In the book Linear Algebra Done Right, it is said that to determine quickly whether a given subset of $V$ is a subspace of $V$, the three conditions, namely additive identity, closed under addition, and closed under scalar multiplication, should be satisfied. The other parts of the definition of a vector space are automatically satisfied.

I think I understand why commutativity, associativity, distributive properties, and multiplicative identity works because their operations are still within the subspace.

But, why don't we need to verify additive inverse, similar to verifying additive identity? Could there be cases where there will be no $v + w = 0$ in the new subspace, $v, w \in U$, $U$ is a subspace?


1 Answer 1


You also need to check that the set is non-empty. In this case closure under scalar multiplication guarantees that the additive inverse of any $v$ in the set is also in the set, since for the scalar $-1$, $(-1)v$ is in the set.

EDIT: Similarly, for the scalar $0$, $0v={\bf 0}$ is in the set (by the closure of scalar multiplication), whenever the set contains an element/vector $v$.

  • 4
    $\begingroup$ To be fair, you will get non-emptiness by verifying that $\vec{0}$ is in the set. I think that is what is meant by additive identity. $\endgroup$ Commented Dec 9, 2018 at 23:15
  • $\begingroup$ I think I understand your logic. But, then I have another question on the definition of Vector Space (not subspace). If there is closure under scalar multiplication, do we need to prove that additive inverse exists? since for scalar -1, it guarantees that there is an additive inverse. $\endgroup$
    – JOHN
    Commented Dec 10, 2018 at 22:39

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