Sufficiency of n eqs. for n unkwns.

  • Thread starter turin
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In summary, the conversation revolved around the confusion when dealing with complex variables, specifically in regards to the number of equations needed to determine the unknowns. Some argued that two equations are needed per complex variable, while others pointed out that each complex equation is actually two equations, one for the imaginary part and one for the real part. Ultimately, the fact that n independent equations are needed for n unknowns was agreed upon as the definition of independence. An example was given to show that while two equations may determine the magnitudes, the phases can still be arbitrary.
  • #1
turin
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I remember from way back in jr high learning that a set of n independent equations is sufficient to determine n unkowns. Lately, I've been a little confused when it comes to complex variables.

My confusion is particularly manifest when dealing with magnitudes and complex conjugation.

I just wanted to get a second oppinion. For example, given two complex valued unkowns w and z, it seems like one should actually consider these as four unkowns, two for the real parts and two more for the imaginary parts. But then again, they can be determined by only two equations. But then again, each equation seems to have two compoents. I don't know how I should think about this.
 
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  • #2
Consider that each complex equation is really two equations, one for the imagiary part, and one for the real part, so if you separate into components, you still get a matching number of equations and unknowns.
 
  • #3
Score 1 for 2 eqs. required per complex variable.
 
  • #4
turin said:
Score 1 for 2 eqs. required per complex variable.

Two real equations (i.e. with real coefficients) per complex variable.

As NateTG said, that is still one complex equation per variable.

The fact that n "independent" equations in n unknowns has a unique solution is pretty much the definition of "independent".
 
  • #5
HallsofIvy said:
The fact that n "independent" equations in n unknowns has a unique solution is pretty much the definition of "independent".
Here's an example.

|z| + |w| = 1
z + w* = 0

This system of equations pins down the magnitudes but leaves the phases arbitrary. Does this show that these equations are not independent?
 

What does "Sufficiency of n eqs. for n unkwns." mean?

This phrase refers to a concept in mathematics and engineering where the number of equations (eqs) is equal to the number of unknown variables (unkwns). It is often used in solving systems of equations to determine if there are enough equations to find a unique solution for each variable.

Why is it important to have a sufficient number of equations for the number of unknowns?

Having a sufficient number of equations ensures that there is only one solution for each unknown variable. Without enough equations, there may be multiple solutions or no solutions at all. This can lead to inaccurate or unreliable results in mathematical and engineering problems.

What happens if there are more equations than unknowns?

In this case, the system of equations is said to be overdetermined. This means that there are more constraints than necessary to find a unique solution for each variable. As a result, some of the equations may be redundant or contradictory, and the system may become unsolvable.

Can a system of equations with n equations and n unknowns always be solved?

No, there are cases where a system of equations may not have a unique solution even if there are an equal number of equations and unknowns. This can occur if the equations are not independent or if there is a lack of information in the equations.

How can one determine if a system of equations has a sufficient number of equations for the number of unknowns?

The sufficiency of equations for unknowns can be determined by using methods such as Gaussian elimination or matrix operations. These methods can help identify if there are enough equations to find a unique solution for each unknown variable.

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