Proof on homogeneous equations

[toxi]

I need some help here... I've got the following assignment to do

Prove that if M>N then any system of N homogeneous equations in M unknowns has many solutions.

I am a bit stuck with this one. I thought about creating a MxN Matrix and to display the determinant with 1's.

and then say about the remaining colums after the rows with leading 1's stop (r = M-N), that they can represented by any value so there are many solutions
is that correct?

 

[ice109]

i don't know what you're saying but yes if you write down a matrix with M columns and N rows you'll see that you have a linearly dependent set of vector spanning the column space hence many solutions.

 

[tacman]

Yes, you are on the right track. Let's break down the proof step by step:

1. Start by assuming that M>N, which means there are more unknowns than equations. This is the scenario we are trying to prove.

2. Create a MxN matrix with 1's in the first N rows and 0's in the remaining rows. This is called an augmented matrix and represents the system of N homogeneous equations in M unknowns.

3. Now, let's look at the determinant of this matrix. By the properties of determinants, we know that if we multiply a row by a scalar, the determinant is also multiplied by that scalar. Since we have N rows with 1's, we can multiply any of the remaining rows by a scalar and the determinant will still be 0.

4. This means that the determinant of this matrix is 0, which implies that the system of equations is consistent (has many solutions). This is because a 0 determinant means that there is at least one row of zeros, which corresponds to a free variable in the solution.

5. Since we have M-N free variables (represented by the remaining columns after the leading 1's), we have many solutions to the system of equations.

In conclusion, if M>N, any system of N homogeneous equations in M unknowns has many solutions due to the fact that there will always be at least M-N free variables in the solution.