[더플러스수학] 울산과고 고급수학 프린트 1

1. 다음 사상 중에서 선형사상인 것을 찾아라.

(1) $\displaystyle T~:~\mathbb{R}^3  \longrightarrow \mathbb{R}^2 ,~T(x_1 ,~x_2 ,~x_3 )=(3x_1+2x_3 ,~x_2 )$

(2) $\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T(x_1 ,~x_2  )=( \left|x_1 \right| ,~x_2 )$

(3) $\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T(x_1 ,~x_2  )=(x_1 +1 ,~x_1 +x_2 )$

(4) $\displaystyle T~:~\mathbb{R_3}[x]  \longrightarrow \mathbb{R}_3 [x] ,~T(a+bx+cx^2  )=1+ax+bx^2$

 

(정답 및 풀이)

(1) 선형사상이다.

$\displaystyle (x_1 ,~x_2 ,~x_3 ) \in \mathbb{R}^3 ,~ k \in \mathbb{R}$에 대하여

$\displaystyle \begin{align} T \left(\textcolor {red}{k} (x_1 ,~x_2 ,~x_3 ) \right) &= T \left(\textcolor {red}{k} x_1 ,~\textcolor {red}{k} x_2 ,~\textcolor {red}{k}x_3 ) \right)\\&=\left(3\textcolor {red}{k} x_1+2 \textcolor {red}{k} x_3 ,~\textcolor {red}{k} x_2 \right) \\&=\textcolor {red}{k} (3x_1+2x_3 ,~x_2 ) = \textcolor {red}{k} T(x_1 ,~x_2 ,~x_3 )\end{align}$

또, $\displaystyle (x_1 ,~x_2 ,~x_3 ) \in \mathbb{R}^3, ~(y_1 ,~y_2 ,~y_3 ) \in \mathbb{R}^3$에 대하여

$\displaystyle \begin{align} T \left(\textcolor {red}{ (x_1 ,~x_2 ,~x_3 )}+\textcolor {blue}{ (y_1 ,~y_2 ,~y_3 )}  \right) &= T \left( x_1 +y_1 ,~x_2 +y_2 ,~x_3 +y_3 \right)\\&=\left(3 (x_1 +y_1 )+2   (x_3 +y_3),~(x_2+y_2 ) \right) \\&=\left(3 x_1+2x_3 ,~ x_2 \right) +\left( 3y_1 +2  y_3,~y_2 \right)\\& =  T \left(\textcolor {red}{ x_1 ,~x_2 ,~x_3 } \right)+T(\textcolor {blue}{ y_1 ,~y_2 ,~y_3 }  ) \end{align}$

(2) 선형사상이 아니다. 반례로

$\displaystyle \begin{align} T(1 ,~3  )+T(-3,~2) &= (1,~3)+(\textcolor{red} {|-3|},~2)=(4,~5) \\& \neq T((1,~3)+(-3,~2))=T(-2,~5)\\&=( \textcolor{red} {|-2|},~5) =( 2,~5) \end{align}$

(3) 선형사상이 아니다. 반례로

$\displaystyle \begin{align} T(1 ,~3  )+T(2,~5) &= (2,~4)+(3,~7)= (5,~11) \\& \neq T((1,~3)+(2,~5))=T(3,~8)\\&=( 4,~11) \end{align}$

(4) 선형사상이 아니다. 반례로

$\displaystyle \begin{align} T( \textcolor {red}{3} (2+3x+4x^2 ) )&=T (6+9x+12x^2 )=1+6x+9x^2  \\&\neq \textcolor {red}{3}T(2+3x+4x^2 )=3 (1+2x+3x^2 )=3+6x+9x^2  \end{align}$

 

2. 체 $\displaystyle F$에 대하여 다음 사상이 선형사상인지를 판정하여라.

