[선형대수학-일차변환] Linear Mappings

Linear Mappings

1. Suppose the mapping $ F~:~R ^ {2} \rightarrow R ^ {2} $ is defined by $ F ( x,~y)= ( x+y,~x) $. Show that F is linear.

pf) We need to show that $ F ( v+w)=F ( v)+F ( w) $ and $ F ( kv)=kF ( v) $, where $ v $ and $ w $ are any elements of $ R ^ {2} $ and $ k $ is any scalar. Let $ v= ( a,~b) $ and $ w= ( a ' ,~b ' ) $. Then

$ v+w= ( a+a ' ,~b+b ' ) $ and $ kv= ( ka,~kb) $

We have $ F ( v)= ( a+b,~a) $ and $ F ( w)= ( a ' +b ' ,~a ' ) $. Thus,

$$ F ( v+w)=F ( a+a ' ,~b+b ' )= ( a+a ' +b+b ' ,~a+a ' ) $$

$$ ( a+b,~a)+ ( a ' +b ' ,~a ' )=F ( v)+F ( w) $$

and

$$ F ( kv)=F ( ka,~kb)= ( ka+kb,~ka)=k ( a+b,~a)=kF ( v) $$

Because $ v,~w,~k $ were arbitrary, $ F $ is linear.

 

2. Suppose $ F~:~R ^ {3} \rightarrow R ^ {2} $ is defined by $ F ( x,~y,~z)= ( x+y+z,~2x-3y+4z) $. Show that F is linear.

pf) We argue via matrices. Writing vectors as columns, the mapping F may be written in the form $ F ( v)=Av $, where $ v=[x,~y,~z] ^ {T} $ and

$$ A= \left[ \matrix {1&1&1\\2&-3&4} \right] $$

Then, using properties of matrices, we have

$$F(v+w)=A(v+w)=Av+Aw=F(v)+F(w)$$

and

$$F(k v)=A(kv)=kF(v) $$

Thus, $ F $ is linear.

 

3. Show that the following mappings are not linear:

(a) $ F~:~R ^ {2} \rightarrow R ^ {2} $ defined by $ F ( x,~y)= ( xy,~x) $

(b) $ F~:~R ^ {2} \rightarrow R ^ {2} $ defined by $ F ( x,~y)= ( x+3,~2y,~x+y) $

(c) $ F~:~R ^ {3} \rightarrow R ^ {2} $ defined by $ F ( x,~y,~z)= ( |x|,~y+z) $

sol) (a) Let $ v= ( 1,~2) $ and $ w= ( 3,~4) $ : then $ v+w= ( 4,~6) $. Also,

$$ F ( v)= ( 1 ( 2),~1)= ( 2,~1) $, $ F ( w)= ( 3 ( 4),~3)= ( 12,~3) $$

Hence,

$$ F ( v+w)= ( 4 ( 6),~4)= ( 24,~6) \neq F ( v)+F ( w) $$

(b) Because $ F ( 0,~0)= ( 3,~0,~0) \neq ( 0,~0,~0) $, $ F $ cannot be linear.

(c) Let $ v= ( 1,~2,~3) $ and $ k=-3 $. Then $ kv= ( -3,~-6,~-9) $. We have

$ F ( v)= ( 1,~5) $ and $ kF ( v)=-3 ( 1,~5)= ( -3,~-15) $

Thus,

$$ F ( kv)=F ( -3,~-6,~-9)= ( 3,~-15) \neq kF ( v) $$

Accordingly, $ F $ is not linear.

