Here is my attempt to answer this guys, i'd really appreciate any corrections.(adsbygoogle = window.adsbygoogle || []).push({});

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

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