Use the Gram-Schmidt Process to construct an orthogonal set of vectors using an inner product [closed]
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This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.
Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:
Column vectors: $[1, 2]$, $[1, 1]$
Inner Product: $langle u, v rangle = u^TAv,$
where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.
linear-algebra gram-schmidt
edited Dec 14 '18 at 6:05
The Pointer
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asked Dec 14 '18 at 4:34
SarahSarah
41
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closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16
This question appears to be off-topic. The users who voted to close gave this specific reason:
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If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.
Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:
Column vectors: $[1, 2]$, $[1, 1]$
Inner Product: $langle u, v rangle = u^TAv,$
where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.
linear-algebra gram-schmidt
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
asked Dec 14 '18 at 4:34
SarahSarah
41
$endgroup$
closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54
add a comment |
$begingroup$
This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.
Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:
Column vectors: $[1, 2]$, $[1, 1]$
Inner Product: $langle u, v rangle = u^TAv,$
where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.
linear-algebra gram-schmidt
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
asked Dec 14 '18 at 4:34
SarahSarah
41
$endgroup$
This question on my Linear Algebra test has been stumping me. Every time I finish this problem I end up with vectors which aren't orthogonal.
Use the Gram-Schmidt Process to construct an orthogonal set of vectors for the given set of vectors using the specified inner product:
Column vectors: $[1, 2]$, $[1, 1]$
Inner Product: $langle u, v rangle = u^TAv,$
where $A= begin{bmatrix}6&4\4&6end{bmatrix}$.
linear-algebra gram-schmidt
linear-algebra gram-schmidt
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
asked Dec 14 '18 at 4:34
SarahSarah
41
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
asked Dec 14 '18 at 4:34
SarahSarah
41
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
edited Dec 14 '18 at 6:05
The Pointer
2,48731641
2,48731641
asked Dec 14 '18 at 4:34
SarahSarah
41
asked Dec 14 '18 at 4:34
SarahSarah
41
asked Dec 14 '18 at 4:34
SarahSarah
41
41
closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0 Dec 14 '18 at 15:16
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheGeekGreek, José Carlos Santos, Riccardo.Alestra, Namaste, The Chaz 2.0
If this question can be reworded to fit the rules in the help center, please edit the question.
2
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If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
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– amd
Dec 14 '18 at 8:54
add a comment |
2
$begingroup$
If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54
2
2
$begingroup$
If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54
$begingroup$
If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54
add a comment |
2 Answers
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Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
$$
u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
$$
We compute
$$
langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
$$
and
$$
langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
$$
so
$$
u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
$$
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
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Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.
Let $v_1 = (1, 2)$.
$v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$
Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.
Check:
$left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.
Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf
answered Dec 14 '18 at 7:46
The PointerThe Pointer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
$$
u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
$$
We compute
$$
langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
$$
and
$$
langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
$$
so
$$
u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
$$
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
$endgroup$
add a comment |
$begingroup$
Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
$$
u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
$$
We compute
$$
langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
$$
and
$$
langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
$$
so
$$
u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
$$
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
$endgroup$
add a comment |
$begingroup$
Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
$$
u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
$$
We compute
$$
langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
$$
and
$$
langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
$$
so
$$
u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
$$
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
$endgroup$
Let $v_1=(1,2)$, $v_2=(1,1)$. We can let the first vector in our orthogonal set be $u_1=v_1=(1,2)$. The second vector $u_2$ is found by subtracting the projection of $v_2$ onto $u_1$ from $v_2$:
$$
u_2 = v_2 - mathsf{proj}_{u_1}(v_2) = v_2 - frac{langle v_2,u_1rangle}{langle u_1,u_1rangle}u_1.
$$
We compute
$$
langle v_2,u_1rangle = pmatrix{1&1}pmatrix{6&4\4&6}pmatrix{1\2}= 30
$$
and
$$
langle u_1,u_1rangle = pmatrix{1&2}pmatrix{6&4\4&6}pmatrix{1\2}= 46,
$$
so
$$
u_2 = (1,1) - frac{15}{23}cdot(1,2) = left(frac8{23}, -frac7{23}right).
$$
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
answered Dec 14 '18 at 7:33
Math1000Math1000
19.4k31746
19.4k31746
add a comment |
add a comment |
$begingroup$
Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.
Let $v_1 = (1, 2)$.
$v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$
Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.
Check:
$left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.
Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
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add a comment |
$begingroup$
Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.
Let $v_1 = (1, 2)$.
$v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$
Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.
Check:
$left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.
Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
$endgroup$
add a comment |
$begingroup$
Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.
Let $v_1 = (1, 2)$.
$v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$
Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.
Check:
$left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.
Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
$endgroup$
Let $V$ be an inner product space, with the inner product defined as $langle u, v rangle = u^T A v$, where $A= begin{bmatrix}6&4\4&6end{bmatrix}$. The set ${ (1, 2), (1, 1) }$ is a basis for $V$, since the vectors are linearly independent and span $V$.
Let $v_1 = (1, 2)$.
$v_2 = (1, 1) - dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} v_1 = dfrac{langle (1 , 1), (1, 2) rangle}{langle (1, 2), (1, 2) rangle} (1, 2) = (1,1) - dfrac{30}{46}(1, 2) = left( dfrac{8}{23} , -dfrac{7}{23} right)$
Therefore, $left { (1, 2), left( dfrac{8}{23} , -dfrac{7}{23} right) right }$ is an orthogonal basis for $V$.
Check:
$left langle (1, 2), left( dfrac{8}{23}, dfrac{-7}{23} right) right rangle = dfrac{112}{23} - dfrac{112}{23} = 0$, as required.
Reference: http://www.math.tamu.edu/~yvorobet/MATH304-503/Lect3-07web.pdf
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
edited Dec 14 '18 at 23:27
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
answered Dec 14 '18 at 7:46
The PointerThe Pointer
2,48731641
2,48731641
add a comment |
add a comment |
2
$begingroup$
If you show your work instead of making everyone guess, then someone could point out where you’re going wrong.
$endgroup$
– amd
Dec 14 '18 at 8:54