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High Schooler (Advanced Linear Algebra Help) [closed]
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My teacher gave us a homework problem that is considered an "advanced topic" that we have not really delved deeply into yet. I'm struggling to connect the dots in the question to formulate my answer.
I assume that I will be constructing a system of linear equations that gives solutions d, -b, -c, and a. My question is how the basis U = (1, t, t^2, t^3) relates to T(at^3 + bt^2 + ct + d) so that I can move forward and solve this problem.
linear-algebra matrices linear-transformations change-of-basis
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
asked Jul 21 at 11:23
SleepApnea
426
closed as off-topic by José Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz Jul 27 at 17:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz
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up vote
2
down vote
favorite
My teacher gave us a homework problem that is considered an "advanced topic" that we have not really delved deeply into yet. I'm struggling to connect the dots in the question to formulate my answer.
I assume that I will be constructing a system of linear equations that gives solutions d, -b, -c, and a. My question is how the basis U = (1, t, t^2, t^3) relates to T(at^3 + bt^2 + ct + d) so that I can move forward and solve this problem.
linear-algebra matrices linear-transformations change-of-basis
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
asked Jul 21 at 11:23
SleepApnea
426
closed as off-topic by José Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz Jul 27 at 17:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
My teacher gave us a homework problem that is considered an "advanced topic" that we have not really delved deeply into yet. I'm struggling to connect the dots in the question to formulate my answer.
I assume that I will be constructing a system of linear equations that gives solutions d, -b, -c, and a. My question is how the basis U = (1, t, t^2, t^3) relates to T(at^3 + bt^2 + ct + d) so that I can move forward and solve this problem.
linear-algebra matrices linear-transformations change-of-basis
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
asked Jul 21 at 11:23
SleepApnea
426
My teacher gave us a homework problem that is considered an "advanced topic" that we have not really delved deeply into yet. I'm struggling to connect the dots in the question to formulate my answer.
I assume that I will be constructing a system of linear equations that gives solutions d, -b, -c, and a. My question is how the basis U = (1, t, t^2, t^3) relates to T(at^3 + bt^2 + ct + d) so that I can move forward and solve this problem.
linear-algebra matrices linear-transformations change-of-basis
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
asked Jul 21 at 11:23
SleepApnea
426
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
edited Jul 21 at 11:24
Parcly Taxel
33.6k136588
33.6k136588
asked Jul 21 at 11:23
SleepApnea
426
asked Jul 21 at 11:23
SleepApnea
426
asked Jul 21 at 11:23
SleepApnea
426
426
closed as off-topic by José Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz Jul 27 at 17:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz
closed as off-topic by José Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz Jul 27 at 17:04
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." РJos̩ Carlos Santos, amWhy, Jyrki Lahtonen, Xander Henderson, Mostafa Ayaz
add a comment |Â
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3 Answers
3
active
oldest
votes
up vote
4
down vote
To find the matrix of your transformation, see what the transformation does to each basis element.
For instance: to find the third column of the matrix, see what $T$ does to the third element of $U$, $t^2$. If $M$ denotes the matrix of the transformation, then we find that (in the notation described here)
$$
M pmatrix0\0\1\0 = M [t^2]_U = [T(t^2)]_B = left[pmatrix0&-1\0&0right]_B = pmatrix0\-1\0\0
$$
So, the matrix we're looking for has the form
$$
M = pmatrix?&?&0&?\?&?&-1&?\?&?&0&?\?&?&0&?
$$
answered Jul 21 at 14:59
Omnomnomnom
121k784170
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up vote
3
down vote
We have the basis $U= u_1, u_2, u_3, u_4 $ and
$B= b_1, b_2, b_3, b_4 $. Then $T$ acts like this:
beginalign
T(u_1) &= T(1) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace1_dcdot 1) =
beginbmatrix
1 & 0 \
0 & 0
endbmatrix
= b_1
\
T(u_2) &= T(t) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace1_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
-1 & 0
endbmatrix
= -b_3
\
T(u_3) &= T(t^2) =
T(underbrace0_acdot t^3 + underbrace1_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & -1\
0 & 0
endbmatrix
= -b_2
\
T(u_4) &= T(t^3) =
T(underbrace1_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
0 & 1
endbmatrix
= b_4
endalign
answered Jul 21 at 11:41
mvw
30.4k22250
add a comment |Â
up vote
0
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The matrix is $$M= begin bmatrix 0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endbmatrix $$
Note that $$Mbegin pmatrix a\b\c\dendpmatrix =begin pmatrix d\-b\-c\aendpmatrix$$
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
2
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
To find the matrix of your transformation, see what the transformation does to each basis element.
