying the defining relations $f_i(v_j)=\delta_{ij}$ ($v_1=(1,0,1)$, $v_2=(1,2,1)$, $v_3=(0,0,1)$).

For $f_1$: $f_1(v_1)=a+c=1$, $f_1(v_2)=a+2b+c=0$, $f_1(v_3)=c=0$ give $c=0$, $a=1$, $b=-\frac12$, so

$$f_1(x,y,z) = x - \frac12 y.$$

For $f_2$: $f_2(v_1)=a+c=0$, $f_2(v_2)=a+2b+c=1$, $f_2(v_3)=c=0$ give $c=0$, $a=0$, $b=\frac12$, so

$$f_2(x,y,z) = \frac12 y.$$

For $f_3$: $f_3(v_1)=a+c=0$, $f_3(v_2)=a+2b+c=0$, $f_3(v_3)=c=1$ give $c=1$, $a=-1$, $b=0$, so

$$f_3(x,y,z) = -x+z.$$

Direct substitution confirms $f_i(v_j)=\delta_{ij}$ for each $i,j$, so $\beta^* = \{f_1,f_2,f_3\}$.

(2) Let $\{e_1^*,e_2^*,e_3^*\}$ be the dual basis of the standard basis. Since $f_1=e_1^*-2e_2^*$, $f_2=e_1^*+e_2^*+e_3^*$, $f_3=e_2^*-3e_3^*$, the matrix whose rows are the coordinates of $f_1,f_2,f_3$ relative to $\{e_1^*,e_2^*,e_3^*\}$ is

$$A = \begin{pmatrix} 1 & -2 & 0 \\ 1 & 1 & 1 \\ 0 & 1 & -3 \end{pmatrix}.$$

Since $\det A = -10 \neq 0$, the rows of $A$ — the coordinate vectors of $f_1,f_2,f_3$ — are linearly independent, hence $f_1,f_2,f_3$ themselves are linearly independent in $V^*$. As $\dim V^* = 3$, $\{f_1,f_2,f_3\}$ is a basis for $V^*$.

To find a basis $\{v_1,v_2,v_3\}$ of $V$ whose dual basis is $\{f_1,f_2,f_3\}$, we solve $f_i(v_j)=\delta_{ij}$, i.e. $A[v_j]=e_j$, so $[v_j]=A^{-1}e_j$; that is, the $j$-th column of $A^{-1}$ gives the coordinates of $v_j$. Computing,

$$A^{-1} = \frac{1}{10}\begin{pmatrix} 4 & 6 & 2 \\ -3 & 3 & 1 \\ -1 & 1 & -3 \end{pmatrix},$$

so

$$v_1 = \left(\frac{2}{5}, -\frac{3}{10}, -\frac{1}{10}\right), \quad v_2 = \left(\frac{3}{5}, \frac{3}{10}, \frac{1}{10}\right), \quad v_3 = \left(\frac{1}{5}, \frac{1}{10}, -\frac{3}{10}\right).$$

Direct substitution confirms $f_i(v_j)=\delta_{ij}$, and since $A^{-1}$ is invertible, $\{v_1,v_2,v_3\}$ is a basis of $V$. By definition, its dual basis is exactly $\{f_1,f_2,f_3\}$.

Solution. (1) $f(p(x)) = 2p'(0)+p''(1)$ is a linear functional. For any $p,q \in P(\mathbb{R})$ and scalar $c$, linearity of differentiation gives $(p+cq)'(0) = p'(0)+cq'(0)$ and $(p+cq)''(1) = p''(1)+cq''(1)$, so

$$f(p+cq) = 2(p'(0)+cq'(0)) + (p''(1)+cq''(1)) = \bigl(2p'(0)+p''(1)\bigr) + c\bigl(2q'(0)+q''(1)\bigr) = f(p)+cf(q).$$

Hence $f$ is a linear functional.

(2) $f(x,y,z)=x^2+y^2+z^2$ is not a linear functional. For instance, $f(1,0,0)=1$ but $f(2,0,0)=4 \neq 2f(1,0,0)=2$, so $f$ fails to preserve scalar multiplication. Hence $f$ is not a linear functional.

(3) $f(x,y)=(2x,4y)$ is not a linear functional. A linear functional on $V$ is by definition a linear transformation from $V$ into the scalar field $F$, but this $f$ maps $\mathbb{R}^2$ into $\mathbb{R}^2$, not into $\mathbb{R}$. The map itself is linear, but its codomain is $\mathbb{R}^2 \neq F$, so by definition $f$ cannot be a linear functional.

Solution. Write $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$, and define

$$f(x,y,z) = \det\begin{pmatrix} x & y & z \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{pmatrix} = (u_2v_3-u_3v_2)x - (u_1v_3-u_3v_1)y + (u_1v_2-u_2v_1)z.$$

Since the determinant is linear in each row, $f$ is a linear functional on $\mathbb{R}^3$. As $\{u,v\}$ is linearly independent, $(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1) = u \times v \neq (0,0,0)$, so $f \neq 0$.

