Preservation of linear independence under field extension

Question

Let $L/K$ be a field extension and $x \in L$ be transcendental over $K$. Let $A, B$ be $K$-matrices of same size. I want to show that if $A$ has linearly independent columns over $K$, then $Ax+B$ has linearly independent columns over $L$. Here's my attempt.

Linear independence of columns $Ax+B$ is equivalent to $(Ax+B)v = 0$ having no non-trivial solution, with $v$ a vector over $L$. Comparing coefficients, we my write $v$ as $v = \cdots v_{1/x} \frac{1}{x} + v_1 + v_x x + \cdots$ for vectors $v_1, v_x, v_{1/x}$ with entries in $K$. Then $(Ax+B)v = 0$ iff. $$ \begin{pmatrix} \ddots \\ & A & B \\ & & A & B\\ & & & & \ddots \end{pmatrix} \begin{pmatrix} \vdots \\ v_{1/x} \\ v_0 \\ v_{x} \\ \vdots \end{pmatrix} = 0. $$ Because $A$ has linearly independent columns, this is the case iff. $v_{1/x}, v_0 = 0$.

Here I'm stuck. Is the statement outright wrong in this generality? And there are a few other issues:

• what about relations $x$ satifies? For example, with $x=\mathrm{i}$ in $L/K = \mathbf{C}/\mathbf{R}$, we have $v_{1/x} = -v_x$. Does the real matrix $\left(\begin{smallmatrix}B & -A \\ A & B\end{smallmatrix}\right)$ have linearly independent columns if $A$ does? If $x^2-1$ (and not $x^2+1$), then certainly not, but of course this gives no field. But still, I hope you get my point that the conclusion is not obvious.
• What to do with the dots? If $\dim_K L$ is finite, then sure, the l.e.s. comes to an end. What if it doesn't?

Answer 1

[Edited to cover the non-square case.]

Suppose $A, B$ are $m\times n$.

As $A$ has $n$ linearly independent columns it has at least at many rows as columns so $m\geqslant n$. We can therefore find $(m-n)$ $K$-columns so that the square $m\times m$ $K$-matrix $A_1$ got by adjoining these extra columns to $A$ is non-singular. In the same way extend $B$ to a square non-singular $B_1$.

$\det A_1x+B_1$ is a polynomial in $K[x]$. If $\det A_1x+B_1=0$ then (as $x$ is transcendental) every coefficient of the polynomial must be $0$. However the leading coefficient is $\det A_1\ne 0$.