Why does row reduction not change the dependence relation between columns?
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I understand why it wouldn't change the dependence relation between rows, but that isn't intuitively clear for the columns (for me at least).
I realize that non-trivial solutions to the equation Ax = 0 implies a dependence relation among the columns of A, but what I don't understand is why elementary matrix operations wouldn't change the solution set.
linear-algebra matrices
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
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add a comment |
$begingroup$
I understand why it wouldn't change the dependence relation between rows, but that isn't intuitively clear for the columns (for me at least).
I realize that non-trivial solutions to the equation Ax = 0 implies a dependence relation among the columns of A, but what I don't understand is why elementary matrix operations wouldn't change the solution set.
linear-algebra matrices
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
$endgroup$
2
$begingroup$
In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last.
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– amd
Dec 4 '18 at 17:21
$begingroup$
See lem.ma/6wy and the subsequent lessons.
$endgroup$
– Lemma
Dec 4 '18 at 18:14
add a comment |
$begingroup$
I understand why it wouldn't change the dependence relation between rows, but that isn't intuitively clear for the columns (for me at least).
I realize that non-trivial solutions to the equation Ax = 0 implies a dependence relation among the columns of A, but what I don't understand is why elementary matrix operations wouldn't change the solution set.
linear-algebra matrices
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
$endgroup$
I understand why it wouldn't change the dependence relation between rows, but that isn't intuitively clear for the columns (for me at least).
I realize that non-trivial solutions to the equation Ax = 0 implies a dependence relation among the columns of A, but what I don't understand is why elementary matrix operations wouldn't change the solution set.
linear-algebra matrices
linear-algebra matrices
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
asked Dec 4 '18 at 17:07
James RonaldJames Ronald
1257
1257
2
$begingroup$
In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last.
$endgroup$
– amd
Dec 4 '18 at 17:21
$begingroup$
See lem.ma/6wy and the subsequent lessons.
$endgroup$
– Lemma
Dec 4 '18 at 18:14
add a comment |
2
$begingroup$
In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last.
$endgroup$
– amd
Dec 4 '18 at 17:21
$begingroup$
See lem.ma/6wy and the subsequent lessons.
$endgroup$
– Lemma
Dec 4 '18 at 18:14
2
2
$begingroup$
In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last.
$endgroup$
– amd
Dec 4 '18 at 17:21
$begingroup$
In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last.
$endgroup$
– amd
Dec 4 '18 at 17:21
$begingroup$
See lem.ma/6wy and the subsequent lessons.
$endgroup$
– Lemma
Dec 4 '18 at 18:14
$begingroup$
See lem.ma/6wy and the subsequent lessons.
$endgroup$
– Lemma
Dec 4 '18 at 18:14
add a comment |
2 Answers
2
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What you call "elementary matrix operations" amount to left-multiplication by an invertible matrix; therefore you are replacing $A$ with $BA$ for some invertible $B$.
But then $BAx=0$ iff $Ax=0$, so the solution set hasn't changed.
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
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$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
There are many technical explanations that prove it, but don't explain why it's true.
What convinces me is to imagine two columns where one is two times the other.
Does swapping 1st and 3rd entries in each column change that relationship?
Obviously no.
Does multiplying 2nd entry by, say, 7 in each column change that relationship?
Obviously no.
Does adding a multiple of the 3rd entry to the 4th entry in each column change that relationship?
Obviously no.
That gives me an insight that convinces me and I hope that you find it helpful.
For more, see https://lem.ma/6wy and the subsequent lessons.
answered Dec 4 '18 at 18:20
LemmaLemma
63529
$endgroup$
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
What you call "elementary matrix operations" amount to left-multiplication by an invertible matrix; therefore you are replacing $A$ with $BA$ for some invertible $B$.
But then $BAx=0$ iff $Ax=0$, so the solution set hasn't changed.
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
$endgroup$
$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
What you call "elementary matrix operations" amount to left-multiplication by an invertible matrix; therefore you are replacing $A$ with $BA$ for some invertible $B$.