 

4. Let $ F~:~R ^ {2} \rightarrow R ^ {2} $ be the linear mapping for which $ F ( 1,~2)= \left ( 2,~3 \right ) $ and $ F ( 0,~1)= ( 1,~4) $. [Note that $ \left\{ \left ( 1,~2 \right ) ,~ \left ( 0,~1 \right ) \right\} $ is a basis of $ R ^ {2} $, so such a linear map $ F $ exists and is unique by Theorem 5.2.] Find a formula for F; that is, find $ F ( a,~b) $.

sol) Write $ ( a,~b) $ as a linear combination of $ ( 1,~2) $ and $ ( 0,~1) $ using unknowns $ x $ and $ y $,

$ \left ( a,~b \right ) =x \left ( 1,~2 \right ) +y \left ( 0,~1 \right ) = \left ( x,~2x+y \right ) $, so $ a=x,~b=2x+y $

Solve for $ x $ and $ y $ in terms of $ a $ and $ b $ to get $ x=a,~y=-2a+b $. Then

$ F \left ( a,~b \right ) =xF \left ( 1,~2 \right ) +yF \left ( 0,~1 \right ) =a \left ( 2,~3 \right ) + \left ( -2a+b \right ) \left ( 1,~4 \right ) = \left ( b,~-5a+4b \right ) $

 

5. Suppose a linear mapping $ F~:~V \rightarrow U $ is one-to-one and onto. Show that the inverse mapping $ F ^ {-1} ~:~U \rightarrow V $ is also linear.

sol) Suppose $ u,~u ' \in U $. Because $ F $ is one-to-one and onto, there exist unique vectors $ v,~v ' \in V $ for which $ F ( u)=u ' $ and $ F ( v)=v ' $. Because F is linear, we also have

$ F ( v+v ' )=F ( v)+F ( v ' )=u+u ' $ and $ F ( kv)=kF ( v)=ku $

By definition of the inverse mapping,

$ F ^ {-1} ( u)=v,~F ^ {-1} ( u ' )=u,~F ^ {-1} ( u+u ' )=v+v ' ,~F ^ {-1} ( ku)=kv $.

Then

$ F ^ {-1} ( u+u ' )=v+v ' =F ^ {-1} ( u)+F ^ {-1} ( u ' ) $, $ F ^ {-1} ( ku)=kv=kF ^ {-1} ( u) $

Thus, $ F ^ {-1} $ is linear.

.

 Matrix Representation of a Linear Operator

 

Let $ T $ be a linear operator (transformation) from a vector space $ V $ into itself, and suppose $ S= \left\{ u _ {1} ,~u _ {2} ,~ \cdots ,~u _ {n} \right\} $ is a basis of $ V $. Now $ T \left ( u _ {1} \right ) ,~T \left ( u _ {2} \right ) ,~ \cdots ,~T ( u _ {n} ) $ are vectors in V, and so each is a linear combination of the vectors in the basis $ S $ say,

$$ T \left ( u _ {1} \right ) =a _ {11} u _ {1} +a _ {12} u _ {2} + \cdots +a _ {1n} u _ {n} $$

$$ T \left ( u _ {2} \right ) =a _ {21} u _ {1} +a _ {22} u _ {2} + \cdots +a _ {2n} u _ {n} $$

$$ \cdots \cdots \cdots \cdots \cdots  \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots $$

$$ T \left ( u _ {n} \right ) =a _ {n1} u _ {1} +a _ {n2} u _ {2} + \cdots +a _ {nn} u _ {n} $$

 

The following definition applies.

DEFINITION: The transpose of the above matrix of coefficients, denoted by $m_S (T)$ or $T_S$, is called the matrix representation of $T$ relative to the basis $S$, or simply the matrix of $T$ in the basis $S$. (The subscript $S$ may be omitted if the basis $S$ is understood.)

Using the coordinate (column) vector notation, the matrix representation of T may be written in the form

$$ m _ {s} \left ( T \right ) =[T] _ {s} = \left [ \left [ T \left ( u _ {1} \right ) \right ] _ {s} , \left [ T \left ( u _ {2} \right ) \right ] _ {s} , \cdots , \left [ T \left ( u _ {n} \right ) \right ] _ {s} \right ] $$

That is, the columns of $ m ( T) $ are the coordinate vectors of $ T \left ( u _ {1} \right ) ,~T \left ( u _ {2} \right ) ,~ \cdots ,~T \left ( u _ {n} \right ) $, respectively.

 

6. Let $ F~:~R ^ {2} \rightarrow R ^ {2} $ be the linear operator defined by $ F ( x,~y)= \left ( 2x+3y,~4x-5y \right ) $.

