Advice for proving existence claims

There is no general recipe or easy shortcut. If you want to get better at this kind of proofs, you should try to read solutions to various theorems first and learn them without even trying to do them on your own, but fully learn them. You need to see some of these ideas before to be able to use them in future problems. This is how I would prove this result: 



We consider two cases: 

(i) The image of $T$ is in the span of $w_2,\ldots, w_n$, i.e. $w_1 \not \in T(V)$. Then for any arbitrary basis $v_1, v_2, \dots, v_m$ we have
\[T(v_k)=0w_1+A_{2,k}w_{2}+\dots A_{k,n}w_n,    k=1,2, \dots m.\]
Hence $A_{1,k}=0$ for $k=1,2, \dots m$,  which means that all entries in the first row of $A$ are zero. 

(ii) $w_1 \in T(V)$. Then $w_1= T(v_1)$ for some $v_1 \in V$. Extend $v_1$ to a basis $v_1, v_2, \dots, v_m$ of $V$. Then 
\[T(v_1)=b_1^{1} w_1+b_1^{2}w_{2}+\dots + b_1^{n} w_n=1w_1+0w_{2}+\dots 0 w_n =w_1\]
and
\[T(v_k)=b_k^{1} w_1+b_k^{2}w_{2}+\dots + b_k^{n} w_n.\]
Now define $\hat{v}_k=v_k-b_k^{1}v_1.$ Then $\hat{v}_k$ are linearly independent and 
\[T(\hat{v}_k)=0w_1+A_{2,k}w_{2}+\dots A_{k,n}w_n.\]
Hence $A_{1,k}=0$ for $k=1,2, \dots m$,  which again means that all entries, except the first one, in the first row of $A$ are zero. 


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This was a particularly challenging problem and it took me about 45 minutes to answer. You should offer much higher bounties in the future, otherwise your questions may not get answered.