But then $BAx=0$ iff $Ax=0$, so the solution set hasn't changed.
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
$endgroup$
$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
What you call "elementary matrix operations" amount to left-multiplication by an invertible matrix; therefore you are replacing $A$ with $BA$ for some invertible $B$.
But then $BAx=0$ iff $Ax=0$, so the solution set hasn't changed.
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
$endgroup$
What you call "elementary matrix operations" amount to left-multiplication by an invertible matrix; therefore you are replacing $A$ with $BA$ for some invertible $B$.
But then $BAx=0$ iff $Ax=0$, so the solution set hasn't changed.
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
answered Dec 4 '18 at 17:11
FedericoFederico
5,014514
5,014514
$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
$begingroup$
Oh that's also a great way of looking at it, forgot that elementary matrix operations is just left multiplying by an identity matrix. Thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
There are many technical explanations that prove it, but don't explain why it's true.
What convinces me is to imagine two columns where one is two times the other.
Does swapping 1st and 3rd entries in each column change that relationship?
Obviously no.
Does multiplying 2nd entry by, say, 7 in each column change that relationship?
Obviously no.
Does adding a multiple of the 3rd entry to the 4th entry in each column change that relationship?
Obviously no.
That gives me an insight that convinces me and I hope that you find it helpful.
For more, see https://lem.ma/6wy and the subsequent lessons.
answered Dec 4 '18 at 18:20
LemmaLemma
63529
$endgroup$
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
There are many technical explanations that prove it, but don't explain why it's true.
What convinces me is to imagine two columns where one is two times the other.
Does swapping 1st and 3rd entries in each column change that relationship?
Obviously no.
Does multiplying 2nd entry by, say, 7 in each column change that relationship?
Obviously no.
Does adding a multiple of the 3rd entry to the 4th entry in each column change that relationship?
Obviously no.
That gives me an insight that convinces me and I hope that you find it helpful.
For more, see https://lem.ma/6wy and the subsequent lessons.
answered Dec 4 '18 at 18:20
LemmaLemma
63529
$endgroup$
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
There are many technical explanations that prove it, but don't explain why it's true.
What convinces me is to imagine two columns where one is two times the other.
Does swapping 1st and 3rd entries in each column change that relationship?
Obviously no.
Does multiplying 2nd entry by, say, 7 in each column change that relationship?
Obviously no.
Does adding a multiple of the 3rd entry to the 4th entry in each column change that relationship?
Obviously no.
That gives me an insight that convinces me and I hope that you find it helpful.
For more, see https://lem.ma/6wy and the subsequent lessons.
answered Dec 4 '18 at 18:20
LemmaLemma
63529
$endgroup$
There are many technical explanations that prove it, but don't explain why it's true.
What convinces me is to imagine two columns where one is two times the other.
Does swapping 1st and 3rd entries in each column change that relationship?
Obviously no.
Does multiplying 2nd entry by, say, 7 in each column change that relationship?
Obviously no.
Does adding a multiple of the 3rd entry to the 4th entry in each column change that relationship?
Obviously no.
That gives me an insight that convinces me and I hope that you find it helpful.
For more, see https://lem.ma/6wy and the subsequent lessons.
answered Dec 4 '18 at 18:20
LemmaLemma
63529
answered Dec 4 '18 at 18:20
LemmaLemma
63529
answered Dec 4 '18 at 18:20
LemmaLemma
63529
answered Dec 4 '18 at 18:20
LemmaLemma
63529
63529
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
$begingroup$
This helps a lot, thank you!
$endgroup$
– James Ronald
Dec 4 '18 at 20:44
add a comment |
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2
$begingroup$
In fact, row reduction does change dependencies among the rows. In the RREF of a matrix, all of the zero rows—the ones that are linearly dependent—always come last.
$endgroup$
– amd
Dec 4 '18 at 17:21
$begingroup$
See lem.ma/6wy and the subsequent lessons.
$endgroup$
– Lemma
Dec 4 '18 at 18:14