臺灣綜合大學系統 115 學年度學士班轉學生聯合招生考試
科目名稱 線性代數
解答:$$\text{A vector space $V$ over a field $F$ is a non-empty set equipped with two operations:} \\ \text{vector addition and scalar multiplication, such that the following eight axioms hold} \\ \text{ for all }u,v,w \in V \text{ and all }a,b\in F:\\ \text{1. Commutativity of addition: }u + v = v + u \\ \text{2. Associativity of addition: }(u + v) + w = u + (v + w)\\ \text{3. Additive identity: There exists an element }0 \in V \text{such that }v + 0 = v\; \forall v \in V. \\ \text{4. Additive inverse: For every } v \in V, \text{there exists an element }-v \in V \text{such that }v + (-v) = 0. \\\text{5.Compatibility of scalar multiplication: }a(bv) = (ab)v\\ \text{6.Identity element of scalar multiplication: }1v = v, \text{where } 1 \text{is the multiplicative identity in }F.\\\text{7. Distributivity of scalar multiplication over vector addition: }a(u + v) = au + av \\ \text{8.Distributivity of scalar multiplication over field addition: }(a + b)v = av + bv$$
解答:$$\cases{x=0\\ y=0} \Rightarrow 0\ne 0+1 \Rightarrow \text{ the zero vector }(0,0) \not \in S \Rightarrow \bbox[red, 2pt]{\text{No, the set does not form a subspace of }R^2}$$
解答:$$\text{Let $V$ be a vector space over a field $F$ (typically $\mathbb{R}$ or $\mathbb{C}$).}\\ \text{ An inner product is a function $\langle \cdot, \cdot \rangle : V \times V \to F$ that satisfies the following}\\ \text{three axioms for all vectors $u, v, w \in V$ and all scalars $a \in F$:} \\ \text{1. Symmetry (or Conjugate Symmetry): $\langle u, v \rangle = \overline{\langle v, u \rangle}$} \\ \text{2. Linearity in the first argument:}\\ \qquad \text{Additivity: $\langle u + v, w \rangle = \langle u, w \rangle + \langle v, w \rangle$} \\ \qquad \text{Homogeneity: }\langle au, v \rangle = a\langle u, v \rangle \\ \text{3. Positive-definiteness: $\langle v, v \rangle \geq 0$ for all $v \in V$, and $\langle v, v \rangle = 0$ if and only if $v = 0$.}$$
解答:$$\text{Let $f,g,h\in C[0,1]$ and let $c \in \mathbb R$.} \\ \text{1. symmetry} \\ \qquad \langle f,g \rangle = \int_0^1 f(x) g(x)\,dx =\int_0^1 g(x)f(x)\,dx = \langle g,f\rangle \\\text{2. Linearity in the first argument} \\ \langle cf+h, g\rangle = \int_0^1 \left[ cf(x)+h(x) \right]g(x)\,dx = \int_0^1 \left[ cf(x)g(x) +h(x) g(x) \right] \,dx \\= c\int_0^1 \left[ f(x)g(x) \right] + \int_0^1 [h(x) g(x) ] \,dx = c\langle f,g \rangle + \langle h,g\rangle \\ \text{3. Positive-definiteness} \\ \langle f,f\rangle = \int_0^1 f(x)f(x)\,dx = \int_0^1 (f(x))^2\,dx \ge 0 \\ \qquad \text{Next, we must show that $\langle f, f \rangle = 0 \iff f = 0$.} \\ \text{If $f(x) = 0$ for all $x$, then $\int_0^1 0^2 dx = 0$.} \\ \text{Suppose } \langle f,f\rangle =0 \Rightarrow \int_0^1 (f(x))^2\,dx =0 \Rightarrow (f(x))^2=0 \Rightarrow f(x)=0, \;\forall x\in [0,1] \\ \text{Since all three axioms are satisfied, $\langle .,. \rangle$ defines an inner product on $C[0,1]$ }\bbox[red, 2pt]{QED.}$$
解答:$$\text{Let $V$ and $W$ be vector spaces over a field $F$.}\\ \text{ A function $T: V \to W$ is defined as a linear transformation if it }\\ \text{satisfies the following two conditions for all vectors $u, v \in V$ and all scalars $c \in F$:} \\\text{1. Additivity: }T(u+v)=T(u)+T(v) \\ \text{2. Homogeneity: }T(cv)= cT(v)$$
