In other words, rule #1 must hold if rule #2 is to hold.Īnd honestly, rule #1 also must hold if rule #3 is to hold. So for rule #2 to hold, the subspace must include the 0 vector. Well suppose we multiply by the scalar 0? We would get the 0 vector. If rule #2 holds, then the 0 vector must be in your subspace, because if the subspace is closed under scalar multiplication that means that vector A multiplied by ANY scalar must also be in the subspace. Well, imagine a vector A that is in your subspace, and is NOT equal to zero. So ( x, y, z ) : | x + y + z | = 1 ( x, y, z ) : | x + y + z | = 1 is not a subspace.Remember the three rules that Sal gave for the definition of a subspace? They were: 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: Now for being closed under multiplication.Which is always true, so we are closed under addition Now, if we recall that 2 x 1 + 3 y 1 − z 1 = 0 2 x 1 + 3 y 1 − z 1 = 0 and 2 x 2 + 3 y 2 − z 2 = 0 2 x 2 + 3 y 2 − z 2 = 0, so what we really have is: 0 = − 0 0 = − 0 Is it closed under addition? To work this out, we can take any two v 1 ∈ S v 1 ∈ S and v 2 ∈ S v 2 ∈ S and show that their sum, v 1 + v 2 ∈ S v 1 + v 2 ∈ S:.Is the zero vector a member of the space? Well: 0 = 2 × 0 + 3 × 0 0 = 2 × 0 + 3 × 0, so yes, it is.So for instance, valid subspace for a three dimensional space might be z = 2 x + 3 y z = 2 x + 3 y, which is a plane:Īnd does ( x, y, z ) : z = 2 x + 3 y ( x, y, z ) : z = 2 x + 3 y satisfy our three propositions? Well, we can prove this with a bit of analysis: It must be closed under scalar multiplication: if v ∈ S v ∈ S then α v ∈ S α v ∈ S, else S S is not a subspace.Īll of that was just a fancy way of saying that a subspace just needs to define some equal or lesser-dimensional space that ranges from positive infinity to negative infinity and passes through the origin.It must be closed under addition: if v 1 ∈ S v 1 ∈ S and v 2 ∈ S v 2 ∈ S for any v 1, v 2 v 1, v 2, then it must be true that ( v 1 + v 2 ) ∈ S ( v 1 + v 2 ) ∈ S or else S S is not a subspace.The formal definition of a subspace is as follows: We also often use letters from the greek alphabet to describe arbitrary constants, for instance alpha α α and beta β β. When we use ( x, y, z ) : z = C ( x, y ) ( x, y, z ) : z = C ( x, y ) we are describing a set containing all three dimensional vectors that satisfy the condition that z z equals some function C ( x, y ) C ( x, y ). To answer that question, it is worth defining what a subspace is in terms of its formal properties, then what it is in laymans terms, then the visual definition, showing why it is that those properties need to be satisfied.įor those unfamiliar, we will be using a little bit of set notation here - ∈ ∈ stands for is a member of. ![]() Given some subset of numbers in an n-dimensional space, one of the questions you might get asked is whether or not those numbers make up a subspace.
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