(a) Find the matrix representation of F relative to the basis $ S= \left\{ u _ {1} ,~u _ {2} \right\} = \left\{ \left ( 1,~2 \right ) ,~ \left ( 2,~5 \right ) \right\} $.

(1) First find $ F ( u _ {1} ) $, and then write it as a linear combination of the basis vectors $ u _ {1} $ and $ u _ {2} $. (For notational convenience, we use column vectors.) We have

$ F ( u _ {1} )=F \left ( \left[ \matrix {1\\2} \right] \right ) = \left[ \matrix {8\\-6} \right] =x \left[ \matrix {1\\2} \right] +y \left[ \matrix {2\\5} \right] $ and $ \left ( \matrix {x+2y=8\\2x+5y=-6} \right) $

Solve the system to obtain $ x=52,~y=-22 $. Hence, $ F ( u _ {1} )=52u _ {1} -22u _ {2} $.

(2) Next find $ F ( u _ {2} ) $, and then write it as a linear combination of $ u _ {1} $ and $ u _ {2} $ :

$ F ( u _ {2} )=F \left ( \left[ \matrix {2\\5} \right] \right ) = \left[ \matrix {19\\-17} \right] =x \left[ \matrix {1\\2} \right] +y \left[ \matrix {2\\5} \right] $ and $ \left ( \matrix {x+2y=19\\2x+5y=-17} \right) $

Solve the system to obtain $ x=129,~y=-55 $. Hence, $ F ( u _ {2} )=129u _ {1} -55u _ {2} $.

Now write the coordinates of $ F ( u _ {1} ) $ and $ F ( u _ {2} ) $ as columns to obtain the matrix

$$ [F] _ {s} = \left[ \matrix {52 & 129\\22 & 55} \right] $$

(b) Find the matrix representation of F relative to the (usual) basis $ E= \left\{ e _ {1} ,~e _ {2} \right\} = \left\{ \left ( 1,~0 \right ) ,~ \left ( 0,~1 \right ) \right\} $.

 

7. Find $ F ( e _ {1} ) $ and write it as a linear combination of the usual basis vectors $ e _ {1} $ and $ e _ {2} $, and then find $ F ( e _ {2} ) $ and write it as a linear combination of $ e _ {1} $ and $ e _ {2} $. We have

$ \left \{ \begin{split}F ( e _ {1} )&=F \left ( 1,~0 \right ) = \left ( 2,~2 \right ) =2e _ {1} +4e _ {2}\\F ( e _ {2} )&=F \left ( 0,~1 \right ) = \left ( 3,~-5 \right ) =3e _ {1} -5e _ {2} \end{split} \right. $ and so $ [F] _ {E} = \left[ \matrix {2 & 3\\4 & -5} \right] $

Note that the coordinates of $ F ( e _ {1} ) $ and $ F ( e _ {2} ) $ form the columns, not the rows, of $ [F] _ {E} $. Also, note that the arithmetic is much simpler using the usual basis of $ R ^ {2} $.

 

Matrix Mappings and Their Matrix Representation

8. Consider the following matrix $ A $, which may be viewed as a linear operator on $ R ^ {2} $, and basis $ S $ of $ R ^ {2} $:

$ A= \left[ \matrix {3 & -2\\4 & -5} \right] $ and $ S= \left\{ u _ {1} ,~u _ {2} \right\} = \left\{ \left[ \matrix {1\\2} \right] ,~ \left[ \matrix {2\\5} \right] \right\} $

(We write vectors as columns, because our map is a matrix.) We find the matrix representation of $ A $ relative to the basis $ S $.

(1) First we write $ A ( u _ {1} ) $ as a linear combination of $ u _ {1} $ and $ u _ {2} $. We have

$ A \left ( u _ {1} \right ) = \left[ \matrix {3 & -2\\4 & -5} \right] \left[ \matrix {1\\2} \right] = \left[ \matrix {-1\\-6} \right] =x \left[ \matrix {1\\2} \right] +y \left[ \matrix {2\\5} \right] $ and so $ \left \{ \matrix {x+2y=-1\\2x+5y=-6} \right. $

Solving the system yields $ x=7,~y=-4 $. Thus, $ A \left ( u _ {1} \right ) =7u _ {1} -4u _ {2} $.