(1) $\displaystyle T~:~\mathrm{Mat}_2 (F)   \longrightarrow \mathrm {Mat}_2(F) ,~T( \begin{bmatrix} a&b \\c&d  \end{bmatrix} )=\begin{bmatrix} b&a\\c&d  \end{bmatrix}$

(2) $\displaystyle T~:~\mathrm{Mat}_2 (F)   \longrightarrow \mathrm {Mat}_2(F) ,~T( \begin{bmatrix} a&b \\c&d  \end{bmatrix} )=\begin{bmatrix} 1&b\\c&d  \end{bmatrix}$

 

(정답 및 풀이)

(1) 선형사상이다.

$\displaystyle a,~b,~c,~d \in F ,~k \in F$와 $\displaystyle \begin{bmatrix} a&b \\c&d  \end{bmatrix} \in \mathrm{Mat}_2 (F)$에 대하여

$\displaystyle \begin{align} T \left( \textcolor {red}{k} \begin{bmatrix} a&b \\c&d  \end{bmatrix}  \right) &= T  \left( \begin{bmatrix} \textcolor {red}{k} a& \textcolor {red}{k} b \\\textcolor {red}{k} c& \textcolor {red}{k} d  \end{bmatrix}  \right)=\begin{bmatrix} \textcolor {red}{k} b & \textcolor {red}{k} a \\\textcolor {red}{k} c& \textcolor {red}{k} d  \end{bmatrix}  \\&=\textcolor {red}{k}\begin{bmatrix}  b &   a \\  c&   d  \end{bmatrix} = \textcolor {red}{k} T \left(  \begin{bmatrix} a&b \\c&d  \end{bmatrix}  \right)  \end{align}$

또, 임의의 $\displaystyle a_1,~b_1,~c_1,~d_1 \in F ,a_2,~b_2,~c_2,~d_2\in F$일 때, $\displaystyle \begin{bmatrix} a_1&b_1 \\c_1&d_1  \end{bmatrix} \in \mathrm{Mat}_2 (F),~ \begin{bmatrix} a_2&b_2 \\c_2&d_2  \end{bmatrix} \in \mathrm{Mat}_2 (F)$에 대하여

$\displaystyle \begin{align} T \left(  \begin{bmatrix} a_1&b_1 \\c_1&d_1  \end{bmatrix} \textcolor {red} {+}  \begin{bmatrix} a_2&b_2 \\c_2&d_2  \end{bmatrix}    \right) &= T  \left( \begin{bmatrix} a_1 \textcolor {red} {+} a_2 & b_1 \textcolor {red}{+} b_2 \\ c_1 \textcolor {red}{+} c_2 & d_1 \textcolor {red}{+} d_2  \end{bmatrix}  \right)\\&=\begin{bmatrix}  b_1 \textcolor {red}{+} b_2 &a_1 \textcolor {red} {+} a_2 \\ c_1 \textcolor {red}{+} c_2 & d_1 \textcolor {red}{+} d_2  \end{bmatrix}   \\&=\begin{bmatrix}  b_1  &a_1  \\ c_1  & d_1  \end{bmatrix} \textcolor {red}{+} \begin{bmatrix}   b_2 & a_2 \\   c_2 &  d_2  \end{bmatrix} \\& =T \left(  \begin{bmatrix} a_1&b_1 \\c_1&d_1  \end{bmatrix} \right) \textcolor {red} {+} T \left(\begin{bmatrix} a_2&b_2 \\c_2&d_2  \end{bmatrix}    \right) \end{align}$