For instance: to find the third column of the matrix, see what $T$ does to the third element of $U$, $t^2$. If $M$ denotes the matrix of the transformation, then we find that (in the notation described here)
$$
M pmatrix0\0\1\0 = M [t^2]_U = [T(t^2)]_B = left[pmatrix0&-1\0&0right]_B = pmatrix0\-1\0\0
$$
So, the matrix we're looking for has the form
$$
M = pmatrix?&?&0&?\?&?&-1&?\?&?&0&?\?&?&0&?
$$
answered Jul 21 at 14:59
Omnomnomnom
121k784170
add a comment |Â
up vote
4
down vote
To find the matrix of your transformation, see what the transformation does to each basis element.
For instance: to find the third column of the matrix, see what $T$ does to the third element of $U$, $t^2$. If $M$ denotes the matrix of the transformation, then we find that (in the notation described here)
$$
M pmatrix0\0\1\0 = M [t^2]_U = [T(t^2)]_B = left[pmatrix0&-1\0&0right]_B = pmatrix0\-1\0\0
$$
So, the matrix we're looking for has the form
$$
M = pmatrix?&?&0&?\?&?&-1&?\?&?&0&?\?&?&0&?
$$
answered Jul 21 at 14:59
Omnomnomnom
121k784170
add a comment |Â
up vote
4
down vote
up vote
4
down vote
To find the matrix of your transformation, see what the transformation does to each basis element.
For instance: to find the third column of the matrix, see what $T$ does to the third element of $U$, $t^2$. If $M$ denotes the matrix of the transformation, then we find that (in the notation described here)
$$
M pmatrix0\0\1\0 = M [t^2]_U = [T(t^2)]_B = left[pmatrix0&-1\0&0right]_B = pmatrix0\-1\0\0
$$
So, the matrix we're looking for has the form
$$
M = pmatrix?&?&0&?\?&?&-1&?\?&?&0&?\?&?&0&?
$$
answered Jul 21 at 14:59
Omnomnomnom
121k784170
To find the matrix of your transformation, see what the transformation does to each basis element.
For instance: to find the third column of the matrix, see what $T$ does to the third element of $U$, $t^2$. If $M$ denotes the matrix of the transformation, then we find that (in the notation described here)
$$
M pmatrix0\0\1\0 = M [t^2]_U = [T(t^2)]_B = left[pmatrix0&-1\0&0right]_B = pmatrix0\-1\0\0
$$
So, the matrix we're looking for has the form
$$
M = pmatrix?&?&0&?\?&?&-1&?\?&?&0&?\?&?&0&?
