If u is orthogonal to both v and w, and u not equal to 0, argue that u is not in the span of v and w. ( Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Orthogonal projection on span $x$Prove the orthogonal complement is equal to the orthogonal complement of the Span.Linear Independence and “Not in the Span”Incomplete Gauss Jordan elimination: what have I left outConsider the following system and find the values of b for which the system has a solutionGeneral form of a matrix that is both centrosymmetric and orthogonalOrthogonal Complement of SpanFind 4x3 matrix B such that AB = IOrthogonal projection on SpanFor what values of k does this system of equations have a unique / infinite / no solutions?
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If u is orthogonal to both v and w, and u not equal to 0, argue that u is not in the span of v and w. (
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Orthogonal projection on span $x$Prove the orthogonal complement is equal to the orthogonal complement of the Span.Linear Independence and “Not in the Span”Incomplete Gauss Jordan elimination: what have I left outConsider the following system and find the values of b for which the system has a solutionGeneral form of a matrix that is both centrosymmetric and orthogonalOrthogonal Complement of SpanFind 4x3 matrix B such that AB = IOrthogonal projection on SpanFor what values of k does this system of equations have a unique / infinite / no solutions?
$begingroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
New contributor
$endgroup$
add a comment |
$begingroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
New contributor
$endgroup$
add a comment |
$begingroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
New contributor
$endgroup$
QN: If u is orthogonal to both v and w, and u ≠ 0, argue that u is not in the span of v and w.
Where I am at:
I get stuck when it comes to solving my augmented matrix with Gauss Jordan Elimination.
I also tried formulating the following steps to solve the problem.
Create instances of u, v and w that pertain to the question. My visualisation in Geogebra can be viewed here: https://ggbm.at/b6xvwhpa
Set u = av + bw = u (where a and b are constants)
- Disprove (2)
However, I could not get past step 1.
Any pointers would be greatly appreciated.
linear-algebra matrices
linear-algebra matrices
New contributor
New contributor
edited 16 mins ago
YuiTo Cheng
2,52341037
2,52341037
New contributor
asked 28 mins ago
Dimen3ionalDimen3ional
82
82
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New contributor
add a comment |
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2 Answers
2
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oldest
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$begingroup$
$defmyvec#1bf#1$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
$endgroup$
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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active
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$begingroup$
$defmyvec#1bf#1$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
$endgroup$
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
add a comment |
$begingroup$
$defmyvec#1bf#1$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
$endgroup$
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
add a comment |
$begingroup$
$defmyvec#1bf#1$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
$endgroup$
$defmyvec#1bf#1$
This is the same as saying
if $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, and $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=myvec 0$.
So, if $myvec u$ is in the span of $myvec v$ and $myvec w$, then $myvec u=amyvec v+bmyvec w$ for some scalars $a,b$. Assuming also $myvec u$ is orthogonal to both $myvec v$ and $myvec w$, this means $myvec ucdot myvec v=0$ and $myvec ucdotmyvec w=0$, so
$$myvec ucdot myvec u=myvec ucdot(amyvec v+bmyvec w)=a(myvec ucdot myvec v)+b(myvec ucdot myvec w)=a0+b0=0 .$$
Since $myvec ucdot myvec u=0$ we have $myvec u=myvec 0$.
answered 20 mins ago
DavidDavid
70.1k668131
70.1k668131
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
add a comment |
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
I thought about this, however, the question specifically states that u can not be zero. If u can't be zero does that not prevent the above proof, or am I looking at this the wrong way?
$endgroup$
– Dimen3ional
13 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
12 mins ago
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
$endgroup$
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
$endgroup$
add a comment |
$begingroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
$endgroup$
If $u$ belongs to the span of $v$ and $w$ the $u=av+bw$ for some scalars $a$ and $b$. Since $langle u, v rangle=0$ and $langle u, w rangle=0$ we get $langle u, (av+bw) rangle=0$ so $langle u, u rangle=0$. This means $u=0$ which is a contradiction.
answered 19 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
75.6k53270
add a comment |
add a comment |
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
Dimen3ional is a new contributor. Be nice, and check out our Code of Conduct.
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