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hivesaeed4

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Given a homogeneous system of n linear equations in m unknowns if m>n (i.e. there are more unknowns than equations) there will be infinitely many solutions to the system.

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- Thread starter hivesaeed4
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- #1

hivesaeed4

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Given a homogeneous system of n linear equations in m unknowns if m>n (i.e. there are more unknowns than equations) there will be infinitely many solutions to the system.

- #2

chiro

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Think about when you have a unique solution (and the system is not inconsistent): for a unique solution you have as many elements in the vector as you have variables.

If m > n then as long as the system is not inconsistent, then many of the rows are going to dissappear and become 0 vectors when you do row reduction (like Gaussian elimination) since some of those vectors will be linearly dependent on the other vectors.

But if m < n it means that as long as you have one non-zero vector it means that at most you will never be able to get a unique solution (since you need minimum m = n for this to happen).

So you will need to introduce a parameter and then relate those free parameters to the final solution.

Geometrically if you want to think of a solution of n n-dimensional plane equations it means that for a unique solution, all points have to intersect at the same point.

Think about the 3D case for the moment: if you have two planes intersect then it means that you either have the two planes being equal (which means the intersection gives the whole plane) or you get an intersection which gives a line as your solution.

If you add one more plane that is not linearly dependent on the other two then your line will reduce to a point and a point is a unique solution.

So until we get m = n independent vectors, we have situations where we can have many solutions instead of 1 and this is the visual explanation of why the statement you are wondering about actually holds.

- #3

HallsofIvy

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1) the equation has a unique solution.

2) the equation has NO solution.

3) the equation has an infinite number of solutions.

If A is invertible, we can multiply both sides of Ax= b by its inverse to get [itex]A^{-1}Ax= x= A^{-1}b[/itex]. That clearly is a solution to the equation and, because x

If A is not invertible, then it has rank, k, less than n. That means that A maps all of R

In that case, if b does not happen to lie in that subspace, the equation, Ax= b, has no solution. For any x, Ax is in that subspace and b is not. If b does happen to lie in that subspace, there is a solution, a vector x such that Ax= b, but it is also true that for any v in the null space of A, A(x+ v)= Ax+ Av= b+ 0= b so there are an infinite number of solutions.

Since that is true for square matrices, what does it have to do with this problem? Well, if we have m equations, in n unknowns, with m< n, we can represent it has the matrix equation Ax= b where A is Not square. It has m rows and n columns. But we can make it square by adding n- m rows of

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Condition for a unique solution of matrix A that have infinite solutions and optimisation of tr(A)

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