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.
Best 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.