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. (
$begingroup$
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
edited 55 mins ago
YuiTo Cheng
2,52341037
asked 1 hour ago
Dimen3ionalDimen3ional
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 55 mins ago
YuiTo Cheng
2,52341037
asked 1 hour ago
Dimen3ionalDimen3ional
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 55 mins ago
YuiTo Cheng
2,52341037
asked 1 hour ago
Dimen3ionalDimen3ional
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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
edited 55 mins ago
YuiTo Cheng
2,52341037
asked 1 hour ago
Dimen3ionalDimen3ional
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 55 mins ago
YuiTo Cheng
2,52341037
asked 1 hour ago
Dimen3ionalDimen3ional
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 55 mins ago
YuiTo Cheng
2,52341037
edited 55 mins ago
YuiTo Cheng
2,52341037
edited 55 mins ago
YuiTo Cheng
2,52341037
2,52341037
asked 1 hour ago
Dimen3ionalDimen3ional
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Dimen3ionalDimen3ional
132
asked 1 hour ago
Dimen3ionalDimen3ional
132
132
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dimen3ional is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$defmyvec#1{{bf#1}}$
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 59 mins ago
DavidDavid
70.1k668131
$endgroup$
$begingroup$
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
52 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 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.
answered 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$defmyvec#1{{bf#1}}$
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 59 mins ago
DavidDavid
70.1k668131
$endgroup$
$begingroup$
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
52 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 mins ago
add a comment |
$begingroup$
$defmyvec#1{{bf#1}}$
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 59 mins ago
DavidDavid
70.1k668131
$endgroup$
$begingroup$
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
52 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 mins ago
add a comment |
$begingroup$
$defmyvec#1{{bf#1}}$
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 59 mins ago
DavidDavid
70.1k668131
$endgroup$
$defmyvec#1{{bf#1}}$
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 59 mins ago
DavidDavid
70.1k668131
answered 59 mins ago
DavidDavid
70.1k668131
answered 59 mins ago
DavidDavid
70.1k668131
answered 59 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
52 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 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
52 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 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
52 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
52 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 mins ago
$begingroup$
Have you studied proof by contradiction or by contrapositive? That's what this is.
$endgroup$
– David
51 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.
answered 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
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.
answered 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
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.
answered 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
$endgroup$
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 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
answered 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
answered 58 mins ago
Kavi Rama MurthyKavi Rama Murthy
75.6k53270
answered 58 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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