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Proving that the solution of Ax=0 is a vector space 
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#1
Jul1608, 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.) 


#2
Jul1608, 11:27 PM

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P: 25,251

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?



#3
Jul1708, 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.



#4
Jul1708, 04:08 AM

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PF Gold
P: 39,285

Proving that the solution of Ax=0 is a vector space
(Think about A(v+ 0)= Av by definition of "0" and use the distributive law.) 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 


#5
Jul1708, 04:21 AM

P: 42

thanks mate



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