Graph Theory+ Number Theory. Problem: Let $G$ be a connected graph. Let $T$ be a spanning tree of $G.$ Prove that $T$ contains a spanning subgraph $T'$ such that, for each vertex $ v$, the degree of $v$ in $G$ and the degree of $v$ in $T'$ are equal modulo $2.$
I tried a couple of examples but was unable to get some intuition for this problem, any hints/suggestions will be much appreciated. 
 A: Here is an algebraic proof.Have you ever learned Data Structure(I major in computer science...)?A kind of matrix named incidence matrix is defined as below:It is a $|V|*|E|$ matrix on $GF(2)$.If edge $e_i=(v_i,v_j)$ then $(v_i,e_i)=1 , (v_i,e_j)=1 $,else $(v,e)=0$.
It is deep night in China and I'm really sleepy,but I really want to offer help.so I draw a picture for you without using Latex to illustrate the matrix and the graph,easy to see,they are corresponding with each other.
$\textbf{Lemma 1}$：The columns of matrix of spanning tree $T$(which is surrounded by the dotted line) is the  maximal linear independent set of column space of incidence matrix of G(denoted by $M$).
$\textbf{Proof}$.They are linear independent because there are no circles in a tree(If columns are linear dependent,those edges will form a circle).And it's maximal because add an edge to the spanning tree will add a circle.
$\textbf{Lemma2}$ There are $|V|-1$ columns in spanning tree matrix.
$\textbf{Proof}$ It's easy.For a tree,$|E|=|V|-1$.
$\textbf{Lemma3}$ If $M$ is a (incidence) matrix of a spanning tree,then it does not contain a zero row.
Why?Because in a spanning tree,every vertex is connected to at least one edge,and only one.
$M$ is denoted as the incidence matrix of $G$.
Then our aim is to:$\textbf{Select columns from spanning tree matrix and form them into a new matrix N,such that}$ $\textbf{for every row of $M,N$,the sum of elements of $M$ equals the sum of elements of $N$.}$The sum here means exclusive or $\oplus$.Because the sum of rows indicate the degree of vertexs.
For example,In the picture provided,$deg(v_1)=3$,and the sum of first row is $3$.If in $GF(2)$,$3$ mod $2$ = $1$,and $1 \oplus 1 \oplus 1 = 1$.
Then how to do?If the first row does not satisfy our aim,choose a column $j$ that $M_{1,j}$=1(As was proved in Lemma 3,no zero row!) and delete that column(So first row must satisfy).And do the same again on the second row...
Sometimes $T^{'}$ has to be an empty graph,for example,$G$ is a triangle.Three vertexs,three edges.