(2) Next we write $ A ( u _ {2} ) $ as a linear combination of $ u _ {1} $ and $ u _ {2} $. We have

$ A \left ( u _ {1} \right ) = \left[ \matrix {3 & -2\\4 & -5} \right] \left[ \matrix {2\\5} \right] = \left[ \matrix {-4\\-7} \right] =x \left[ \matrix {1\\2} \right] +y \left[ \matrix {2\\5} \right] $ and so $ \left \{ \matrix {x+2y=-4\\2x+5y=-7} \right. $

Solving the system yields $ x=-6,~y=1 $. Thus, $ A \left ( u _ {2} \right ) =-6u _ {1} +u _ {2} $. Writing the coordinates of $ A \left ( u _ {1} \right ) $ and $ A \left ( u _ {2} \right ) $ as columns gives us the following matrix representation of $ A $:

$$ [A] _ {S} = \left[ \matrix {7 & -6\\-4 & 1} \right] $$

 

9. Consider the linear mapping $ F~:~R ^ {2} \rightarrow R ^ {2} $ defined by $ F ( x,~y)= \left ( 3x+4y,~2x-5y \right ) $ and the following bases of $ R ^ {2} $:

$ E= \left\{ e _ {1} ,~e _ {2} \right\} = \left\{ ( 1,~0),~ ( 0,~1) \right\} $ and $ S= \left\{ u _ {1} ,~u _ {2} \right\} = \left\{ ( 1,~2),~ ( 2,~3) \right\} $

(a) Find the matrix $ A $ representing F relative to the basis $ E $.

(b) Find the matrix $ B $ representing F relative to the basis $ S $.

 

(a) Because $ E $ is the usual basis, the rows of $ A $ are simply the coefficients in the components of $ F ( x,~y) $; that is, using $ \left ( a,~b \right ) =ae _ {1} +be _ {2} $, we have

$ \left \{ \begin{split} F ( e _ {1} )&=F \left ( 1,~0 \right ) &= \left ( 3,~2 \right ) &=3e _ {1} +2e _ {2}\\F ( e _ {2} )&=F \left ( 0,~1 \right ) &= \left ( 4,~-5 \right ) &=4e _ {1} -5e _ {2} \end{split}  \right. $ and so $ A= \left[ \matrix {3 & 4\\2 & -5} \right] $

Note that the coefficients of the basis vectors are written as columns in the matrix representation.

(b) First find $ F ( u _ {1} ) $ and write it as a linear combination of the basis vectors $ u _ {1} $ and $ u _ {2} $. We have

$ F ( u _ {1} )=F \left ( 1,~2 \right ) = \left ( 11,~-8 \right ) =x \left ( 1,~2 \right ) +y \left ( 2,~3 \right ) $ and so    $ \left \{ \matrix {x+2y=11\\2x+3y=-8} \right. $

Solve the system to obtain $ x=-49,~y=30 $. Therefore

$$ F ( u _ {1} )=-40u _ {1} +30u _ {2} $$

Next find $ F ( u _ {2} ) $ and write it as a linear combination of the basis vectors $ u _ {1} $ and $ u _ {2} $. We have

$ F ( u _ {2} )=F \left ( 2,~3 \right ) = \left ( 18,~-11 \right ) =x \left ( 1,~2 \right ) +y \left ( 2,~3 \right ) $ and so $ \left \{ \matrix {x+2y=18\\2x+3y=-11} \right. $

Solve for $ x $ and $ y $ to obtain $ x=-76,~y=47 $. Hence

$$ F ( u _ {2} )=-76u _ {1} +47u _ {2} $$

Write the coefficients of $ u _ {1} $ and $ u _ {2} $ as columns to obtain $ B= \left[ \matrix {-49 & -76\\30 & 47} \right] $

(b') Alternatively, one can first find the coordinates of an arbitrary vector $ \left ( a,~b \right ) $ in $ R ^ {2} $ relative to the basis $ S $. We have

