I was asked to prove that the set of all linear transformations from the vector space $V$ to the vector space $W$ is a vector space. I was going through the axioms, and I am stuck on the multiplicative identity part.

The function $T(x)=1$ is not a linear transformation since $T(\alpha x)=1$ whereas $\alpha T(x)=\alpha$.

Which linear transformation is the multiplicative identity then?

I asked this to one of my classmates and he said it is the function $T(x)=x$. If we are taking multiplication to mean composition, I understand why this is true. (In that case, I don't understand the multiplicative associativity. Also, I feel like this should have been stated.)

However, if we are talking about multiplication in the normal sense, I don't understand how this is true. For example, if we let $V=W=\mathbb{R}$, all the linear transformations are of the form $T(x)=ax$ for some $a$. If we multiply this by $x$, we get $ax^2$ which is not even a linear transformation, let alone equal to $T$.

I feel like I am asking a very easy question. If that is the case, I apologize. However, I don't understand what I am missing.

  • 3
    $\begingroup$ Vector spaces have additive identities, not multiplicative identities. $f(x)=0$ is the additive identity. $\endgroup$
    – lulu
    Jul 4, 2020 at 20:58
  • 3
    $\begingroup$ If you meant the multiplicative identity for scalar multiplication, then of course that is the multiplicative identity for the underlying field. $\endgroup$
    – lulu
    Jul 4, 2020 at 21:01

1 Answer 1


Product of linear maps is defined as composition:

(ST)(u) = S(T(u))

hence multiplicative identity T(x)=x makes sense.

Multiplicative associativity (ST=TS), where S,T are linear maps, is not a requirement for a vector space. In order for linear maps to form vector space they should only adhere to scalar multiplicative associativity, i.e. (ab)S = a(bS), where a,b are scalars (R, C, etc).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .