Mixing
Show that the following two lines intersect at a point with $z neq 0$
$begingroup$
Consider the following two lines in complex projective space $mathbf{P^2}$:
$a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{2}z = 0$.
We suppose further that $a_{1}b_{2} - b_{1}a_{2} neq 0$. We must show these two lines intersect at a point with $z neq 0$.
I understand that if there is a point $(x:y:z)$ that lies on both lines, it must follow that $z neq 0$, since if $z = 0$ we have "linearly independent" equations and so the only solution would be $(0:0:0)$ which is not a point in $mathbf{P^2}$.
So to show that there is a solution with $z neq 0$ i'm not sure what to do. I think by doing gaussian elimination on the two equations above, we can find conditions on $x$ and $y$ in terms of $z$ to get that a point of the form $(alpha z, beta z, z)$ is a solution to both equations and so the two lines intersect at $(alpha z: beta z: z) in mathbf{P^2}$ where $z neq 0$.
Is this the correct way to approach this problem?
algebraic-geometry algebraic-curves
asked Jan 21 at 10:30
trynalearntrynalearn
662314
$endgroup$
add a comment |
$begingroup$
Consider the following two lines in complex projective space $mathbf{P^2}$:
$a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{2}z = 0$.
We suppose further that $a_{1}b_{2} - b_{1}a_{2} neq 0$. We must show these two lines intersect at a point with $z neq 0$.
I understand that if there is a point $(x:y:z)$ that lies on both lines, it must follow that $z neq 0$, since if $z = 0$ we have "linearly independent" equations and so the only solution would be $(0:0:0)$ which is not a point in $mathbf{P^2}$.
So to show that there is a solution with $z neq 0$ i'm not sure what to do. I think by doing gaussian elimination on the two equations above, we can find conditions on $x$ and $y$ in terms of $z$ to get that a point of the form $(alpha z, beta z, z)$ is a solution to both equations and so the two lines intersect at $(alpha z: beta z: z) in mathbf{P^2}$ where $z neq 0$.
Is this the correct way to approach this problem?
algebraic-geometry algebraic-curves
asked Jan 21 at 10:30
trynalearntrynalearn
662314
$endgroup$
add a comment |
$begingroup$
Consider the following two lines in complex projective space $mathbf{P^2}$:
$a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{2}z = 0$.
We suppose further that $a_{1}b_{2} - b_{1}a_{2} neq 0$. We must show these two lines intersect at a point with $z neq 0$.
I understand that if there is a point $(x:y:z)$ that lies on both lines, it must follow that $z neq 0$, since if $z = 0$ we have "linearly independent" equations and so the only solution would be $(0:0:0)$ which is not a point in $mathbf{P^2}$.
So to show that there is a solution with $z neq 0$ i'm not sure what to do. I think by doing gaussian elimination on the two equations above, we can find conditions on $x$ and $y$ in terms of $z$ to get that a point of the form $(alpha z, beta z, z)$ is a solution to both equations and so the two lines intersect at $(alpha z: beta z: z) in mathbf{P^2}$ where $z neq 0$.
Is this the correct way to approach this problem?
algebraic-geometry algebraic-curves
asked Jan 21 at 10:30
trynalearntrynalearn
662314
$endgroup$
Consider the following two lines in complex projective space $mathbf{P^2}$:
$a_{1}x + b_{1}y + c_{1}z = 0$ and $a_{2}x + b_{2}y + c_{2}z = 0$.
We suppose further that $a_{1}b_{2} - b_{1}a_{2} neq 0$. We must show these two lines intersect at a point with $z neq 0$.
I understand that if there is a point $(x:y:z)$ that lies on both lines, it must follow that $z neq 0$, since if $z = 0$ we have "linearly independent" equations and so the only solution would be $(0:0:0)$ which is not a point in $mathbf{P^2}$.
So to show that there is a solution with $z neq 0$ i'm not sure what to do. I think by doing gaussian elimination on the two equations above, we can find conditions on $x$ and $y$ in terms of $z$ to get that a point of the form $(alpha z, beta z, z)$ is a solution to both equations and so the two lines intersect at $(alpha z: beta z: z) in mathbf{P^2}$ where $z neq 0$.