(2) 선형사상이 아니다. 반례로

$\displaystyle \begin{align} T \left( \textcolor {red}{3} \begin{bmatrix} 1&2 \\3&4  \end{bmatrix}  \right) &= T  \left( \begin{bmatrix} \textcolor {red}{3}\times  1& \textcolor {red}{3}\times  2 \\ \textcolor {red}{3}\times 3& \textcolor {red}{3}\times 4  \end{bmatrix}  \right)=T \left(\begin{bmatrix}3 & 6 \\ 9 & 12 \end{bmatrix}   \right)  = \begin{bmatrix}  \textcolor {red}{1} &   6 \\  9&   12  \end{bmatrix} \\& \neq  \textcolor {red}{3} T \left(  \begin{bmatrix} 1&2 \\3&4  \end{bmatrix}  \right)   =\textcolor {red}{3}   \begin{bmatrix} 1&2 \\3&4  \end{bmatrix} = \begin{bmatrix} \textcolor {red}{3}&6 \\9&12  \end{bmatrix}     \end{align}$

 

 

3. 다음과 같이 정의된 선형사상 $\displaystyle T$에 대하여 $\displaystyle T (1,~0)$과 $\displaystyle T(0,~1)$을 구하고 또, $\displaystyle T$의 행렬을 구하여라. 여기서 $\displaystyle k$는 실수이다.

(1) $\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T( x_1 ,~x_2 )=(x_1 ,~-x_2 )$

(2) $\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T(x_1 ,~x_2  )=( -x_1  ,~x_2 )$

(3) $\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T(x_1 ,~x_2  )=(-x_1 ,~- x_2 )$

(4) $\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T(x_1 ,~x_2  )=(kx_1 ,~k x_2 )$

 

(정답 및 풀이)

(1) $\displaystyle T (1,~0)=(1,~0),~T(0,~1)=(0,~-1)$이다.

$\displaystyle \begin{bmatrix} 1&0\\0&-1 \end{bmatrix}$

(2) $\displaystyle T (1,~0)=(-1,~0),~T(0,~1)=(0,~1)$이다.

$\displaystyle \begin{bmatrix} -1&0\\0&1 \end{bmatrix}$

(3) $\displaystyle T (1,~0)=(-1,~0),~T(0,~1)=(0,~-1)$이다.

$\displaystyle \begin{bmatrix} -1&0\\0&-1 \end{bmatrix}   =-I$

(4) $\displaystyle T (1,~0)=(k,~0),~T(0,~1)=(0,~k)$이다.

$\displaystyle \begin{bmatrix} k&0\\0&k \end{bmatrix}=k I$

 

 

 

4. 다음과 같이 정의된 선형사상 $\displaystyle T$에 대하여 아래 물음에 답하여라.

$\displaystyle T~:~\mathbb{R}^4  \longrightarrow \mathbb{R}^3 ,~T( x_1 ,~x_2 ,~x_3 ,~x_4 )=(y_1 ,~y_2 ,~y_3 )$

$\displaystyle \begin{cases} y_1 =4  x_1 +2x_2 -x_3 +x_4\\ y_2 =  2x_2 +x_3 +x_4 \\y_3 =  x_1 +x_4 \end{cases}$

(1) $\displaystyle T(1,~0,~0,~0),~ T(0,~1,~0,~0),~T(0,~0,~1,~0),~T(0,~0,~0,~1)$을 구하여라.

(2) $\displaystyle T(1,~1,~0,~0),~ T(2,~1,~0,~1),~T(1,~1,~1,~1)$을 구하여라.

 

(정답 및 풀이)

(1) $\displaystyle T(1,~0,~0,~0)=(4,~0,~1)$, $\displaystyle T(0,~1,~0,~0)=(2,~2,~0)$, $\displaystyle  T(0,~0,~1,~0)=(-1,~1,~0)$, $\displaystyle T(0,~0,~0,~1)=(1,~1,~1)$

(2)

$\displaystyle \begin{align} T(1,~1,~0,~0)&= T ((1,~0,~0,~0)+(0,~1,~0,~0))\\&=T(1,~0,~0,~0)+T(0,~1,~0,~0)\\&=(4,~0,~1)+(2,~2,~0)=(6,~2,~1)\end{align}$,

