Why are all linear maps that sends one basis to another basis or to 0 still linear?

Question

I notice that in a lot of linear transformation proofs involves making assumptions about mapping a basis vector to another basis vector or to 0. I was wondering why this is even allowed and is this true for all linear maps?
So for example , take an arbitrary map L(U,W). I often see proofs that says, "assume (u1, u2, u3...um) are basis vectors for U and take a linear map where T(u1) is mapped to w1 where w1 is a basis vector for W and T(v2) is mapped to 0 in W."
I was wondering why you are can freely allowed to do this.

Morgan Rodgers · Accepted Answer

We can in general have that a linear map in $mathcal{L}(U,W)$ sends some vectors to $0$. But as long as we assume that $T$ is not the trivial map (ie we are assuming that $T(u) neq 0$ for at least one vector in $U$), then we can call one such vector $u_{1}$ and extend it to a basis $u_{1}, ldots, u_{n}$ for $U$ such that $T(u_{1}) neq 0$; then since $T(u_{1}) in W setminus {0}$, we can define $w_{1} = T(u_{1})$ and extend ${w_{1}}$ to a basis ${w_{1}, ldots, w_{m}}$ for $W$.
This is why it is common to 1: Assume that $T$ is not the zero map, 2: take bases for $U$ and $W$ for which $T(u_{1}) = w_{1}$. If you start assuming that, for instance $T(u_{1}) = w_{1}$, $T(u_{2}) = w_{2}$, etc are linearly independent vectors in $W$, then you are making more complicated assumptions about the rank of $T$; but assuming $T$ is nontrivial is usually reasonable (because the trivial map can be rather easily considered separately).

Andrea Mori · Answer

Let $V$ and $W$ be two vector spaces.
For the sake of simplicity assume that $dim(V)=n$ and let ${v_1,...,v_n}$ be a basis.
Also, let $w_1,...,w_n$ just any choice of $n$ vectors in $W$, possibly with repetitions.
Then there is a unique linear map $T:Vrightarrow W$ such that $T(v_i)=w_i$ for all $i=1,...,n$. Since any vector $vin V$ can be expressed in a unique way as a linear combination $v=a_1v_1+cdots+a_nv_n$ the map $T$ can be defined as
$$
T(v)=a_1w_1+cdots+a_nw_n.
$$
It is a standard exercise to check that $T$ is well defined, linear and unique in the sense that if $T^prime$ is another linear map such that $T^prime(v_i)=w_i$ for all $i$ then $T=T^prime$.

Wuestenfux · Answer

A linear mapping $f:Urightarrow W$ is usually defined by assigning images to a basis $u_1,ldots,u_n$ of $U$.
Then the mapping is well-defined, i.e., each vector $u=sum_i a_iu_i$, with $a_i$ scalars, is then assigned an image vector $f(u) = sum_i a_i f(u_i)$.
The image of $f$ is generally a subspace of $W$ generated by the vectors $f(u_1),ldots,f(u_n)$.

Why are all linear maps that sends one basis to another basis or to 0 still linear?

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