Which are linear transformations?

Question

1) Define $T:V\rightarrow V$ for $T(v)=v$, $\forall$ $v \in V$.

For this one I have:

Let $T(v)=v$ be the identity transformation. Because $T(u+v)=T(u)+T(v)$ and $T(cu)=cu=cT(u)$. $T$ is a linear transformation.

But I don't know if is that quite simple. The other alternative I have is:

If $v={v_1,v_2,...,v_n}$ and is a basis for $V$, then any $v\in V$ can be written as the linear combination $v=a_1v_1+a_2v_2+...+a_nv_n$ with $a_i$ scalar. Then

$$T(v)=T(a_1v_1+a_2v_2+...+a_nv_n)$$ $$T(v)=a_1T(v_1)+a_2T(v_2)+...+a_nT(v_n)$$ $$T(v)=a_1v_1+a_2v_2+...+a_nv_n$$ $$T(v)=v$$

Therefore, $T$ is a linear transformation.

Can be any of them?

2) Let $V$, $W$ vector spaces. Define $T:V\rightarrow W$ for $T(v)=0$, $\forall$ $v \in V$.

For this one I have:

Let $v=c_1v_1+c_2v_2+...+c_nv_n$ be an arbitrary vector in $v$. Then, $$T(v)=T(c_1v_1+c_2v_2+...+c_nv_n)$$ $$T(v)=c_1T(v_1)+c_2T(v_2)+...+c_nT(v_n)$$ $$T(v)=0+0+...+0$$ $$T(v)=0$$

Therefore, $T$ is a linear transformation.

Is this right?

3) Define $T: C[0,1]\rightarrow R$ for $T(f)=f(0)+1$.

For this last one I have not clue where to start. Help please!

Answer 1

1. Your first approach if fine. The second one isn't, since it assumes that the mape is linear.
2. Again, you are assuming that $T$ is linear! All you have to do is$$T(u+v)=0=0+0=T(u)+T(v)\text{ and }T(cu)=0=c\times0=cT(u).$$
3. It is not linear, since $T(0)\neq0$.

Answer 2

You have rightly concluded the first two ones. For the third one is not a linear mapping. For any linear mapping $T$ we must have $T(0)=0$. Let $f(x)=0$. Therefore $f$ lies inside the domain of $T$ since it is always continuous but $$T(0)=T(f(x))=f(0)+1=1\ne 0$$furthermore if $$T(f)=f(0)+1$$then $$T(kf)=kf(0)+1\ne kf(0)+k=kT(f)\qquad\qquad k\ne 1$$