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Proving that the solution of Ax=0 is a vector space

by mohdhm
Tags: proving, solution, space, vector
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mohdhm
#1
Jul16-08, 11:14 PM
P: 42
Here is my attempt to answer this guys, i'd really appreciate any corrections.


a vector space has the 0 vector
the vector space is closed under vector addition and scalar multiplication (AKA for every vector u, v in the subspace, there exists a vector u + v in the subspace)

Here we go.

The trivial solution is an indication that the 0 vector exists. I don't know how to prove that one, unless i write out the linear combination and set all the scalars to 0

The trivial solution is not the only solution: free variables and possibly 0 columns or rows exist. (also an indication that the 0 vector exists).

so if we row reduce the matrix to row echalon form and solve for [x1 ... xn] by moving the free variables to the right side of the homogenius equation then that would be the same as the additive property i mentioned above right? but how would i even know that u or v are in the vector space to begin with ? (remindeR: u and v are vector columns of the matrix A.)
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Dick
#2
Jul16-08, 11:27 PM
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You don't have to solve anything. If u and v are in the null space N={x:Ax=0} that just means Au=0 and Av=0. To show it's closed under vector addition, for example, you just need to show (u+v) is in N. That is, show A(u+v)=0. Can you do that?
mohdhm
#3
Jul17-08, 01:39 AM
P: 42
how can you show that when you are not given u or v, unless its super simple and just happens to be worth a lot of marks.

HallsofIvy
#4
Jul17-08, 04:08 AM
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Proving that the solution of Ax=0 is a vector space

Quote Quote by mohdhm View Post
Here is my attempt to answer this guys, i'd really appreciate any corrections.


a vector space has the 0 vector
the vector space is closed under vector addition and scalar multiplication (AKA for every vector u, v in the subspace, there exists a vector u + v in the subspace)

Here we go.

The trivial solution is an indication that the 0 vector exists. I don't know how to prove that one, unless i write out the linear combination and set all the scalars to 0
No, you can't set the scalars to 0, they are given in "A". What you want to do is the other way around: show that the 0 vector satisfies the equation A0= 0. Since A is a linear transformation, what do you know about linear transformations? In particular, what is A0?
(Think about A(v+ 0)= Av by definition of "0" and use the distributive law.)

The trivial solution is not the only solution: free variables and possibly 0 columns or rows exist. (also an indication that the 0 vector exists).

so if we row reduce the matrix to row echalon form and solve for [x1 ... xn] by moving the free variables to the right side of the homogenius equation then that would be the same as the additive property i mentioned above right? but how would i even know that u or v are in the vector space to begin with ? (remindeR: u and v are vector columns of the matrix A.)
You want to show that if u and v are in the space, (that is, they satisfy Au= 0, Av= 0) then au+ bv is also in the space, for any numbers a and b (What is A(au+ bv)?).

DON'T think about "row echelon form" or anything like that. Just use the properties of linear transformations (matrix multiplication) themselves: A(u+ v)= Au+ Av and A(bu)= b(Au).

Finally, u and v are not "vector columns of the matrix A". They are any vectors of the correct dimension to be multiplied by A
mohdhm
#5
Jul17-08, 04:21 AM
P: 42
thanks mate


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