# A field $\mathbb K$ is commutative iff for every matrix on $\mathbb K$ the rank of its column vector is equal to the one of its transpose

I have a little tricky question if anyone could help me see through it , it would be really appreciated :

We have $$\mathbb K$$ a field such that : $$\alpha a \neq a\alpha$$ I don't see why the rank of the column matrix : $$\begin{pmatrix} a&\alpha a \\ 1&\alpha \end{pmatrix}$$

is $$=2$$ , and that the rank of the column matrix (#)

$$\begin{pmatrix} a& 1 \\ \alpha a&\alpha \end{pmatrix}$$

is $$=1$$

(PS: How can we deduce from this that a field $$\mathbb K$$ is commutative iff for every matrix on $$\mathbb K$$ the rank of its column vector is equal to the one of its transpose)

I know that the column vectors are a vector space with the external multiplication ( right side ) and to prove (#) I tried to write the second column as the first times a factor scalar $$x$$ I don't know what I should do next, maybe eliminate $$x$$ from the equations but isn't a contradiction? Any help would be welcomed thanks in advance.

## 1 Answer

I assume that $$K^2$$ is viewed as a right-vector space over $$K$$.

Then, in the second matrix the second row is the first one times$$a$$, so the rank is one (since the matrix is nonzero, it cannot be zero).

To prove that the rank of the first one is $$2$$, we need to prove that the columns are linearly independent. It is equivalent to show that the kernel of the matrix is zero (note that multiplying a matrix by a vector yields a right linear combination of the columns and vice versa).

So we have to solve $$ax+ \alpha ay=0$$ and $$x+\alpha y=0$$. Hence $$x=-\alpha y$$ and $$-a \alpha y + \alpha a y=(-a\alpha +\alpha a) y=0$$. By assumption on $$a$$ ana $$\alpha$$, $$-a\alpha +\alpha a$$ is a nonzero element of $$K$$, hence invertible. Consequently, $$y=0$$ and $$x=0$$. QED

This proves the desired equivalence: if $$K$$ is commutative, then it is known that a matrix and its transpose have same rank. If $$K$$ is not commutative, then by definition yo ucan find $$a,\alpha\in$$ such that $$a\alpha\neq \alpha a$$. The matrix of your post is then an example of a matrix which has not the same rank as its transpose.

• Sir .. you're answer is perfectly clear .. thank you
– user730480
Dec 21, 2020 at 15:02