Find transformation matrix in terms of basis b and c

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Linear transformation: $[x,y]to[4y-x, 2x, 8x-2y]$
with respect to bases:

$B=[e_1,e_1+e_2]$

$C=[e_1,e_2,e_3]$

edited Sep 9 at 6:56

giannispapav

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asked Sep 9 at 6:27

Kcurse

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up vote
1
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Linear transformation: $[x,y]to[4y-x, 2x, 8x-2y]$
with respect to bases:

$B=[e_1,e_1+e_2]$

$C=[e_1,e_2,e_3]$

edited Sep 9 at 6:56

giannispapav

1,340323

asked Sep 9 at 6:27

Kcurse

add a commentÂ |Â

up vote
1
down vote

favorite

Linear transformation: $[x,y]to[4y-x, 2x, 8x-2y]$
with respect to bases:

$B=[e_1,e_1+e_2]$

$C=[e_1,e_2,e_3]$

edited Sep 9 at 6:56

giannispapav

1,340323

asked Sep 9 at 6:27

Kcurse

Linear transformation: $[x,y]to[4y-x, 2x, 8x-2y]$
with respect to bases:

$B=[e_1,e_1+e_2]$

$C=[e_1,e_2,e_3]$

linear-algebra linear-transformations

edited Sep 9 at 6:56

giannispapav

1,340323

asked Sep 9 at 6:27

Kcurse

edited Sep 9 at 6:56

giannispapav

1,340323

asked Sep 9 at 6:27

Kcurse

edited Sep 9 at 6:56

giannispapav

1,340323

edited Sep 9 at 6:56

giannispapav

1,340323

edited Sep 9 at 6:56

giannispapav

1,340323

asked Sep 9 at 6:27

Kcurse

asked Sep 9 at 6:27

Kcurse

asked Sep 9 at 6:27

Kcurse

add a commentÂ |Â

2 Answers
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Let $E = e_1, e_2$ be the canonical basis for $mathbbR^2$. Your map, call it $A$, has this matrix with respect to canonical bases for $mathbbR^2$ and $mathbbR^3$:

$$A_(C,E) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrix$$

To get $A_(C,B)$, the transformation matrix is the matrix of the identity map with respect to bases $B$ and $E$, namely $I_(E, B) = beginbmatrix 1 & 1 \ 0 & 1endbmatrix$. Now we have

$$A_(C,B) = A_(C,E)I_(E,B) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrixbeginbmatrix 1 & 1 \ 0 & 1endbmatrix = beginbmatrix -1 & 3 \ 2 & 2 \ 8 & 6endbmatrix$$

answered Sep 9 at 7:26

mechanodroid

24.7k62245

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Hint

$e_1=[1,0]mapsto [-1,2,8]=-1e_1+2e_2+8e_3$

$e_2=[1,1]mapsto [3,2,6]=3e_1+2e_2+6e_3$

Let's call the transformation $f$ then $(f:B,C)=beginpmatrix -1 & 3 \
2 & 2\
8 & 6 endpmatrix$

(I wrote the coefficients vertically)

edited Sep 9 at 8:21

answered Sep 9 at 6:35

giannispapav

1,340323

I understand that but how do I form the transformation matrix?
â€“Â Kcurse
Sep 9 at 6:46

add a commentÂ |Â

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2 Answers
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2 Answers
2

active

oldest

votes

up vote
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Let $E = e_1, e_2$ be the canonical basis for $mathbbR^2$. Your map, call it $A$, has this matrix with respect to canonical bases for $mathbbR^2$ and $mathbbR^3$:

$$A_(C,E) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrix$$

To get $A_(C,B)$, the transformation matrix is the matrix of the identity map with respect to bases $B$ and $E$, namely $I_(E, B) = beginbmatrix 1 & 1 \ 0 & 1endbmatrix$. Now we have

$$A_(C,B) = A_(C,E)I_(E,B) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrixbeginbmatrix 1 & 1 \ 0 & 1endbmatrix = beginbmatrix -1 & 3 \ 2 & 2 \ 8 & 6endbmatrix$$

answered Sep 9 at 7:26

mechanodroid

24.7k62245

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up vote
0
down vote

Let $E = e_1, e_2$ be the canonical basis for $mathbbR^2$. Your map, call it $A$, has this matrix with respect to canonical bases for $mathbbR^2$ and $mathbbR^3$:

$$A_(C,E) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrix$$

To get $A_(C,B)$, the transformation matrix is the matrix of the identity map with respect to bases $B$ and $E$, namely $I_(E, B) = beginbmatrix 1 & 1 \ 0 & 1endbmatrix$. Now we have

$$A_(C,B) = A_(C,E)I_(E,B) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrixbeginbmatrix 1 & 1 \ 0 & 1endbmatrix = beginbmatrix -1 & 3 \ 2 & 2 \ 8 & 6endbmatrix$$

