# What is the dimension of kernel of the linear transformation if the field is the set of rational numbers?

Let $$X$$ be the set of tetrahedron’s edges and $$Y$$ the set of its faces. To any function $$f$$ on $$X$$ with values in a field $$K$$, we assign a function $$g$$ on $$Y$$ defined as

$$g(y)$$ = $$\sum_{x⊂y} f(x)$$

That is, the value of $$g$$ on a face equals the sum of the values of $$f$$ on the sides of this face.

The above assignment defines a linear map $$φ : F(X, K) → F(Y, K)$$, where $$F(X, K)$$ denotes the space of function from $$X$$ to $$K$$ and similarly for $$F(Y, K)$$

Is The kernel of $$φ$$ has dimension two if $$K$$ is the field of rational numbers?

Why the dimension of $$F(X,K)$$ and $$F(Y,K)$$ are $$6$$ and $$4$$ respectively?

You can consider the function $$1_{l_i}$$ that assigns to an edge $$x$$ the value $$0$$ if $$x\neq l_i$$ and $$1$$ otherwise, for each $$i=1, \dots 6$$.

If $$f$$ is a function on $$X=\{l_1, \dots , l_6\}$$, then you get

$$f=\sum_{i=1}^6 f(l_i)\cdot 1_{l_i}$$

and so $$\{1_{l_1}, \dots , 1_{l_6}\}$$ is a base of $$F(X, K)$$

This is trivial if one observe that $$F(X,K)$$ corresponds to the free $$K-algebra$$ generated by the set $$X$$.

Thus we get the dimension of $$F(X,K)$$ is equal to $$6$$, that is exactly the number of tethraedron edges.

An analogous reasoning can be done for $$F(Y,K)$$, that has dimension $$4$$ equal to the number of faces.

At this point you can observe $$\phi$$ is a surjective map. If $$g$$ is a function on $$Y$$, then we want to find $$f$$ such that $$\phi(f)=g$$, i.e. it is satisfied the following linear system:

$$g(F_1)=f(l_1)+f(l_2)+f(l_3)$$

$$g(F_2)=f(l_3)+f(l_4)+f(l_5)$$

$$g(F_3)=f(l_4)+f(l_2)+f(l_6)$$

$$g(F_4)=f(l_1)+f(l_5)+f(l_6)$$

and so

$$k_1=g(F_1)-g(F_2)+g(F_3)-g(F_4)=2[f(l_2)-f(l_5)]$$

that implies, if $$K$$ has not characteristic $$2$$, that

$$f(l_2)=f(l_5)+\frac{1}{2}k_1$$

If we continue the reasoning, we get

$$g(F_1)-g(F_2)=f(l_1)+\frac{1}{2}k_1-f(l_4)$$

and so

$$f(l_1)=f(l_4)+(g(F_1)-g(F_2)-\frac{1}{2}k_1)=f(l_4)+k_2$$

Moreover

$$f(l_3)=g(F_2)-(f(l_4)+f(l_5))$$

$$f(l_6)=g(F_4)-k_2-(f(l_4)+f(l_5))$$

Thus a general solution of the system will be

$$f(l_1):=a+ g(F_1)-g(F_2)-\frac{1}{2}k_1$$

$$f(l_2):= b+\frac{1}{2}k_1$$

$$f(l_3):=g(F_2)-(a+b)$$

$$f(l_4):=a$$

$$f(l_5):=b$$

$$f(l_6):=g(F_4)-k_2-(a+b)$$

for some fixed $$a,b\in K$$.

In other words $$\phi$$ is surjective and so, by nullity rank theorem, we get

$$\dim( \ker(\phi))=6-4=2$$

Note that $$|X|=6$$ and $$|Y|=4$$ and $$F(X,K)=K^6$$, while $$F(Y,K)=K^4$$.

For clarity, let us assume $$X=\{(e_1,e_2),(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4))\}$$ and $$Y=\{(e_1,e_2,e_3),(e_1,e_2,e_4),(e_1,e_3,e_4),(e_2,e_3,e_4)\}$$ and come to the description of $$\varphi$$.

It can be described as

$$\varphi(\alpha(e_1,e_2,e_3)+\beta(e_1,e_2,e_4)+\gamma(e_1,e_3,e_4)+\delta(e_2,e_3,e_4))\\ =\alpha((e_1,e_2)+(e_2,e_3)+(e_1,e_3))+\beta((e_1,e_2)+\cdots)+\cdots\\ =(\alpha+\beta)(e_1,e_2)+(\alpha+\gamma)(e_1,e_3)+(\beta+\gamma)(e_1,e_4)+(\alpha+\delta)(e_2,e_3)+(\beta+\delta)(e_2,e_4)+(\delta+\gamma)(e_3,e_4)$$.

I hope you will be able to conclude from here. Good luck!