$ \left ( a,~b \right ) =x \left ( 1,~2 \right ) +y \left ( 2,~3 \right ) = \left ( x+2y,~2x+3y \right ) $, and so $ \left\{\begin{split} x+2y&=a\\ 2x+3y&=b \end{split} \right.$

Solve for $ x $ and $ y $ in terms of $ a $ and $ b $ to get $ x=-3a+2b $, $ y=2a-b $. Thus,

$$ \left ( a,~b \right ) = \left ( -3a+2b \right ) u _ {1} + \left ( 2a-b \right ) u _ {2} $$

Then use the formula for $ \left ( a,~b \right ) $ to find the coordinates of $ F ( u _ {1} ) $ and $ F ( u _ {2} ) $ relative to $ S $:

$\left \{ \begin{split} F ( u _ {1} )&=F ( 1,~2)&= \left ( 11,~-8 \right ) &=-49u _ {1} +30u _ {2} \\ F ( u _ {2} )&=F ( 2,~3)&= \left ( 18,~-11 \right ) &=-76u _ {1} +47u _ {2} \end{split} \right. $ and so $ B= \left[ \matrix {-49 & -76\\30 & 47} \right] $

 

10. Consider the following $ 2 \times 2 $ matrix $ A $ and basis $ S $ of $ R ^ {2} $:

$ A= \left[ \matrix {2 & 4\\5 & 6} \right] $ and $ S= \left\{ u _ {1} ,~u _ {2} \right\} = \left\{ \left[ \matrix {1\\-2} \right] ,~ \left[ \matrix {3\\-7} \right] \right\} $

The matrix $ A $ defines a linear operator on $ R ^ {2} $. Find the matrix $ B $ that represents the mapping $ A $ relative to the basis $ S $.

First find the coordinates of an arbitrary vector $ \left ( a,~b \right ) ^ {T} $ with respect to the basis $ S $. We have

$ \left[ \matrix {a\\b} \right] =x \left[ \matrix {1\\-2} \right] +y \left[ \matrix {3\\-7} \right] $ or $ \left \{ \matrix {x+3y=a\\-2x-7y=b} \right. $

Solve for $ x $ and $ y $ in terms of $ a $ and $ b $ to get $ x=7a+3b $, $ y=-2a-b $. Thus,

$$ \left ( a,~b \right ) ^ {T} = \left ( 7a+3b \right ) u _ {1} + \left ( -2a-b \right ) u _ {2} $$

Then use the formula for $ \left ( a,~b \right ) ^ {T} $ to find the coordinates of $ Au _ {1} $ and $ Au _ {2} $ relative to $ S $:

$$ Au _ {1} = \left[ \matrix {2 & 4\\5 & 6} \right] \left[ \matrix {1\\-2} \right] = \left[ \matrix {-6\\-7} \right] =-63u _ {1} +19u _ {2} $$

$$ Au _ {2} = \left[ \matrix {2 & 4\\5 & 6} \right] \left[ \matrix {3\\-7} \right] = \left[ \matrix {-22\\-27} \right] =-23u _ {1} +71u _ {2} $$

Writing the coordinates as columns yields

$$ B= \left[ \matrix {-63 & -235\\19 & 71} \right] $$

 

11. Find the matrix representation of each of the following linear operators $ F $ on $ R ^ {3} $ relative to the usual basis $ E= \left\{ e _ {1} ,~e _ {2} ,~e _ {3} \right\} $ of $ R ^ {3} $ that is, find $ [F]=[F] _ {E} $ ;

(a) $ F $ defined by $ F ( x,~y,~z)= \left ( x+2y-3z,~4-5y-6z,~7x+8y+9z \right ) $

(b) $ F $ defined by the $ 3 \times 3 $ matrix $ A= \left[ \matrix {1 & 1 & 1\\2 & 3 & 4\\5 & 5 & 5} \right] $.