Since $f(u)$ and $f(v)$ are determinants with a repeated row, $f(u)=f(v)=0$, so $\operatorname{span}\{u,v\} \subseteq \operatorname{Ker}(f)$. On the other hand, since $f \neq 0$, $\dim(\operatorname{Ker} f) = 3 - \operatorname{rank}(f) = 3-1=2$, and since $\{u,v\}$ is linearly independent, $\dim(\operatorname{span}\{u,v\})=2$. As the two subspaces have equal dimension and one contains the other,

$$\operatorname{Ker}(f) = \operatorname{span}\{u,v\},$$

so this $f$ is the desired functional.

Proof. For each $i=0,\ldots,n$, define the Lagrange interpolating polynomial

$$p_i(x) = \prod_{j\neq i} \frac{x-c_j}{c_i-c_j} \in P_n(F).$$

If $j \neq i$ then $p_i(c_j)=0$, and $p_i(c_i)=1$, so $f_i(p_j) = p_j(c_i) = \delta_{ij}$.

We first show that $\{p_0,\ldots,p_n\}$ is a linearly independent subset of $P_n(F)$. Suppose $\sum_{j=0}^n a_j p_j = 0$. Applying $f_i$ to both sides gives

$$0 = f_i\Bigl(\sum_{j=0}^n a_j p_j\Bigr) = \sum_{j=0}^n a_j f_i(p_j) = \sum_{j=0}^n a_j \delta_{ij} = a_i$$

for every $i$, so all $a_i=0$. Hence $\{p_0,\ldots,p_n\}$ is linearly independent. Since $\dim P_n(F) = n+1$ and we have $n+1$ linearly independent vectors, $\gamma = \{p_0,\ldots,p_n\}$ is a basis for $P_n(F)$.

Now $f_i(p_j)=\delta_{ij}$ is precisely the defining relation for $\{f_0,\ldots,f_n\}$ to be the dual basis $\gamma^*$ of $\gamma$, so $\{f_0,\ldots,f_n\} = \gamma^*$. Since the dual basis of a basis of a finite-dimensional vector space is always a basis of the dual space, $\{f_0,\ldots,f_n\}$ is a basis for $V^* = P_n(F)^*$.

Solution. (1) By the definition of the transpose, $T^t(f) = f \circ T$, so

$$T^t(f)(x,y) = f(T(x,y)) = f(3x+2y, x) = 2(3x+2y)+x = 7x+4y.$$

Hence $T^t(f)(x,y) = 7x+4y$.

(2) Let $\beta = \{e_1,e_2\}$ be the standard basis. Since $T(e_1)=T(1,0)=(3,1)$ and $T(e_2)=T(0,1)=(2,0)$,

$$[T]_\beta = \begin{pmatrix} 3 & 2 \\ 1 & 0 \end{pmatrix}.$$

Let $\beta^* = \{e_1^*,e_2^*\}$ be the standard dual basis, so $e_1^*(x,y)=x$ and $e_2^*(x,y)=y$. Then

$$(T^t(e_1^*))(x,y) = e_1^*(3x+2y,x) = 3x+2y = 3e_1^*(x,y)+2e_2^*(x,y),$$

$$(T^t(e_2^*))(x,y) = e_2^*(3x+2y,x) = x = e_1^*(x,y),$$

so $T^t(e_1^*) = 3e_1^*+2e_2^*$ and $T^t(e_2^*) = e_1^*$. Hence

$$[T^t]_{\beta^*} = \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}.$$

Since $([T]_\beta)^t = \begin{pmatrix} 3 & 1 \\ 2 & 0 \end{pmatrix}$, we confirm that

$$[T^t]_{\beta^*} = ([T]_\beta)^t.$$

Proof. Let $\dim V = n$ and let $\beta = \{v_1,\ldots,v_n\}$ with dual basis $\beta^* = \{f_1,\ldots,f_n\}$. Since $V$ is finite-dimensional, $V^*$ is finite-dimensional as well and $\beta^*$ is a basis for $V^*$, so $\beta^*$ in turn has a dual basis $\beta^{**} = \{f_1^*,\ldots,f_n^*\} \subseteq V^{**}$, uniquely characterized by $f_i^*(f_j) = \delta_{ij}$ $(1 \leq i,j \leq n)$.

To show $\psi(\beta) = \{\psi(v_1),\ldots,\psi(v_n)\}$ equals $\beta^{**}$, it suffices to show that $\psi(v_i)$ satisfies the defining relation of $f_i^*$ for each $i$. For any $j=1,\ldots,n$, by the definition of $\psi$,

$$\psi(v_i)(f_j) = f_j(v_i) = \delta_{ji} = \delta_{ij}.$$

(The second equality holds because $\beta^*$ is the dual basis of $\beta$, i.e., $f_j(v_i)=\delta_{ji}$.) This is exactly the relation that $f_i^*$ must satisfy, and since a dual basis is uniquely determined by this relation,

$$\psi(v_i) = f_i^* \quad (i=1,\ldots,n).$$

Therefore

$$\psi(\beta) = \{\psi(v_1),\ldots,\psi(v_n)\} = \{f_1^*,\ldots,f_n^*\} = \beta^{**}.$$