![]() Yes, this vector set is closed under addition because when any two vectors in the set are added to each other, they produce another vector that will be located inside the vector space too. Yes, the origin is inside the shaded area on the graph, therefore the vector space contains the zero vector. So ( x, y, z ) : | x + y + z | = 1 ( x, y, z ) : | x + y + z | = 1 is not a subspace.The first thing we have to do in order to comprehend the concepts of subspaces in linear algebra is to completely understand the concept of R n R^ R 2 are met: Well, this is actually a non-linear function, and we can show that it is not a subspace pretty easily through a counterexample, by showing that the set is not closed under addition. What about things that are not subspaces? For instance: ( x, y, z ) : | x + y + z | = 1 ( x, y, z ) : | x + y + z | = 1 Therefore, as the subset defined by ( x, y, z ) : 2 x + 3 y = z ( x, y, z ) : 2 x + 3 y = zis a subspace of the volume defined by the three-dimensional real numbers. So if we multiply through α α: 2 α x + 3 α y = α z 2 α x + 3 α y = α zĪnd this is always true, since it is just a scalar. Again, we need to prove this by analysis:
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