$$
answered Jul 21 at 14:59
Omnomnomnom
121k784170
answered Jul 21 at 14:59
Omnomnomnom
121k784170
answered Jul 21 at 14:59
Omnomnomnom
121k784170
answered Jul 21 at 14:59
Omnomnomnom
121k784170
121k784170
add a comment |Â
add a comment |Â
up vote
3
down vote
We have the basis $U= u_1, u_2, u_3, u_4 $ and
$B= b_1, b_2, b_3, b_4 $. Then $T$ acts like this:
beginalign
T(u_1) &= T(1) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace1_dcdot 1) =
beginbmatrix
1 & 0 \
0 & 0
endbmatrix
= b_1
\
T(u_2) &= T(t) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace1_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
-1 & 0
endbmatrix
= -b_3
\
T(u_3) &= T(t^2) =
T(underbrace0_acdot t^3 + underbrace1_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & -1\
0 & 0
endbmatrix
= -b_2
\
T(u_4) &= T(t^3) =
T(underbrace1_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
0 & 1
endbmatrix
= b_4
endalign
answered Jul 21 at 11:41
mvw
30.4k22250
add a comment |Â
up vote
3
down vote
We have the basis $U= u_1, u_2, u_3, u_4 $ and
$B= b_1, b_2, b_3, b_4 $. Then $T$ acts like this:
beginalign
T(u_1) &= T(1) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace1_dcdot 1) =
beginbmatrix
1 & 0 \
0 & 0
endbmatrix
= b_1
\
T(u_2) &= T(t) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace1_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
-1 & 0
endbmatrix
= -b_3
\
T(u_3) &= T(t^2) =
T(underbrace0_acdot t^3 + underbrace1_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & -1\
0 & 0
endbmatrix
= -b_2
\
T(u_4) &= T(t^3) =
T(underbrace1_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
0 & 1
endbmatrix
= b_4
endalign
answered Jul 21 at 11:41
mvw
30.4k22250
add a comment |Â
up vote
3
down vote
up vote
3
down vote
We have the basis $U= u_1, u_2, u_3, u_4 $ and
$B= b_1, b_2, b_3, b_4 $. Then $T$ acts like this:
beginalign
T(u_1) &= T(1) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace1_dcdot 1) =
beginbmatrix
1 & 0 \
0 & 0
endbmatrix
= b_1
\
T(u_2) &= T(t) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace1_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
-1 & 0
endbmatrix
= -b_3
\
T(u_3) &= T(t^2) =
T(underbrace0_acdot t^3 + underbrace1_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & -1\
0 & 0
endbmatrix
= -b_2
\
T(u_4) &= T(t^3) =
T(underbrace1_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
0 & 1
endbmatrix
= b_4
endalign
answered Jul 21 at 11:41
mvw
30.4k22250
We have the basis $U= u_1, u_2, u_3, u_4 $ and
$B= b_1, b_2, b_3, b_4 $. Then $T$ acts like this:
beginalign
T(u_1) &= T(1) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace1_dcdot 1) =
beginbmatrix
1 & 0 \
0 & 0
endbmatrix
= b_1
\
T(u_2) &= T(t) =
T(underbrace0_acdot t^3 + underbrace0_bcdot t^2 + underbrace1_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
-1 & 0
endbmatrix
= -b_3
\
T(u_3) &= T(t^2) =
T(underbrace0_acdot t^3 + underbrace1_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & -1\
0 & 0
endbmatrix
= -b_2
\
T(u_4) &= T(t^3) =
T(underbrace1_acdot t^3 + underbrace0_bcdot t^2 + underbrace0_ccdot t + underbrace0_dcdot 1) =
beginbmatrix
0 & 0 \
0 & 1
endbmatrix
= b_4
endalign
answered Jul 21 at 11:41
mvw
30.4k22250
edited Jul 21 at 12:30
answered Jul 21 at 11:41
mvw
30.4k22250
answered Jul 21 at 11:41
mvw
30.4k22250
answered Jul 21 at 11:41
mvw
30.4k22250
30.4k22250
add a comment |Â
add a comment |Â
up vote
0
down vote
The matrix is $$M= begin bmatrix 0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endbmatrix $$
Note that $$Mbegin pmatrix a\b\c\dendpmatrix =begin pmatrix d\-b\-c\aendpmatrix$$
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
2
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
add a comment |Â
up vote
0
down vote
The matrix is $$M= begin bmatrix 0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endbmatrix $$
Note that $$Mbegin pmatrix a\b\c\dendpmatrix =begin pmatrix d\-b\-c\aendpmatrix$$
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
2
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The matrix is $$M= begin bmatrix 0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endbmatrix $$
Note that $$Mbegin pmatrix a\b\c\dendpmatrix =begin pmatrix d\-b\-c\aendpmatrix$$
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
The matrix is $$M= begin bmatrix 0&0&0&1\0&-1&0&0\0&0&-1&0\1&0&0&0endbmatrix $$
Note that $$Mbegin pmatrix a\b\c\dendpmatrix =begin pmatrix d\-b\-c\aendpmatrix$$
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
answered Jul 21 at 12:00
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
2
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
add a comment |Â
2
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
2
2
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
The (ordered) basis is $(1, t, t^2, t^3)$, not $(t^3,t^2,t,1)$.
– Ennar
Jul 21 at 15:11
add a comment |Â