# The multiplicative identity of the vector space formed by the set of all linear transformations

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.

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