Is this the correct way to approach this problem?
algebraic-geometry algebraic-curves
algebraic-geometry algebraic-curves
asked Jan 21 at 10:30
trynalearntrynalearn
662314
asked Jan 21 at 10:30
trynalearntrynalearn
662314
asked Jan 21 at 10:30
trynalearntrynalearn
662314
asked Jan 21 at 10:30
trynalearntrynalearn
662314
asked Jan 21 at 10:30
trynalearntrynalearn
662314
662314
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: When working in the subspace $zneq0$ of $Bbb P^2$, you can assume $z=1$, and thus project that subspace isomorphically onto the $xy$-plane.
answered Jan 21 at 10:53
ArthurArthur
117k7116200
$endgroup$
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Hint:
put it into matrix terms $$bf {Ax} = z, bf{c}$$
You are supposing that the determinant of $bf A$ is not null, so it is invertible:
so for whichever $z$ and $bf c$ you get a unique solution for $bf x$.
And if $z ne 0$ ...
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
$endgroup$
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
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
$begingroup$
Hint: When working in the subspace $zneq0$ of $Bbb P^2$, you can assume $z=1$, and thus project that subspace isomorphically onto the $xy$-plane.
answered Jan 21 at 10:53
ArthurArthur
117k7116200
$endgroup$
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Hint: When working in the subspace $zneq0$ of $Bbb P^2$, you can assume $z=1$, and thus project that subspace isomorphically onto the $xy$-plane.
answered Jan 21 at 10:53
ArthurArthur
117k7116200
$endgroup$
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Hint: When working in the subspace $zneq0$ of $Bbb P^2$, you can assume $z=1$, and thus project that subspace isomorphically onto the $xy$-plane.
answered Jan 21 at 10:53
ArthurArthur
117k7116200
$endgroup$
Hint: When working in the subspace $zneq0$ of $Bbb P^2$, you can assume $z=1$, and thus project that subspace isomorphically onto the $xy$-plane.
answered Jan 21 at 10:53
ArthurArthur
117k7116200
answered Jan 21 at 10:53
ArthurArthur
117k7116200
answered Jan 21 at 10:53
ArthurArthur
117k7116200
answered Jan 21 at 10:53
ArthurArthur
117k7116200
117k7116200
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
$begingroup$
Ah right. So in doing so, we choose a particular value of z, and so the family of solutions becomes a single solution. Do we obtain this solution in the way I described above by gaussian elimination? (eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Hint:
put it into matrix terms $$bf {Ax} = z, bf{c}$$
You are supposing that the determinant of $bf A$ is not null, so it is invertible:
so for whichever $z$ and $bf c$ you get a unique solution for $bf x$.
And if $z ne 0$ ...
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
$endgroup$
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Hint:
put it into matrix terms $$bf {Ax} = z, bf{c}$$
You are supposing that the determinant of $bf A$ is not null, so it is invertible:
so for whichever $z$ and $bf c$ you get a unique solution for $bf x$.
And if $z ne 0$ ...
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
$endgroup$
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
Hint:
put it into matrix terms $$bf {Ax} = z, bf{c}$$
You are supposing that the determinant of $bf A$ is not null, so it is invertible:
so for whichever $z$ and $bf c$ you get a unique solution for $bf x$.
And if $z ne 0$ ...
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
$endgroup$
Hint:
put it into matrix terms $$bf {Ax} = z, bf{c}$$
You are supposing that the determinant of $bf A$ is not null, so it is invertible:
so for whichever $z$ and $bf c$ you get a unique solution for $bf x$.
And if $z ne 0$ ...
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
edited Jan 21 at 11:13
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
answered Jan 21 at 10:57
G CabG Cab
19.8k31340
19.8k31340
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
$begingroup$
My linear algebra is weak (I am learning it at the same time). Do we obtain this solution in the way I described above by gaussian elimination? (set z = 1, eliminate x term from 2nd equation, solve for y, and then plug y and z=1 into first equation to find x?).
$endgroup$
– trynalearn
Jan 21 at 10:59
add a comment |
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