You can represent the cubic as a matrix multiplication as (x² 1) M (x, 1), where M is a 2x2 matrix and (x, 1) is a column vector. When the determinant of M is 0, it means the rows are linearly dependent, lets say the second r=c/a times the first row. You can then write M as a product of two vectors: (1, r)(a b), where (1, r) is again a column vector. (In general, this can only be done for rank 1 or 0 matrices, but since we only have a 2x2 matrix, M is rank
Consider writing the matrix A as: [ a c ] [ b d ] (note that this is the transpose of the normal orientation). And consider the two vectors v = (x, 1) and u = (x², 1) Then, the expression Au · v = 0, expands to the standard cubic expression, so solving this matrix expression is equivalent. Thinking about what this means... for a dot product to be zero, either one of the two vectors being dotted is zero, or the two are perpendicular. Obviously v is nonzero, so our two options are that Au is zero, or Au and v are perpendicular. In the case where v is perpendicular to Au, then it cannot be the case that swapping x to -x will give another solution (assuming x is nonzero)... as this does not change Au, but v does change to a different vector that isn't parallel to the original v, so they can't both be perpendicular to Au. On the other hand, if Au is zero, then swapping x for -x _does_ give another solution, as u doesn't change, so Au will still be zero. But it is only possible for a matrix times a nonzero vector to be zero, if that matrix has zero determinant. Proving the reverse direction is a bit more complicated, but handwavily, if A is singular, then there exists a vector (x,y) such that A(x,y) = 0, and then if y is nonzero we can also say A(x/y,1) = 0, and so ±sqrt(x/y) are solutions to the cubic. Figuring out what the "y is not 0" condition implies (and "x/y > 0" if you want real results) is a bit fiddlier, but doable.
Very nice! I wouldn't have thought to introduce the vectors (x, 1), (x^2, 1), and use the dot product like that, but this reveals some interesting connections!
❤ necessary part - sum of roots = -b/a beta = - b/a putting in equation - b^3/a^2+b^3/a^2- bc/a+ d = 0 or bc = ad putting d = bc/a in equation ax^3+bx^2+cx +bc/a = 0 ax^2(x+b/a)+c(x+b/a) = 0 (x+b/a) (x^2+c/a) = 0 for real roots c/a < = 0 hence necessary conditions are bc = ad & c/a < = 0 sufficient part - we will prove that above mentioned conditions are sufficient also bc = ad hence d = bc/a putting in equation (x+b/a) (x^2+c/a) = 0 also c/a < = 0 hence roots are √( - c/a) , - √( - c/a) , - b/a it proves the sufficient part.
That was pretty cool as some math game, but grouping, if possible, would be quite obvious in the first glance and remembering this trick is way far from a shortcut.
When I'm tutoring students in algebra, I tell them that the vast majority of cubic equations are unsolvable with non-analytic methods, so if you're being asked to solve one, there's a good chance it's "groupable", and so check whether ad = bc and try factoring by grouping.
You can represent the cubic as a matrix multiplication as (x² 1) M (x, 1), where M is a 2x2 matrix and (x, 1) is a column vector.
When the determinant of M is 0, it means the rows are linearly dependent, lets say the second r=c/a times the first row.
You can then write M as a product of two vectors: (1, r)(a b), where (1, r) is again a column vector. (In general, this can only be done for rank 1 or 0 matrices, but since we only have a 2x2 matrix, M is rank
Beautiful! :)
Amazing, this makes so much sense!
This representation is interesting. you can define a ring structure on space of cubic polynomials with it. Idk if thats anything useful though.
We usually write such cubics with a > 0, so the c/a < 0 condition reduces to c < 0.
Consider writing the matrix A as:
[ a c ]
[ b d ]
(note that this is the transpose of the normal orientation). And consider the two vectors v = (x, 1) and u = (x², 1)
Then, the expression Au · v = 0, expands to the standard cubic expression, so solving this matrix expression is equivalent.
Thinking about what this means... for a dot product to be zero, either one of the two vectors being dotted is zero, or the two are perpendicular. Obviously v is nonzero, so our two options are that Au is zero, or Au and v are perpendicular.
In the case where v is perpendicular to Au, then it cannot be the case that swapping x to -x will give another solution (assuming x is nonzero)... as this does not change Au, but v does change to a different vector that isn't parallel to the original v, so they can't both be perpendicular to Au.
On the other hand, if Au is zero, then swapping x for -x _does_ give another solution, as u doesn't change, so Au will still be zero. But it is only possible for a matrix times a nonzero vector to be zero, if that matrix has zero determinant.
Proving the reverse direction is a bit more complicated, but handwavily, if A is singular, then there exists a vector (x,y) such that A(x,y) = 0, and then if y is nonzero we can also say A(x/y,1) = 0, and so ±sqrt(x/y) are solutions to the cubic. Figuring out what the "y is not 0" condition implies (and "x/y > 0" if you want real results) is a bit fiddlier, but doable.
Apparently you missed the comment by barutjeh?
Very nice! I wouldn't have thought to introduce the vectors (x, 1), (x^2, 1), and use the dot product like that, but this reveals some interesting connections!
❤
necessary part -
sum of roots = -b/a
beta = - b/a
putting in equation
- b^3/a^2+b^3/a^2- bc/a+ d = 0
or bc = ad
putting d = bc/a in equation
ax^3+bx^2+cx +bc/a = 0
ax^2(x+b/a)+c(x+b/a) = 0
(x+b/a) (x^2+c/a) = 0
for real roots c/a < = 0
hence necessary conditions are bc = ad & c/a < = 0
sufficient part -
we will prove that above mentioned conditions are sufficient also
bc = ad hence d = bc/a
putting in equation
(x+b/a) (x^2+c/a) = 0
also c/a < = 0
hence roots are
√( - c/a) , - √( - c/a) , - b/a
it proves the sufficient part.
Great video!
I just want to point out that it seems these roots may not be unique with this argument. Not an issue, just wanted to note that.
Sorry, I don't understand what you mean exactly... could you please explain this in more detail?
@ just that it’s possible for alpha and beta to be the same, or minus alpha and beta to be the same, or even for alpha=- alpha if alpha=0
Is it just me or this equivalent to classifying all cubics that can be directly factored by grouping?
That was pretty cool as some math game, but grouping, if possible, would be quite obvious in the first glance and remembering this trick is way far from a shortcut.
When I'm tutoring students in algebra, I tell them that the vast majority of cubic equations are unsolvable with non-analytic methods, so if you're being asked to solve one, there's a good chance it's "groupable", and so check whether ad = bc and try factoring by grouping.