answered Sep 9 at 7:26

mechanodroid

24.7k62245

add a commentÂ |Â

up vote
0
down vote

Let $E = e_1, e_2$ be the canonical basis for $mathbbR^2$. Your map, call it $A$, has this matrix with respect to canonical bases for $mathbbR^2$ and $mathbbR^3$:

$$A_(C,E) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrix$$

To get $A_(C,B)$, the transformation matrix is the matrix of the identity map with respect to bases $B$ and $E$, namely $I_(E, B) = beginbmatrix 1 & 1 \ 0 & 1endbmatrix$. Now we have

$$A_(C,B) = A_(C,E)I_(E,B) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrixbeginbmatrix 1 & 1 \ 0 & 1endbmatrix = beginbmatrix -1 & 3 \ 2 & 2 \ 8 & 6endbmatrix$$

answered Sep 9 at 7:26

mechanodroid

24.7k62245

Let $E = e_1, e_2$ be the canonical basis for $mathbbR^2$. Your map, call it $A$, has this matrix with respect to canonical bases for $mathbbR^2$ and $mathbbR^3$:

$$A_(C,E) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrix$$

To get $A_(C,B)$, the transformation matrix is the matrix of the identity map with respect to bases $B$ and $E$, namely $I_(E, B) = beginbmatrix 1 & 1 \ 0 & 1endbmatrix$. Now we have

$$A_(C,B) = A_(C,E)I_(E,B) = beginbmatrix -1 & 4 \ 2 & 0 \ 8 & -2endbmatrixbeginbmatrix 1 & 1 \ 0 & 1endbmatrix = beginbmatrix -1 & 3 \ 2 & 2 \ 8 & 6endbmatrix$$

answered Sep 9 at 7:26

mechanodroid

24.7k62245

answered Sep 9 at 7:26

mechanodroid

24.7k62245

answered Sep 9 at 7:26

mechanodroid

24.7k62245

answered Sep 9 at 7:26

mechanodroid

24.7k62245

add a commentÂ |Â

up vote
0
down vote

Hint

$e_1=[1,0]mapsto [-1,2,8]=-1e_1+2e_2+8e_3$

$e_2=[1,1]mapsto [3,2,6]=3e_1+2e_2+6e_3$

Let's call the transformation $f$ then $(f:B,C)=beginpmatrix -1 & 3 \
2 & 2\
8 & 6 endpmatrix$

(I wrote the coefficients vertically)

edited Sep 9 at 8:21

answered Sep 9 at 6:35

giannispapav

1,340323

I understand that but how do I form the transformation matrix?
â€“Â Kcurse
Sep 9 at 6:46

add a commentÂ |Â

up vote
0
down vote

Hint

$e_1=[1,0]mapsto [-1,2,8]=-1e_1+2e_2+8e_3$

$e_2=[1,1]mapsto [3,2,6]=3e_1+2e_2+6e_3$

Let's call the transformation $f$ then $(f:B,C)=beginpmatrix -1 & 3 \
2 & 2\
8 & 6 endpmatrix$

(I wrote the coefficients vertically)

edited Sep 9 at 8:21

answered Sep 9 at 6:35

giannispapav

1,340323

I understand that but how do I form the transformation matrix?
â€“Â Kcurse
Sep 9 at 6:46

add a commentÂ |Â

up vote
0
down vote

Hint

$e_1=[1,0]mapsto [-1,2,8]=-1e_1+2e_2+8e_3$

$e_2=[1,1]mapsto [3,2,6]=3e_1+2e_2+6e_3$

Let's call the transformation $f$ then $(f:B,C)=beginpmatrix -1 & 3 \
2 & 2\
8 & 6 endpmatrix$

(I wrote the coefficients vertically)

edited Sep 9 at 8:21

answered Sep 9 at 6:35

giannispapav

1,340323

Hint

$e_1=[1,0]mapsto [-1,2,8]=-1e_1+2e_2+8e_3$

$e_2=[1,1]mapsto [3,2,6]=3e_1+2e_2+6e_3$

Let's call the transformation $f$ then $(f:B,C)=beginpmatrix -1 & 3 \
2 & 2\
8 & 6 endpmatrix$

(I wrote the coefficients vertically)

edited Sep 9 at 8:21

answered Sep 9 at 6:35

giannispapav

1,340323

edited Sep 9 at 8:21

answered Sep 9 at 6:35

giannispapav

1,340323

answered Sep 9 at 6:35

giannispapav

1,340323

answered Sep 9 at 6:35

giannispapav

1,340323

I understand that but how do I form the transformation matrix?
â€“Â Kcurse
Sep 9 at 6:46

add a commentÂ |Â

I understand that but how do I form the transformation matrix?
â€“Â Kcurse
Sep 9 at 6:46

I understand that but how do I form the transformation matrix?
â€“Â Kcurse
Sep 9 at 6:46

add a commentÂ |Â

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