(c) $ F $ defined by $ F ( e _ {1} )= \left ( 1,~3,~5 \right ) $, $ F ( e _ {2} )= \left ( 2,~4,~6 \right ) $, $ F ( e _ {3} )= \left ( 7,~7,~7 \right ) $. (Theorem 5.2 states that a linear map is completely defined by its action on the vectors in a basis.)

sol) (a) Because $ E $ is the usual basis, simply write the coefficients of the components of $ F ( x,~y,~z) $ as rows:

$$ [F]= \left[ \matrix {1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9} \right] $$

(b) Because $ E $ is the usual basis, $ [F]=A $, the matrix $ A $ itself.

(c) Here

$\begin{split}  F ( e _ {1} )&= \left ( 1,~3,~5 \right ) &=e _ {1} +3e _ {2} +5e _ {3} \\ F ( e _ {2} )&= \left ( 2,~4,~6 \right ) &=2e _ {1} +4e _ {2} +6e _ {3} \\ F ( e _ {3} )&= \left ( 7,~7,~7 \right ) &=7e _ {1} +7e _ {2} +7e _ {3} \end{split} $ and so $ [F]= \left[ \matrix {1 & 2 & 7\\3 & 4 & 7\\5 & 6 & 7} \right] $

 

That is, the columns of $ [F] $ are the images of the usual basis vectors.

 

Consider the following $ 3 \times 3 $ matrix $ A $ and basis $ S $ of $ R ^ {3} $:

$ A= \left[ \matrix {1 & -2 & 1\\3 & -1 & 0\\1 & 4 & -2} \right] $ and $ S= \left\{ u _ {1} ,~u _ {2} ,~u _ {3} \right\} = \left\{ \left[ \matrix {1\\1\\1} \right] ,~ \left[ \matrix {0\\1\\1} \right] , \left[ \matrix {1\\2\\3} \right] \right\} $

The matrix $ A $ defines a linear operator on $ R ^ {3} $. Find the matrix $ B $ that represents the mapping $ A $

relative to the basis $ S $. (Recall that $ A $ represents itself relative to the usual basis of $ R ^ {3} $.)

The matrix $ A $ defines a linear operator on $ R ^ {3} $. Find the matrix $ B $ that represents the mapping $ A $

relative to the basis S. (Recall that $ A $ represents itself relative to the usual basis of $ R ^ {3} $.)

First find the coordinates of an arbitrary vector $ \left ( a,~b,~c \right ) $ in $ R ^ {3} $ with respect to the basis $ S $. We have

$ \left[ \matrix {a\\b\\c} \right] =x \left[ \matrix {1\\1\\1} \right] +y \left[ \matrix {0\\1\\1} \right] +z \left[ \matrix {1\\2\\3} \right] $     or      $ \left \{ \begin{split} &x&+z&=a\\&x+y&+2z&=b\\&x+y&+2z&=c \end{split} \right. $

Solve for $ x,~y,~z $ in terms of $ a,~b,~c $ to get

$$ x=a+b-c,~y=-a+2b-c,~z=c-b $$

thus $ ( a,~b,~c) ^ {T} = ( a+b-c)u _ {1} + ( -a+2b-c)u _ {2} + ( c-b)u _ {3} $

 

Then use the formula for $ ( a,~b,~c) ^ {T} $ to find the coordinates of $ Au _ {1} ,~Au _ {2} ,~Au _ {3} $ relative to the basis $ S $:

$\left \{ \begin{split} A ( u _ {1} )&=A ( 1,~1,~1) ^ {T} = \left ( 0,~2,~3 \right ) ^ {T} &=-u _ {1} +u _ {2} +u _ {3} \\ A ( u _ {2} )&=A ( 1,~1,~0) ^ {T} = \left ( -1,~-1,~2 \right ) ^ {T} &=-4u _ {1} -3u _ {2} +3u _ {3} \\ A ( u _ {3} ) &=A ( 1,~2,~3) ^ {T} = \left ( 0,~1,~3 \right ) ^ {T} &=-2u _ {1} -u _ {2} +2u _ {3} \end{split} \right. $ so $ B= \left[ \matrix {-1 & -4 & -2\\1 & -3 & -1\\1 & 3 & 2} \right] $