$\displaystyle \begin{align} T(2,~1,~0,~1)&= T ((2,~0,~0,~0)+(0,~1,~0,~0)+(0,~0,~0,~1))\\&=T (\textcolor{red}{2}(1,~0,~0,~0)+(0,~1,~0,~0)+(0,~0,~0,~1)) \\&=\textcolor{red}{2}T(1,~0,~0,~0)+T(0,~1,~0,~0)+T(0,~0,~0,~1)\\&=\textcolor{red}{2} (4,~0,~1)+(2,~2,~0)+(1,~1,~1)=(11,~3,~2)\end{align}$,

$\displaystyle \begin{align} T(1,~1,~1,~1)&= T ((1,~0,~0,~0)+(0,~1,~0,~0)+(0,~0,~1,~0)+(0,~0,~0,~1)\\&=T(1,~0,~0,~0)+T(0,~1,~0,~0)+T(0,~0,~1,~0)+T(0,~0,~0,~1)\\&=(4,~0,~1)+(2,~2,~0)+(-1,~1,~0)+(1,~1,~1)=(5,~3,~2)\end{align}$

 

 

 

5. 다음 두 선형변환에 대하여 $\displaystyle S \circ T$와 $\displaystyle T \circ S$를 정의하여라.

$\displaystyle T~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~T( x_1 ,~x_2 )=(x_1 +2x_2 ,~-x_1 +4x_2 )$

$\displaystyle S~:~\mathbb{R}^2  \longrightarrow \mathbb{R}^2 ,~S( x_1 ,~x_2 )=(2x_1 -x_2 ,~x_1 +3x_2 )$

 

(정답 및 풀이)

$\displaystyle \begin{align} (S \circ T )(x_1 ,~x_2)&=S \left(T(x_1 ,~x_2 ) \right) =S( x_1 +2x_2 ,~-x_1 +4x_2 ) \\&=2(x_1 +2x_2)-(-x_1 +4x_2 ),~(x_1 +2x_2)+3(-x_1 +4x_2 )\\&=(3x_1+8x_2 ,~ -2x_1+14x_2 )  \end{align}$

$\displaystyle \begin{align} (T \circ S )(x_1 ,~x_2)&=T \left(S(x_1 ,~x_2 ) \right) =T(2x_1 -x_2 ,~x_1 +3x_2 ) \\&=(2x_1 -x_2)+2(x_1 +3x_2 ),~-(2x_1 -x_2)+4(x_1 +3x_2 )\\&=(4x_1+5x_2 ,~ 2x_1+13x_2 )  \end{align}$

 

 

 

6. 체 $\displaystyle F$ 위의 벡터공간 $\displaystyle V ,~W$에 대하여 사상 $\displaystyle T~:~V  \longrightarrow W$가 다음 조건을 만족시킬 때, $\displaystyle T$는 선형사상임을 증명하여라.

$\displaystyle T (av +b w)=a T(v)+b T(w)$    $\displaystyle v,~w \in V ,~a,~b \in F$

 

(정답 및 풀이)

선형사상의 정의는

체 $\displaystyle F$ 위의 벡터공간 $\displaystyle V ,~W$에 대하여 사상 $\displaystyle T~:~V  \longrightarrow W$가 다음 조건을 만족한다. 
$\displaystyle v,~w \in V ,~a,~b \in F$일 때, 
(1) $\displaystyle T (a v)=a T(v)$    
(2) $\displaystyle T ( v+w)= T(v)+T(w)$

이다.

$\displaystyle T (av +b w)=a T(v)+b T(w)$    $\displaystyle v,~w \in V ,~a,~b \in F$    $\displaystyle \ast$

을 만족한다고 가정하면 $\displaystyle \ast$ 식에 $\displaystyle b=0$을 대입하면

$\displaystyle T (a v)=a T(v)$  

또 $\displaystyle \ast$ 식에 $\displaystyle a=1,~b=1$을 대입하면

$\displaystyle T ( v+w)= T(v)+T(w)$  

따라서 선형사상의 정의를 다 만족하므로 이 사상은 선형사상이다.