Three equations of planes, dimension should be 1?

[JonnyG]

I always thought that one independent equation cuts down the dimension by 1, so if we had two planes, say x - y - z = 1 and x + y + z = 1, then because these are two independent equations, the dimension of the intersection should be 1 because each plane is cutting down the dimension by 1.

Using this same logic, the intersection of three planes should be 0 dimensional, i.e. a point, but it is possible to have three independent planes whose intersection is a line, which is 1-dimensional. What's wrong with my reasoning?

 

[PeroK]

The x-y plane and the x-z plane intersect in the x-axis. There are other planes that contain the x-axis.

 

[JonnyG]

My question is, in the case of three independent planes intersecting at a line, why isn't the dimension of the intersection cut down by 3, because we have 3 independent equations?

 

[PeroK]

My question is, in the case of three independent planes intersecting at a line, why isn't the dimension of the intersection cut down by 3, because we have 3 independent equations?

What do you mean by "independent" planes?

 

[JonnyG]

What do you mean by "independent" planes?

Let's say that any two intersecting planes are independent if they are non-coincident.

For example, two lines in R^2 are linearly independent if they are not parallel. The solution set for their intersection is cut down by 1 dimension for each equation, so the solution set is 0-dimensinonal, i.e. a point.

Similarly, two intersecting planes in R^3 are either the same, or they're independent, in which case the solution set for the intersection is 1-dimensional (a line), because each plane equation cuts down the dimension of the solution set by `1 each. I hope I am being clear.

 

[PeroK]

Let's say that any two intersecting planes are independent if they are non-coincident.

For example, two lines in R^2 are linearly independent if they are not parallel. The solution set for their intersection is cut down by 1 dimension for each equation, so the solution set is 0-dimensinonal, i.e. a point.

Similarly, two intersecting planes in R^3 are either the same, or they're independent, in which case the solution set for the intersection is 1-dimensional (a line), because each plane equation cuts down the dimension of the solution set by `1 each. I hope I am being clear.

Yes, but clearly there are infinitely many planes containing a line. Not just two.

 

[etotheipi]

If your three equations $a_i x + b_i y + c_i z = d_i$, where $i = 1,2,3$, are linearly independent, then you will get a unique solution for the intersection (i.e. a point). You can check if they're linearly independent by taking the determinant of the matrix of the coefficients.

If the equations are linearly dependent, then you can no longer obtain a unique solution. You can't say that an equation that is not linearly independent of one you already have will reduce the dimensionality of the solution set.

 

[JonnyG]

If your three equations $a_i x + b_i y + c_i z = d_i$, where $i = 1,2,3$, are linearly independent, then you will get a unique solution for the intersection (i.e. a point). You can check if they're linearly independent by taking the determinant of the matrix of the coefficients.

If the equations are linearly dependent, then you can no longer obtain a unique solution. You can't say that an equation that is not linearly independent of one you already have will reduce the dimensionality of the solution set.

What about the three equations:

$$2x - y + z = 1 $$
$$3x - 5y + 4z = 3$$
$$3x + 2y - z = 0$$

They intersect in the line $$\{ (-t/7 + 2/7, 5t/7 - 3/7, t) \}$$.

The equations are linearly independent, so each equation should cut down the dimension of the solution set by 1, yet the solution set is 1-dimensional.

 

[etotheipi]

They're not linearly independent, since $$
\begin{vmatrix}
2 & -1 & 1\\
3 & -5 & 4\\
3 & 2 & -1
\end{vmatrix} = 0$$Specifically, multiply the first equation by $3$, and the second by $-1$, then$$2x - y + z = 1 \implies 6x - 3y + 3z = 3$$and$$3x - 5y + 4z =3 \implies -3x +5y -4z = -3$$Add these equations,$$3x + 2y - z = 0$$which is the third equation. Thus, since you only have two linearly independent equations, you only reduce the dimensionality of the solution set by one, i.e. your solution set is a line.

 

[JonnyG]

Oh man, I was comparing two equations at a time to check for linearly independence. My mistake. Thank you for the help.

 

[mfb]

That would be called "pairwise independence". Linear independence of the whole set is a stronger condition.