解答:$$\cases{f(x_1)=x_1^2\\ f(x_2)=x_2^2 \\f(x_1+x_2)=(x_1+x_2)^2 =x_1^2+2x_1x_2+x_2^2} \Rightarrow f(x_1)+f(x_2) \ne f(x_1+x_2), \mbox{ for }x_1x_2\ne 0\\ \Rightarrow \bbox[red, 2pt]{No, f \text{ is NOT linear}}$$
解答:$$\text{Since $T$ is a linear transformation, }T(0)=0 \Rightarrow 0 \in N(T) \\ u,v\in N(T) \Rightarrow T(u+v)=T(u)+T(v)=0+0=0 \Rightarrow u+v\in N(T) \\ u\in N(T) \text{ and $c \in F$} \Rightarrow T(cu)=cT(u)=c\cdot 0=0 \Rightarrow cu\in N(T) \\ \Rightarrow N(T) \text{ is a subspace of }V.\; \bbox[red, 2pt]{QED.}$$
解答:$$\text{A vectro }v\in V \text{ is an eigenvector of $T$ if $v\ne 0$ and there exists $\lambda \in F$ such that} \\\qquad T(v)= \lambda v$$
解答:$$\textbf{(i) } A= \begin{bmatrix}1 & 1 & 1 \\0 & 1 & 0 \\ 0 & 1 & 2 \end{bmatrix} \Rightarrow \det(A-\lambda I) =-(\lambda-1)^2(\lambda-2)=0 \Rightarrow \lambda= \bbox[red, 2pt]{1,2} \\ \textbf{(ii) }\lambda_1=1 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow \begin{bmatrix} 0 & 1 & 1 \\0 & 0 & 0 \\0 & 1 & 1\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow x_2=-x_3 \Rightarrow v = x_1 \begin{bmatrix}1\\0\\0 \end{bmatrix}+ x_3 \begin{bmatrix}0\\-1\\1 \end{bmatrix} \\ \qquad \Rightarrow \text{ choosing }v_1 = \begin{bmatrix}1\\0\\0 \end{bmatrix}, v_2= \begin{bmatrix}0 \\-1\\1 \end{bmatrix} \\ \lambda_2=2 \Rightarrow (A-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix} -1 & 1 & 1 \\0 & -1 & 0 \\ 0 & 1 & 0\end{bmatrix} \begin{bmatrix}x_1\\ x_2\\ x_3 \end{bmatrix}=0 \Rightarrow \cases{x_1=x_3\\ x_2=0} \Rightarrow v= x_3 \begin{bmatrix}1\\0\\1 \end{bmatrix} \\ \quad \Rightarrow \text{ choosing }v_3= \begin{bmatrix}1\\0\\1 \end{bmatrix} \\ \Rightarrow Q= [v_1\; v_2\; v_3] = \bbox[red, 2pt]{\begin{bmatrix}1 & 0 & 1 \\0 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix}}$$
解答:$$\text{The standard ordered basis for $P_2(R)$ is $\beta = \{1, x, x^2\}$.} \\ T(f(x)) = f(1) + f'(0)x + (f'(0) + f''(0))x^2 \Rightarrow \cases{f(x)=1 \Rightarrow T(f(x))=1\\ f(x)=x \Rightarrow T(f(x)) =1+x+x^2 \\f(x)=x^2 \Rightarrow T(f(x))=1+2x^2} \\ \Rightarrow [T]_\beta = \bbox[red, 2pt] {\begin{bmatrix} 1& 1& 1\\1& 1& 1\\1& 0& 2\end{bmatrix}}$$
解答:$$\text{Let $A$ be an $n \times n$ real symmetric matrix and $A=A^T$. } \\\text{Let $\lambda$ be an eigenvalue of $A$ with a corresponding non-zero eigenvector $v$.} \\ Av=\lambda v \Rightarrow \overline{Av} = \overline{\lambda v} \Rightarrow A\bar v= \bar \lambda \bar v \Rightarrow (A\bar v)^T= (\bar \lambda \bar v)^T \Rightarrow \bar v^T A^T = \bar \lambda \bar v ^T \Rightarrow \bar v^T A = \bar \lambda \bar v ^T \\\Rightarrow \bar v^T A v = \bar \lambda \bar v ^Tv \Rightarrow \bar v^T \lambda v = \bar \lambda \bar v ^Tv \Rightarrow (\overline{\lambda} - \lambda) \overline{v}^T v = 0 \Rightarrow \bar \lambda=\lambda \Rightarrow \lambda \text{ is real}\; \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(i) }\det(AB)= \det(A)\times \det (B)=3\times (-2)= \bbox[red, 2pt]{-6} \\ \textbf{(ii) }\det(B^{-1}) ={1\over \det(B)}={1\over -2}= \bbox[red, 2pt]{-{1\over 2}}$$
解題僅供參考，其他轉學考試題及詳解
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